Category: Class 11

These MP Board Class 11th Biology Notes for Chapter 17 Breathing and Exchange of Gases help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 17 Breathing and Exchange of Gases

→ Process if oxidation of food inside the cells of the living organisms in presence of specific enzymes to release energy is called as Respiration.

→ Respiration which occurs in presence of oxygen is called as aerobic respiration and which occurs in absence of oxygen is called as anaerobic respiration.

→ In modem birds, the lungs are highly modified by the presence of air sacs which facilitate the circulation of air through the lungs.

→ Inflammation of the pleural membrane called pleurisy causes friction during breathing which is quite painful.

→ In males, the larynx increases considerably in size and becomes much larger than that of the females. The enlarged vocal cords are responsible for lower pitch of the masculine voice.

→ Normally, trachea bifurcates into two bronchi but in snakes, there is only one bronchus while in certain ruminants, pigs and whales there are three bronchi.

→ The human lungs have about 750,000,000 alveoli covering the total area of 100 sq. meter which is 100 times than the surface area of the skin.

→ All tissues are not equally affected by the shortage of oxygen. The more active tissues are more affected. Skeletal muscles, heart and brain are much more affected than skin, intestine and bones.

→ Turtles can respire inside water due to the presence of air sacs in their body.

→ O₂ is transported by haemoglobin in our body.

→ R.B.Cs. do not respire aerobically because of the absence of mitochondria in them.

→ Tfre rate of human breathing is 16-20 per minute at rest.

→ Hyperpnoea is a condition, when breathing rate increases above normal.

→ Hypopnoea is a condition, when breathing rate decreases below normal.

→ In some turtles, cloaca acts as respiratory organ.

→ Respiration is a catabolic process.

→ Gills are the respiratory organs of aquatic animals like fish.

→ Spiracles are small apertures in abdominal segments of insects for gaseous exchange.

→ Asthma : Narrowing of bronchi due to spasm in bronchial muscles. It is an allergic response.

→ Haemocyanin : Copper containing blue coloured pigment found in blood plasma of molluscs.

These MP Board Class 11th Biology Notes for Chapter 16 Digestion and Absorption help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 16 Digestion and Absorption

→ Amala is the richest source of vitamin C.

→ Vitamin B (biotin) is an exception, it is regarded as insoluble in water and fat.

→ Gail bladder is not found in horse, rat, whale.

→ Most vertebrates lack proteases for digesting some fibrous animal proteins like hair, keratin, silk fibroin and wool protein.

→ Peristalsis is the rhythmical contractions of alimentary canal by its muscular action which forces food through the canal.

→ Bolus is the rounded mass of masticated food and saliva.

→ The digestibility of bread is increased by toasting because some of its starch is broken into dextrines during the process.

→ In ruminants (e.g., Cow), fermentation is particularly facilitated by rumination or chewing of the cud. Boluses of food are drawn from the rumen to the mouth, chewed thoroughly to break the plant fibres more finely, mixed with saliva to promote further fermentation and swallowed again into the rumen; where it undergoes further fermentation.

→ The conversion of glucose (blood sugar) into glycogen is called glycogenesis.

→ The conversion of glycogen into glucose is called glycogenolysis.

→ The conversion of excess of carbohydrates (blood sugar) into fats is known as lipogenesis.

→ The gastric juice of infant mammals contains an enzyme called rennin. Rennin in the presence of calcium, acts on milk protein known as casein to form curd called calcium paracaseinate.

→ Heparin is secreted by liver, which prevents the coagulation of blood.

→ The synthesis of fibrinogen and prothrombin take place in liver.

→ Vermiform appendix is a vestigial organ attached with the alimentary canal of human beings.

→ HC1 is the activator of pepsinogen.

→ Lunin (1881) discovered vitamins.

→ Stomach absent in Labeo, lamprey and hagfish.

→ Odontology : Study of teeth.

→ Choleresis : Process of bile secretion.

→ Vomiting centre : Is located in the medulla oblongata.

→ World food day : 17th October.

→ Cholecystostasis : Formation of stone in gall bladder.

→ Coprophagy : Eating of faeces by the rabbit, guinea pig etc.

→ Hernia : Choking of intestine.

→ Osteomalacia : A disease of bones in adults caused due to deficiency of vitamin D.

→ Ruminants : Cud-chewing mammals e.g., cattle.

These MP Board Class 11th Biology Notes for Chapter 15 Plant Growth and Development help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 15 Plant Growth and Development

→ Growth is the permanent change in size and increase in dry weight.

→ Growth is measured by Auxanometer.

→ F. W. Went (1928) discovered the auxins and perform Avena coleoptile test for the bioassay of auxins.

→ Kogl and Haagen Smit (1931) isolated auxin-a (auxentriolic acid, C₁₈H₃₂O₅) from human urine.

→ Auxin-b (auxenolinic acid, C₁₈H₃₀O₄) was isolated by Kogl, Erxleben and Haagen Smit from com germ oil.

→ Heteroauxin (Indole-3 acetic acid, C₁₀H₉O₂N) was isolated from human urine.

→ Auxin initiates and promote cell division in cambium cells (play important role in secondary growth).

→ Auxin prevents premature leaf fall by arresting the formation of abscission layer.

→ Auxin is useful in tissue culture experiments and callus formation.

→ Indolebutyric acid (IBA) is used to induce foot formation in cuttings.

→ Auxin is used to induce parthenocarpy in Tomato, Apples, Cucumbers and other plants.

→ 2,4-D (2,4-Dichlorophenoxyacetic acid) is a synthetic auxin which is used in eradication of weeds (Herbicide or Weedicide).

→ Gibberellin was discovered by Kurosawa (1929), a Japanese.

→ IAA is the principal naturally occurring auxin of all higher plants.

→ The term auxin was coined by Kogl and Haagen Smit (1933). It is a Greek word derived from ‘auxin’ which means to grow.

→ The term hormone was used for the first time by Beylis and Starling (1904).

→ Foolish seedling disease of paddy is caused by a fungus Gibberella fujikuroi which is now known as Fusarium heterosporum.

→ Intemodal elongation is caused by gibberellins.

→ Yabuta and Sumiki isolated the gibberellin for the first time from fungus Gibberella fujikuroi.

→ Gibberellin is used for the elongation of genetically dwarf plants.

→ Gibberellins can substitute low temperature treatments (Vernalization) and long day treatments.

→ Cytokinins are hormones that induce cell division by inducing cytokinesis.

→ Ethylene is a volatile gas present in atmosphere as a component of smoke and other industrial gases.

→ According to Denny (1924), ethylene gas is highly effective in inducing fruit ripening.

→ Degradation of proteins can be prevented by application of cytokinin.

→ Miller et al isolated cytokine from yeast DNA for the first time.

→ Fruits can be made to ripe artificially by treating them with ethylene.

→ Abscisic acid is a powerful growth inhibitor.

→ Cams and Addicott (1963) discovered that Abscisin-II is responsible for shedding of cotton balls.

→ According to Wareing and Cornforth, ‘dormin’ induces bud dormancy.

→ ABA induces leaf abscission.

→ The term photoperiodism was coined by Garner and Allard (1920) to designate the response of organisms to the relative length of the day and night and the term photoperiod to designate the favourable length of day for each plant.

→ Pigment responsible for flowering, seed germination and leaf development is phytochrome.
→ ABA inhibits GA stimulated growth.

→ Lysenko (1938) used the term ‘vernalization’ for low temperature promotion of flowering in plants.

→ Vernalization is a method of inducing early flowering in plants by pretreatment of their seeds at very low temperature.

→ Florigen is the hormone which causes flowering.

→ Vernalin is a hormone which induces flowering after low temperature treatment.

→ Growth shows a characteristic S-shaped growth curve.

→ The growth and development of plants are regulated by many factors.

→ Growth regulators of phytohormones vary chemically and also differ in their specific biological actions.

→ Auxins, gibberellins, cytokinins, abscisic acid and ethylene are the main hormonal substances synthesized in the plants.

→ Photoperiodism has a marked influence on the behaviour of plants particularly flowering and plants are categorized into long day, short day and day neutral plants.

→ Phytochromes are the photoreceptive forms of two kinds : P_R and P_FR.

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 2 Structure of Atom which are most likely to be asked in the exam.

MP Board Class 11th Chemistry Important Questions Chapter 2 Structure of Atom

Structure of Atom Class 11 Important Questions Very Short Answer Type

Question 1.
Write the complete symbol of the atoms with the following atomic number (Z) and atomic mass (A):
(i) Z = 17,A = 35
(ii) Z = 92, A = 233
(iii) Z = 4,A = 9
Solution:
(i) \({ }_{17}^{35} \mathrm{Cl}\)
(ii) \({ }_{92}^{233} \mathrm{U}\)
(iii) \({ }_{4}^{9} \mathrm{Be}\)

Question 2.
Which of the following are Iso-electronic species ue., which have same number of electrons :
Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2-, Ar
Solution:

Thus, iso-electronic species are as follows :
(i) Na⁺ and Mg²⁺
(ii) K⁺,Ca²⁺,S^2- and Ar

Question 3.
What is Threshold frequency?
Answer:
For every metal there is a specific minimum frequency υ₀ which is called Threshold frequency. On frequency less than this, photoelectric effect is not observed.

Question 4.
What is work function?
Answer:
For ejecting an electron, the minimum energy required is hυ₀. This is also known as work function (w₀).

Question 5.
Arrange the following types of radiations in increasing order of their fre-quencies :
(a) Radiations by microwave oven.
(b) Amber light by traffic signal.
(c) Radiations obtained by F.M. Radio.
(d) Cosmic rays by outer dip.
(e) X-rays.
Answer:
Increasing order of frequency of radiations is as follows :
c < a < b < e < d

Question 6.
Write two characteristics of X-rays.
Answer:
Characteristics :

1. Cathode rays travel in a straight line. If an opaque solid object is placed in their path then its shadow is formed on the other side of the glass tube.
2. These rays strike on metals and produce X-rays.

Question 7.
How was neutron discovered? Write its two characteristics.
Answer:
James Chadwick discovered Neutron in the year 1932, by the bombardment of a-particles on Be, a neutral particle was obtained whose mass is nearly equal to the mass of proton and it was called neutron.
Characteristics of Neutron:
Neutron is a neutral particle present in the nucleus of an atom whose mass is nearly equal to the mass of proton i.e., 1.00893 a.m.u or 1.6749 × 10^-27 kg. Its symbol is \({ }_{0}^{1} n[/latex

Question 8.
Write the difference between Atomic number and Mass number.
Answer:
Differences between Atomic number and Mass number

Atomic number	Mass number
1. Atomic number is the number of neutrons present in the nucleus.	Mass number is the sum of the number of protons and neutrons.
2. It is represented by Z.	It is represented by A.

Question 9.
What is Atomic orbital?
Answer:
The three dimensional region in space around the nucleus in which the probability of presence of electron is maximum is called atomic orbital.

Question 10.
What are α-particles?
Answer:
α-particles are nuclei of helium atom which contains 2 unit positive charge. Mass of these particles are 4 unit. These are represented by [latex]\left({ }_{2}^{4} \mathrm{He}\right)^{2+}\)

Question 11.
What conclusions are drawn for Rutherford’s Atomic model by α-particle scattering?
Answer:
Conclusions of Rutherford’s experiment:

1. Most of the α-particles passed through the gold foil undeflected which concludes that there is a lot of empty space in an atom.
2. Few α-particles were deflected through small angles. Thus, the central part of an atom consists of positive charge.
3. Out of the 20,000 particles bombarding only one particle gets deflected and returns back. Thus, the size of nucleus is very small and rigid as compared to the size of the atom.

Question 12.
Write defects of Rutherford’s atomic model.
Answer:
In 1913 Niel Bohr pointed out some defects in proposed Rutherford’s atomic model:
(i) According to electromagnetic theory whenever an electric charge is subjected to acceleration, it emits radiation and loses energy. Bohr argued that due to rotation around positively charged nucleus, electron should continuously emit radiation and lose energy. As a result of this its orbit should become smaller and finally it should drop into the nucleus.
This however does not happen.
(ii) According to this model the atomic spectrum should be continuous but it is line spectrum.
(iii) This model does not determine the number of electrons rotating in different orbits.

Question 13.
In 3p orbital of AI and Si unpaired electrons are present. Which electron experience higher effective nuclear charge by the nucleus?
Solution:
₁₃Al = 1s²,2s²2p⁶,3s²3p¹
₁₄Si = 1s²,2s²2p⁶,3s²3p²
In both, number of orbitals are same i.e., Si (+ 4) and Al (+ 3). Nuclear charge is higher in Si due to which nuclear charge is more effective.

Question 14.
(a) How many subshells are in n = 4?
(b) How many electrons will be present in that subshell for which m_s = –\(\frac{1}{2}\) and n = 4.
Solution:
(a) For n =4 l_max= (n -1)
= (4 – 1) = 3
l = 0,1,2,3
Four subshells which are 4s,4p, Ad, 4f

(b) n = 4 Number of orbitals = n² = 4² =16
Maximum number of electrons in an orbital will be = 2
In every orbital m_s = –\(\frac{1}{2}\) No. of spinning electron = 1
Thus, in 16 orbitals m_s = –\(\frac{1}{2}\)  No. of spinning electrons = 16.

Question 15.
Quantum number of six electrons are given below. Arrange them in in-creasing order of their energy. Is the energy of any of them same :
(a) n = 4,l = 2,m_t= -2,m_s = –\(\frac{1}{2}\)
(b) n = 3,l = 2,m_l, = 1,m_s = +\(\frac{1}{2}\)
(c) n = 4,l = 1,m₁ = 0,m_s = +\(\frac{1}{2}\)
(d) n = 3,l = 2,m_l, = -2,m_s = –\(\frac{1}{2}\)
(e) n = 3,l = 1,m_l = -1,m_s = +\(\frac{1}{2}\)
(f) n = 4,l= 1,m = 0,m_s = +\(\frac{1}{2}\)
Solution:
Energy of Polyelectron atom is equal to the sum of (n + l).
(a) Subshell of set 4d = (n + l) = 4 + 2 = 6
(b) Subshell of set 3d = (n + l) = 3 + 2 = 5
(c) Subshell of set 4p = (n + l) = 4 + 1 = 5
(d) Subshell of set 3d = (n + l) – 3 + 2 = 5
(e) Subshell of set 3p = (n + l) = 3 +1 = 4
(f) Subshell of set 4p = (n + l) = 4 +1 = 5
Thus, (e) < (b) = (d) < (c) = (f) < (a) ‘
3P < 3d = 3d < 4p = 4p < Ad (Increasing order of energy of orbitals).

Question 16.
What are Electromagnetic waves?
Answer:
Radiant energy propagatic through space in the form of waves which travel with the velocity of light and are produced due to oscillations of electric and magnetic field. Therefore, they are known as Electromagnetic waves.

Question 17.
Nucleus of an atom repels α-particles. Why?
Answer:
Each α-particle carry a charge of+2 unit. Every atom contains a positively charged nucleus at its centre. We know that similar charge repel each other. Therefore, nucleus of atom repels α-particle.

Question 18.
Write the difference between Anode rays and Cathode rays.
Answer:
Differences between Anode rays and Cathode rays

Anode Rays	Cathode Rays
1. These are made up of positively charged particles.	These are made up of negatively charged particles.
2. Nature of Anode rays depend on the nature of gas taken in the discharge tube.	Nature of Anode rays do not depend on the nature of gas taken in the discharge tube.
3. Mass of particles present in Anode rays is equal to the mass of gas taken  in the discharge tube.	Mass of particles present in Cathode rays is negligible.

Question 19.
Which of the following orbital will experience high effective nuclear charge:
(a) 2s and 3s,
(b) 4d and 4f
(c) 3d and 3p.
Answer:
(a) 2s orbital is more closer to nucleus, thus it will experience higher effective nuclear charge.
(b) 4d orbital will experience higher effective nuclear charge.
(c) 3p orbital will experience higher effective nuclear charge.

Question 20.
Determine the total number of angular node and radial nodes present in 3p orbital.
Answer:
For 3p orbital: Principal Quantum number n = 3
and Azimuthal Quantum number l = 1
Number of angular nodes = l = 1
Number of Radial nodes = n – l – 1 = 3 – 1 – 1 = 1.

Question 21.
What is the difference between Atom and Ion?
Answer:
Differences between Atom and Ion

Atom	Ion
1. Atoms are completely neutral.	Ions are positively or negatively charged.
2. It contains same number of protons and Electrons.	In it number of protons and neutrons are not same.
3. Atoms are unstable and take part in chemical reaction.	Ion is stable in solution state.

Question 22.
What is Electromagnetic Spectrum?
Answer:
The graph obtained by the arrangement of different types of electromagnetic radiations in the increasing or decreasing order of their frequency or wavelength is called Electromagnetic spectrum.

Question 23.
Matter waves are different from Electromagnetic waves. Explain.
Answer:
Matter waves differ from Electromagnetic waves as follows :

• Value of their wavelength is less than Electromagnetic waves.
• Their velocity is less than that of Electromagnetic waves.
• Matter waves do not emit by particles but are related to particles.

Question 24.
Electrons move around the nucleus in a definite energy orbit. Why?
Answer:
Jumping from one energy level to another energy level by any moving charged particle like electron, will cause absorption or emission of energy radiation which will make it unstable. To provide stability to atom, electron rotates in specific energy orbits.

Question 25.
A moving electron represents both particle and wave nature. Explain the cause.
Answer:
Due to its particle nature an electron has a definite momentum and due to its motion it represents wave character. In this way, electron represents both the characters.

Question 26.
Tell the number of Neutrons and Protons present in the following nuclei:

Answer:

Question 27.
Show that the circumfereiice of the Bohr orbit for the hydrogen atom is an integral multiple of the de-Broglie wavelength associated with the electron revolving around the orbit.
Solution:
According to Bohr model of H atom, in the given state, angular momentum of electron:
mνr = \(\frac{n h}{2 \pi}\)
2πr = \(\frac{n h}{m v}\)
according to de-Broglie
Wavelength λ = \(\frac{h}{m v}\)
2πr = nλ
Thus, circumference (2πr) of Bohr’s orbit of H atom is whole number multiple of deBroglie wavelength.

Question 28.
It is impossible to determine the position and velocity of a fast moving electron simultaneously. Why?
Answer:
Electron is an extremely minute particle. Thus, it can only be seen by smaller wavelength light like y-rays or x-rays. For seeing an electron, it is necessary that photon should return back after striking the electron. But after striking the photon, velocity of electron changes. Therefore, it is not possible to determine the position and velocity of electron simultaneously.

Question 29.
Electronic configuration of Chromium is written as 4s¹3d⁵ instead of 4s²3d⁴. Why?
Answer:
Electronic configuration of Chromium in ground state should be [Ar] 4s²3d⁴ but due to presence of 4 electrons in 3d orbital it is incomplete and unstable. If 1 electron of 4s gets excited and goes to 3d orbital then both 4s and 3d orbitals become half filled and achieve stability. Therefore, electronic configuration of Chromium is written as 4s¹ 3d⁵ instead of 4s² 3d⁴.

Question 30.
Electronic configuration of Copper is written as 4s¹3d¹⁰ instead of 4s¹3d⁵.
Answer:
Electronic configuration of Copper in ground state is [Ar] 4s²3d⁹ in d orbital, there are 9 electrons which is an incomplete orbital and unstable. If an electron of 4s get excited and goes to 3d orbital such that 4s is half filled and 3d orbital is a completely filled orbital which is more stable. Therefore, electronic configuration of Copper is written as 4s¹3d¹⁰ in place of 4s²3d⁹.

Question 31.
Write Max Planck’s Quantum Theory.
Answer:
Max Planck’s Quantum Theory:

• Energy emitted or absorbed by an atom or molecule is not continuous but discontinuous in the form of bundles or packets which are called quantam.
• Energy related to each quanta is directly proportional to the frequency of radiation.
  E ∝ υ
  E = hυ, where h = Planck’s constant.
• Any molecule or atom absorbs or emits energy in simple multiples of quantum.

Question 32.
What are the basic particles of an atom. Write their name, mass and charge.
Answer:
Atom is made up of three particles which are known as fundamental particles or
basic particles.

Question 33.
What are Photons?
Answer:
Energy particles radiated are called photons i.e. for light, quantum ho of radiation is called one photon. These are packets or bundles of negligible mass.

Question 34.
Write main characteristics of Sommerfield atomic model.
Answer:
Main characteristics of Sommerfield atomic model are :

• Electron revolve in spherical orbits along with bigger spherical orbits where there are two radii.
• Orbits are made up of orbitals.
• Number of subshells in a shell is equal to quantum number n.

Question 35.
What is Quantum?
Answer:
According to Max Planck’s Quantum theory, emission or absorption of energy by an atom or molecule is not continuous but is discontinuous in the form of minute packets or quanta and unit of this energy emitted or absorbed is quantum.

Question 36.
Write ‘Aufbau’s principle’.
Answer:
Aufbau’s principle: Aufbau is German word meaning ‘building up’. The building up orbitals means the filling up of orbitals with electrons. Thus, this rule gives us a sequence in which various orbitals are filled up depending upon the relative-order of their energies.
According to this principle, the electrons in various or¬bitals are added progressively in the order of increasing energies starting with the orbital of lowest energy. The increasing order of energies of various orbitals is :

1s, 2s, 2p, 3s, 3p, 4s, 3d,4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f ……………

Question 37.
What is Zeeman Effect?
Answer:
According to Zeeman: “The splitting up of spectral lines under the influence of a magnetic field in fine lines is called Zeeman effect.”

Question 38.
Write the values of all the four quantum numbers for the last electron and unpaired electron of Chlorine.
Answer:

Cl₁₇ = 1s² 2s² 2p⁶ 3s² 3p⁵
For last electron,
n = 3,l=1,m = 0, s = –\(\frac{1}{2}\)
For unpaired electron,
n = 3,l=1,m = +1, s = +\(\frac{1}{2}\)

Question 39.
Why is Emission spectrum known as Line Spectrum?
Answer:
On passing electric current of high voltage through hydrogen gas in a discharge tube, radiations are emitted. If this radiation is allowed to pass through a prism, it splits into various thin lines. If the spectrum of these fine lines are observed then series of coloured lines are obtained, and each line has a definite wavelength, therefore, it is known as line spectrum.

Question 40.
What information is obtained by Azimuthal Quantum number?
Answer:
Azimuthal Quantum number helps to determine the angular momentum. It repre¬sents the shape of the subshells.

Question 41.
What are the symbols of orbitals of p, d and f subshells?
Answer:

• Three orbitals ofp subshell are represented as p_x, p_y and p_z.
• d subshell has five orbitals which are represented as d_xy, d_yz, d_zx, d _{x² – y²} , d_z²
• f subshell has seven orbital? which are represented as f_x³ f_y³ f_z³,f_{x(z²– y²)}’ f_{y,(x²– z²)}’f_{z(y² – x²})’f_xyz.

Question 42.
Determine the mass of one mole electron.
Solution:
Mass of Electron =9.1 × 10^-28 gm
1 mole Electron = 6.023 × 10²³ Electron
Mass of 1 mole Electron =9.1 × 10^-28 × 6.023 × 10²³
= 5.48 × 10^-4gm.

Question 43.
Determine the charge of 1 mole electron.
Solution:
Charge of Electron = 1.6 × 10^-19 coulomb
1 mole Electron = 6.023 × 10²³
Charge of 1 mole Electron .= 1.6 × 10^-19 × 6.023 × 10²³
= 9.63 × 10⁴coulomb.

Question 44.
Which subshell is represented by each group of quantum number in the following:
(a) n = 1 l = 0
(b) n = 2 l= 1
(c) n = 3 l = 2
(d) n = 4 l = 3.
Answer:
On the basis of the given quantum numbers subshells are as follows :
(a) 1s,
(b) 2p,
(c) 3d,
(d) 4f.

Question 45.
On the basis of Bohr’s atomic model what are stable orbits?
Answer:
According to Bohr’s principle, electrons revolve around the nucleus in certain specific orbits where it does not lose any energy. In these orbits, momentum of electrons is \(\frac{h}{2 \pi}\).
These orbits are known as stable orbits.

Question 46.
What are Isotopes and Isobars? Explain with example.
Answer:
Isotopes : Atoms of the same element which have same atomic number but different mass number are called Isotopes.
Example : Hydrogen has three isotopes : \({ }_{1}^{1} \mathrm{H},{ }_{1}^{2} \mathrm{H},{ }_{1}^{3} \mathrm{H}\)
Isobars : Atoms of different elements with same mass number but different atomic number are called Isobars. Example : \({ }_{20}^{40} \mathrm{Ca},{ }_{18}^{40} \mathrm{Ar}\) and \({ }_{19}^{40} \mathrm{~K}\)
are isobars as they have same mass number but different atomic numbers.

Question 47.
Write the values of n and l according to their orbital:
(i) 3s
(ii) 4d
(iii) 5p
(iv) 6f.
Answer:
(i) n = 3, l = 0,
(ii) n = 4, l = 2,
(iii) n = 5, l = 1,
(iv) n = 6, l = 3.

Question 48.
What will be the maximum number of lines emitted obtained when an excited electron n = 6 jumps to ground state in hydrogen atom?
Solution:
Number of lines emitted on moving from n shell to ground state = \(\frac{n(n-1)}{2}\)
∴ n = 6, number of lines obtained = \(\frac{6(6-1)}{2}=\frac{6 \times 5}{2}\) = 15

Question 49.
Write down the electronic configuration of elements with following atomic numbers:
(i) 17
(ii) 24
(iii) 29
(iv) 35.
Answer:
(i) 1s²,2s²,2p⁶,3s²,3p⁵
(ii) 1s²,2s²,2p⁶,3s²,3p⁶,3d⁵,4s¹
(iii) 1s²,2s²,2p⁶,3s²,3p⁶,3d¹⁰,4s¹
(iv) 1s²,2s²,2p⁶,3s²,3 p⁶,3d¹⁰,4s²,4p⁵.

Structure of Atom Class 11 Important Questions Short Answer Type 

Question 1.
Write Pauli’s Exclusion Principle with example. Write its importance.
Answer:
Pauli’s Exclusion principle : No two electrons of an atom can have the same values of all the four quantum numbers.
Example:

The three quantum numbers are same for these two electrons but one quantum number is different, i.e., spin quantum number. Which proves the Pauli’s exclusion principle.

Importance:
From this principle, number of electrons in a shell or main energy level can be known because the number of electrons in an orbit is equal to the number of groups of four quantum numbers in different combinations.

Question 2.
Explain Hund’s rule of maximum multiplicity with example.
Answer:
Hund’s rule :
This rule deals with degenerate orbitals (orbitals having similar energy). It states as, “pairing of electron in the orbitals of a particular subshell (p, d,f) does not take place until all the orbitals of the subshell are singly occupied.”

Question 3.
How de-Broglie explained dual nature of particle?
Answer:
de-Broglie describes the wave nature of electron. According to him, ‘all moving small particles show wave nature’. Behaviour of electron is governed by following funda-mental equations. According to Planck’s wave theory,
E = hυ …(1)
From Einstein’s energy mass relation,
E = mc² …(2)
Where, m is mass of particle and c is velocity of light.
From both eqns. (1) and (2),
mc² = hυ
or
mc² = h\(\frac{c}{\lambda}\), (∵ υ = \(\frac{c}{\lambda}\))
or
λ = \(\frac{h}{m c}\)
Where, λ = wavelength and me = momentum.
This is de Broglie’s equation.
Thus, it is clear that wavelength of any moving particles depends upon its momentum.

Question 4.
Write the main characteristics of Cathode Rays.
Answer:
Characteristics of Cathode Rays :

• Cathode rays travel in straight line with the velocity of light.
• If an electric field is applied in the path of Cathode rays, then these deflect towards positive region.
• A light paddle wheel when kept in the path of Cathode rays start moving which is due to high kinetic energy particles present in the rays.
• These rays ionize the gases.
• These rays strike metals like tungsten and produce X-rays.
• These rays affect the photographic plate.

Question 5.
Write main properties of Electrons.
Answer:
Properties of Electrons :
(i) All gases consist of negatively charged particles which are produced from the Cathode on passing a high voltage current at 10^-4 atmospheric pressure through a gas in the discharge tube.
(ii) Electrons are represented as _-1e⁰
(iii) Mass of electron is \(\frac{1}{1837}\) parts the* mass of hydrogen atom and mass of every electron is 9.1 × 10^-28 gm
or
9.1 × 10^-31kg .
(iv) Charge on an electron is 1.60 × 10^-19 coulomb.

Question 6.
How are anode rays produced? Write characteristics of anode rays.
Answer:
In 1886, Goldstein used perforated and found that some rays liberates from anode and passes through perforated cathode if electric arc is produced in the discharge tube at low pressure and high voltage. These rays deflects in electrical field and magnetic field in opposite direction than electrons. These rays are called anode rays or positive rays.
Four properties of anode rays :

• Anode rays travel in straight line.
• Anode rays consist of positive residues of the gas left when electrons are knocked out from the atoms of the gas.
• Anode rays are positively charged and deflect in electrical field towards negative plate.
• The charge and mass ratio (e/m) are different from electron. These particles are called protons.

Question 7.
Write the characteristics of Protons.
Answer:
Characteristics of protons :
1. Proton is the fundamental particle of proton by passing an electric discharge through a gas in a discharge tube H⁺ ions obtained are called protons or bombarding electron on H atom electrons is released and H⁺ is obtained.
H+ e^– → H⁺+ 2e^–
1. By taking different gases in the discharge tube positive particles of different masses are obtained. Positive particle (proton) obtained from hydrogen is the least. Thus, proton is the fundamental particle of an atom.
2. Mass of proton is 1.00753 a.m.u. or 1.67 x 10^-27 kg.
3. Proton is represented by \({ }^{1} \mathrm{H}\) or \({ }^{1} \mathrm{P}\).
4. Charge on the proton is 1.6 x 10 ^-19 coulomb of positive charge. Magnitude of charge on proton is equal to that of the charge on an electron. Thus, this charge is called unit positive charge.

Question 8.
Write Bohr Bury Scheme of filling of electrons in various shells of atoms.
Answer:
Scientist Bohr Bury proposed the arrangement of filling of electrons in various shells:

• Maximum number of electrons in orbit n is 2n².
• Last shell contains 8 electrons and next inner shell will contain not more than 18 electrons.
• Last shell will contain more than 2 and next to the last will contain more than 9 electrons only when inner shells are filled up.
• Electrons are filled up in the next shell only when there are 8 electrons in the last shell.
• It is not necessary that electrons will be filled in the next shell till the electrons are not filled up according to 2n² rule.

Question 9.
What do you understand by Stability of orbitals?
Answer:
Distribution of half filled and fully filled electrons is symmetrical and this symmetry provides extra stability to electrons. Exchange of electrons in orbitals of similar energy takes place in an atom and during this process energy is released as a result the orbital achieves stability. In half filled and fully filled orbitals possibilities of exchange of electrons is maximum. Therefore, these orbitals are more stable.

Question 10.
Write the difference between Isotopes and Isobars.
Answer:
Differences between Isotopes and Isobars

Isotopes	Isobars
1. Atoms of the same element which have same atomic number but different mass number are called Isotopes.	Atoms of different elements which have same atomic mass but different atomic numbers are called Isobars.
2. Number of protons in the nucleus is same but number of neutrons are different.	Number of protons and neutrons present in the nucleus are different.
3. Chemical properties are same.	Chemical properties are different.
4. Number of electrons present in the outermost shell is same.	Number of electrons present in the outermost shell is different.

Question 11.
Write any four postulates of Bohr’s Atomic model.
Answer:
Bohr’s model of the atom is based on the following postulates :
(i) The electron in the atom moves around the nucleus in a fixed circular path called orbits. These orbits are associated with definite energies and are called energy shells or
energy levels. These are numbered as 1,2, 3, 4, , etc., (from the nucleus) or these are
designated as K, L, M, N, ……….. etc. shells.

(ii) The electron can move only in those orbits for which its angular momentum is an integral multiple of the quantity h/2π.
i. e., Angular momentum = mνr = \(\frac{n h}{2 \pi}\)
Where, mass of electron = m, speed = ν, orbit or radius = r, Planck’s constant = h, an integer (1, 2, 3, 4, ) = n.

(iii) As long as the electron moves in a given orbit, it does not radiate energy. Hence, the energy of the electron is constant as long as it does not change its orbit. Because of this the orbits are also known as stationary states or states of fixed energy.

(iv) The electron absorbs energy when it jumps from an orbit of lower energy to an orbit of higher energy. The energy is emitted when the electron returns to lower energy state. The change of energy Δ E, i.e., energy absorbed or emitted, when the electron moves from one orbit to another, is given by Bohr frequency equation,
ΔE = E₂ -E₁ = hυ.

Question 12.
Show the difference between Orbit and Orbital.
Answer:
Differences between Orbit and Orbital

Orbit	Orbital
1. Spherical path outside the nucleus in which electron revolves is called as orbit.	The three-dimensional area in space around nucleus in which probability of presence of electron is maximum.
2. The position of mobile electron in shell and momentum is decided more accurately in case of orbit.	In orbital the position and momentum of electron is decided by Heisenberg uncertainty principle.
3. Shows planar motion of electron.	Shows three-dimensional motion of electron.
4. All orbits are circular.	Shapes of orbitals are different e.g., s = spherical.
5. Orbits are of scalar nature.	Except s orbital all shows directional property.
6. In any orbit the maximum No. of electron is 2n2 where n is No. of orbit.	In any orbital there are maximum two electrons.

Question 13.
What is (n + l) rule?
Answer:
1.(n + l) rule : Subshell in which sum of ‘n’ and ‘l’ quantum numbers is less, energy of that subshell is also less and electrons will be filled first in that subshell.
Example: For 3d → n = 3, l = 2
∴ (n + l) = 3 + 2 = 5
and 4s → n = 4, l = 0
∴ (n + l) = 4 + 0 = 4
So, the energy of 4s subshell will be less than 3d and electrons will be filled in 4s subshell and after this 3d will be filled.

2. (n + l) value is equal for 2 or more subshells, the electrons will be filled in the subshell having minimum principal quantum number (n) value.
Example: For2p → n = 2,p = 1
∴ (n +l) = 2+1 = 3
and 3s —> n = 3, l = 0
∴ (n +l) = 3 + 0 = 3
But the value of n for 2p subshell is 2 which is less than that of 3s (n = 3), so energy of 2p will be less and electrons will be filled in 2p.

Question 14.
What are the main defects of Bohr’s theory?
Answer:
Main defects of Bohr’s theory are as follows :

• Bohr’s theory could not explain the spectrum of polyelectron atoms.
• According to Bohr’s theory orbits of electrons are considered to be spherical but they are bigger spheres also.
• It does not explain Heisenberg’s uncertainty principle.
• It does not explain dual nature of electron.

Question 15.
What is Heisenberg’s uncertainty principle? Write its mathematical form.
Answer:
Heisenberg’s uncertainty principle : Heisenberg pointed out uncertainty in determination of two conjugate properties simultaneously. According to this principle, it is impossible to determine charge momentum and charge position of electron simultaneously. Mathematical form of uncertainty principle,
Δx × Δp ≥ \(\frac{h}{4 \pi}\)
or
Δx × mΔν ≥ \(\frac{h}{4 \pi}\)
Where, Δx = Uncertainty in position, Δp = Uncertainty in momentum, h = Planck’s constant.
On the basis of wave nature and uncertainty principle of an electron structure of atom can be explained.

Question 16.
Write the ground state electronic configuration of the following on the basis of Aufbau’s principle.
(i) Neon (Z = 10)
(ii) Phosphorus (Z = 15)
(iii) Chlorine (Z = 17)
(iv) Potassium (Z = 19).
Answer:
(i) Neon (Z= 10) : 1s² 2s² 2p_x²,2p_y² 2p_z²:
(ii) Phosphorus (Z = 15) : 1s² 2s² 2p⁶ 3s² 3p_x¹ 3p_y¹ 3p_z¹:
(iii) Chlorine (Z = 17) : 1s² 2s² 2p⁶ 3s². 3p_x²3p_y² 3 p_z¹.
(iv) Potassium (Z = 19) : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹.

Question 17.
Represent Hund’s rule for the following electronic configuration.
(i) \({ }_{8} \mathrm{O}^{+2}\),
(ii) \({ }_{7} \mathbf{N}^{-3}\),
(iii) \({ }_{6} \mathrm{C}^{-1}\)
Answer:
(i) \({ }_{8} \mathrm{O}^{+2}\),
Electronic configuration of 8^O = 1s² 2s² 2p⁴
Electronic configuration of \({ }_{8} \mathrm{O}^{+2}\)

(ii) \({ }_{7} \mathbf{N}^{-3}\),
Electronic configuration of 7^N = 1s² 2s² 2p³
Electronic configuration of \({ }_{7} \mathbf{N}^{-3}\)

(iii) \({ }_{6} \mathrm{C}^{-1}\)
Electronic configuration of 6^C = 1s² 2s² 2p²
Electronic configuration of \({ }_{6} \mathrm{C}^{-1}\)

Question 18.
If one-one electrons are present in 3p_x, 4p_y,and 5P_z then determine value of
four quantum numbers for them.
Answer:
3p_x n = 3, l = 1,m = -1,s = +\(\frac{1}{2}\)
4p_y n = 4, l = 1, m = 0,s = +\(\frac{1}{2}\)
5P_z n = 5, l = 1, m = +1,s = +\(\frac{1}{2}\)

Question 19.
Write the values of the four quantum numbers for 17 ^thelectron of chlorine atom and 26^th electron of Fe.
Answer:
Cl₁₇ : 1s² 2s² 2p⁶ 3s² 3p⁵

Last electron is in 3p subshell n = 3, l = 1, m = 0, s = –\(\frac{1}{2}\)
Fe₂₆ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

Last electron is in 3d subshell. Thus,
n = 3, l = 2, m = -2,s = –\(\frac{1}{2}\)

Question 20.
Who determine the e/m value of an electron and how?
Answer:
Sir J.J. Thomson determined the charge to mass ratio of electron by measuring the deflection under the simultaneous application of electric and magnetic field, applied perpendicular to each other. The apparatus is shown in figure. It is possible to adjust the two fields in such a way that the cathode rays strike the fluorescent screen at the same position as they do when no magnetic field is applied. From the magnitudes of electric and magnetic field thus applied, charge to mass (e/m) ratio of electron can be calculated. The value of elm was found to be 1 .76 × 10⁸ coulomb per gram for electron.

Question 21.
What is natural Radioactivity? Explain.
Answer:
Professor Henry Becquerel observed in the year 1898 that Uranium and its com-pounds possess a unique property. Invisible rays are emitted by these compounds, which produce glow on striking the zinc sulphide screen, affect the photographic plate like X-rays and ionize the gases and these rays also possess the ability to penetrate thin metal sheets. These rays are known as Becquerel rays. Madam Curie called these rays as radioactive rays and this property as Radioactivity.

Thus, “Substances which possess the property of emitting active invisible rays are called radio-active substance and this property is known as Radioactivity.”

Question 22.
Explain the difference between shell, subshell and orbital.
Answer:
Shell:
Shell are circular path around the nucleus in which the electron revolve. Electrons can stay only in those shells in which the value of their angular momentum (mνr)
is \(\frac{h}{2 \pi}\) or its simple integral multiple.

According to Bohr’s principle shells are spherical. Principal Quantum number of shell is represented by (n). Maximum number of electrons in a shell is 2n².

Sub shell or Sub-orbit:
Each shell is made up of various sub-orbits. Sub-orbits are designated by s,p,d,f Sub orbits are determined by values of l.
For s sub-orbit l = 0, No. of electrons = 2
For p sub-orbit l = 1,No. of electrons = 6
For d sub-orbit l = 2,No. of electrons = 10
For/sub-orbit l = 3, No. of electrons = 14
Maximum number of electrons in various sub-orbits are as follows 2 in s, 6 in p, 10 in d and 14 in f.

Orbital :
Every sub-orbit is made up of orbitals. The three-dimensional region in ‘ space around the nucleus where the probability of the presence of electron is maximum is called orbital. There is 1 orbital ins sub-orbit, 3 inp sub-orbit, 5 in d sub-orbit and 7 orbitals in/sub-orbit. Each orbital can have 2 electrons.
Shape of s-orbital is spherical,p orbital is dumb-bell, d orbital is double dumb-bell and/orbitals are complex.

Question 23.
Explain why?
(i) Particles present in cathode rays are negatively charged.
(ii) Cathode rays rotate the wheel placed in its path.
Answer:
(i) If cathode rays are allowed to pass through an electric field then they deflect towards the positive plates, which proves that cathode rays are made up of negatively charged particles.
(ii) If a light paddle wheel is placed in the path of cathode rays, it starts rotating which represents that cathode rays are made up of particles of high kinetic energy. Thus, cathode rays are made up of material particles.

Question 24.
Explain the principle of filling of electrons in various Energy levels.
Answer:
Aufbau’s principle:
Aufbau is German word meaning ‘building up’. The building up orbitals means the filling up of orbitals with electrons. Thus, this rule gives us a sequence in which various orbitals are filled up depending upon the relative-order of their energies.
According to this principle, the electrons in various or¬bitals are added progressively in the order of increasing energies starting with the orbital of lowest energy. The increasing order of energies of various orbitals is :

1s, 2s, 2p, 3s, 3p, 4s, 3d,4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f ……………

1.(n + l) rule : Subshell in which sum of ‘n’ and ‘l’ quantum numbers is less, energy of that subshell is also less and electrons will be filled first in that subshell.
Example: For 3d → n = 3, l = 2
∴ (n + l) = 3 + 2 = 5
and 4s → n = 4, l = 0
∴ (n + l) = 4 + 0 = 4
So, the energy of 4s subshell will be less than 3d and electrons will be filled in 4s subshell and after this 3d will be filled.

2. (n + l) value is equal for 2 or more subshells, the electrons will be filled in the subshell having minimum principal quantum number (n) value.
Example: For2p → n = 2,p = 1
∴ (n +l) = 2+1 = 3
and 3s → n = 3, l = 0
∴ (n +l) = 3 + 0 = 3
But the value of n for 2p subshell is 2 which is less than that of 3s (n = 3), so energy of 2p will be less and electrons will be filled in 2p.

Question 25.
What do you mean by dual nature of electron?
Answer:
Momentum and wave are generally not associated with some object or energy medium. Particles have momentum due to mass while energy produces waves. Electron is a microscopic particle in motion. Being a particle it has momentum and due to motion it behaves like waves also. Thus, particle and wave character of electron are its dual nature. Other particles like proton and neutron also exhibits dual nature when they are accelerated.

Wavelength of waves associated with microscopic particles can be determined by de Broglie’s equation
λ = \(\frac{h}{m v}=\frac{h}{p}\)
or
λ ∝ \(\frac{1}{p}\) (mν = p)
Where λ is the wavelength of microscopic particles, h = Planck constant, m = mass of particle, ν = velocity of particle and p is momentum of particle.

Question 26.
What is Balmer formula? How does it explain hydrogen spectrum?
Answer:
In 1885 J.J. Balmer studied emission spectrum of hydrogen. He provided a for- v mula for the frequencies ofhydrogen spectrum in visible region.
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)\)
Where, value of n is higher than 2, R_H = Rydberg constant.
Putting the value of n as 3,4,5, 6, lines of Balmer series are obtained which are in
visible region. After this in UV region Lyman series, while Paschen series, Brackett series and Pfund series in IR region, are obtained. To explain these lines, Balmer formula is modified as,
\(\overline{\mathrm{v}}=\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)
Where, n₂ > n₁.
For Lyman series, n₁ = 1, n₂ = 2, 3,4,…
∴ \(\bar{v}=R_{H}\left(\frac{1}{1^{2}}-\frac{1}{n_{2}^{2}}\right)\) (In ultraviolet region)
In the near and far infrared region following series of spectral lines are obtained :
Paschen series \(\bar{v}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{3^{2}}-\frac{1}{n_{2}^{2}}\right)\), n₂ = 4,5,6,…
Brackett series \(\bar{v}=R_{H}\left(\frac{1}{4^{2}}-\frac{1}{n_{2}^{2}}\right)\),
n₂ = 5,6,7,…
Pfund series
\(\bar{v}=R_{H}\left(\frac{1}{5^{2}}-\frac{1}{n_{2}^{2}}\right)\),
n₂ = 6,7,8,…

Question 27.
Atomic spectrum is known as line spectrum. Why?
Answer:
This type of spectrum is given by atoms of all elements. These consist of various series of thin bright lines of every colour which are separated by black rays. Every element gives a special type of line spectrum and these spectrum gives a specific identity to those elements.
Example : In spectrum of sodium two clear yellow lines of wavelength 5890A and 5896A are observed.If electric current is passed at low pressure through hydrogen gas filled in a discharge tube then a red light is observed. If light emitted in this way is analysed by the help of spectr6scope then a continuous spectrum containing four main lines is observed which are known as H_α,H_β,H_γ, and H_δ. Such type of spectrum in which only lines are present is called line spectrum.

Question 28.
Write the main conclusions of Rutherford’s scattering experiment.
Answer:
Conclusions of Rutherford’s experiment:
1. Most of the α-particles passed through the gold foil undeflected which concludes that there is a lot of empty space in an atom.
2. Few α-particles were deflected through small angles. Thus the central part of an atom consist of positive charge which is called nucleus.
3. Out of the 20,000 particles bombarding only one particle gets deflected and returns back. Thus the volume of nucleus is a minute part of the atom. Rutherford determined that the radius of the atom is 10^-8 cm (1 Å) and radius of nucleus is 10^-13cm.
4. Electrons revolve in great speed around the nucleus. Due to this speed the centrifugal force produced balanced the electrostatic force of attraction between the positively charged nucleus and the electrons. Due to this balance electron do not fall into the nucleus.

Question 29.
What are Iso-electronic ions? Give examples.
Answer:
Such ions of different elements which differ in charge but the number of electrons are same in ionic state are Called Iso-electronic Ions.
Example 1.

Example 2.
K⁺ and Ca²⁺ are iso-electronic ions.
No. of electrons in K⁺ = 18 Electronic configuration = 2, 8, 8
No. of electrons in Ca²⁺ = 18 Electronic configuration = 2, 8, 8.

Structure of Atom Class 11 Important Questions Long Answer Type

Question 1.
Derive an expression for radius of an orbit on the basis of Bohr’s principle.
Answer:
Bohr’s Radius :
In a hydrogen atom an electron revolves around the centre of spherical orbit. If the radius of this orbit is r and atomic number of hydrogen is Z, then total charge on nucleus is Ze. If a particle moves with constant velocity in the spherical orbit then the centripetal force (force towards the centre) and centrifugal force (force away from the centre) are equal.
Centripetal Force = Electrostatic attractive force between nucleus and electron.
\(=\frac{Z e \times e}{r^{2}}=\frac{Z e^{2}}{r^{2}}\)
and centrifugal force = \(\frac{m v^{2}}{r}\)
Thus \(\frac{Z e^{2}}{r^{2}}=\frac{m v^{2}}{r}\)
∴ r = \(\frac{Z e^{2}}{m v^{2}}\) …(1)
According to Quantum theory, Angular momentum is equal to n\(\left(\frac{h}{2 \pi}\right)\)
Thus, mνr = n\(\frac{h}{2 \pi}\)
or
ν = \(\frac{n h}{2 \pi m r}\) …(2)
From eqns. (1) and (2),
r = \(\frac{n^{2} h^{2}}{4 \pi^{2} m Z e^{2}}\)
Here h, n, m, Z and e are constant.
On substituting the values
r = n² × 0.529 × 10^-8cm
This way radius (r) of First Bohr orbit (n = 1) of hydrogen is 0.529 × 10^-8 cm.

Question 2.
How is a light ray separates into constituent components by pris? Explain with figure.
Answer:

When a white light is allowed to pass through a prism, then its components gets separated into bands of different colours and wavelength. If a photographic plate is placed on the side of the prism, then the seven colours of light form seven bands which is known as continuous spectrum. In this band, one colour band partially overlaps the other colour band. In it, wavelength of violet colour is minimum but refraction is maximum due to which it bends more and gets refracted and its frequency is maximum. Wavelength of red colour is maximum and its frequency is minimum.

Question 3.
What is Heisenberg’s uncertainty principle? Write its mathematical ex-pression.
Answer:
Heisenberg pointed out uncertainty in determination of two conjugate properties simultaneously. According to this principle, it is impossible to determine charge momentum and charge position of electron simultaneously.
Mathematical form of uncertainty principle,
Δx × Ap≥\(\frac{h}{4 \pi}\)
Where, Δx = Uncertainty in position, Δp = Uncertainty in momentum, h = Planck constant.
This equation shows that uncertainty in position (Δx) and uncertainty in momentum (Δp) are inversely proportional to each other i.e.,
Δx ∝ \(\frac{1}{\Delta p}\)
If the uncertainty in position is reduced any way, its uncertainty in velocity or momentum will increase definitely.

Physical concept of uncertainty principle :
It can be understood by the study of effect of light on micro objects like electrons.
Effect of light on position and momentum of micro objects :

In order to observe an object, the wavelength of the incident light should be shorter than its size. If we use light radiations of higher wavelengths, the light reflected does not fall in the visible region. Hence, these are not visible and exact position cannot be determined. But, if we use radiations of high frequency or wavelength shorter than size of object, the electrons will be visible but the photons of high frequency change the momentum. This is because, the photons transfer their energy to the electrons during collision, as a result the electrons change their path.

Question 4.
What do you mean by quantum numbers? Explain the informations obtained by each quantum number. [Most Imp.]
Answer:
There are four types of quantum numbers :
1. Principal quantum number :
It is represented by n. This number informs about the main shell of the atom. It expresses the energy level and its distance from the nucleus in which electron revolves i.e., it determines the size of atom. The value of n are 1, 2, 3, 4, , means electron exist in K, L, M, N,…. energy levels.

2. Azimuthal quantum number :
It is represented by l. It expresses the angular momentum of electron in an elliptical state around the nucleus i.e., the shape of subshell is indicated.
Value of l may be from 0 to (n – 1) i.e.,
For n = 1 value of l = 0 subshell is one s
For n = 2 value of l = 0, 1, subshells are two s and p.

3. Magnetic quantum number :
According to this under magnetic influence the fundamental line for a quantum number m is oriented in various directions. This orientation means an electron of a subshell due to change in energy revolve in different directions. These orientations are called orbitals and are represented by number m. The possible values of m depend upon value of l i.e., for any value of l the value of m may be – l to + l passing through zero.
For 5 electron, l = 0, then m = 0
For/? electron, l = 1, then m = -1, 0, +1
For d electron, l =2, then m = – 2, -1, 0, +1, +2.

4. Spin quantum number :
It is represented by s. From the observation of various spectral evidences it has become clear that each electron in addition to its motion around the nucleus spins like a top about its own axis. As a result of it, angular momentum of electron is increased hence its energy changes. This spinning energy is represented by spin quantum number.
When there is a single electron in the orbital it spins clockwise and its magnetic field
is expressed as +\(\frac{1}{2}\) or [↑] but when another electron enters the same orbital it spins anticlockwise and its magnetic field is expressed as – \(\frac{1}{2}\) or [↑]

Question 5.
Describe Bohr model by different spectral lines present in hydrogen spectrum.
Or,
How is Bohr’s atomic theory helpful to explain line spectrum of hydrogen atom? Describe.
Answer:
Bohr’s model of hydrogen spectrum: In line spectrum of hydrogen atom five series of many lines are obtained. In ultraviolet region Lyman series, in visible region Balmer series and in infrared region Paschen, Brackett and Pfund series are found.

In hydrogen atom only one electron is present. Neil Bohr in 1913 explain the pres-ence of different series. According to Bohr, different energy levels of definite energy are found in atom in which various electrons rotate around the nucleus. Many hydrogen atoms are present in hydrogen gas. When same energy is supplied various electrons jumps to higher energy orbitals by absorbing different amount of energy but these are unstable there. These electrons return back to lower energy levels by emitting photons of different frequencies. Due to these emissions, various lines are obtained in spectrum.

When electrons return to n = 1 energy level from higher energy levels like 2, 3, 4; first, second or third lines of Lyman series are obtained and Balmer series obtained when electron return to n = 2 energy orbit from the higher orbits 3, 4, 5 etc. In such a way Paschen, Brackett and Pfund series are also obtained.
Suppose an electron of energy level n,, having energy Ej, jumps to energy level n2 having energy E2. If the frequency of spectral lines is u then the energy difference will be
ΔE = E₂ – E₁

This is Rydberg’ eqation.
Where R_H = \(\frac{2 \pi^{2} Z^{2} e^{4} m}{c h^{3}}\) = Rydberg constant = 109677.8cm^-1
The value of Rydberg constant is equal to its experimental value.
If the value of n1 is being put 1, 2, 3, 4 or 5, Lyman, Balmer, Paschen, Brackett,
Pfund series of spectral lines are obtained which are accordingly with Bohr’s model.

Question 6.
What is the shape of s,p and d orbital?
Answer:
s-orbital:
Azimuthal quantum number determines the shape of an orbital. If l = 0 then value of angular momentum of the orbital is zero. Thus, s orbital is non-directional and spherically symmetrical. Thus, the probability of the presence of electron up to a definite distance is same in all directions. With the increase in principaL quantum number, size of s-orbital also increases. If value of principal quantum number is n, then in s-orbital n is symmetrically spherical and with the increase in the value of n, energy of s-orbital alsoincreases and number of nodal planes for principal quantum number n is n -1.

Shape of p – orbital : If n = 2 then l=0, 1. For l = 1, value of m are three-1,0,+ 1. It
means p-subshell has three orbitais represented as p_x‚p_y and p_z whose energies are same.
But their orientation is different. p_x‚p_y, p_z are symmetrical in the directions of X, Y and Z axis, p-orbital is dumb-bell in shape. Both the lobes of every orbitai is separated by a plane where electron density is zero and it is called nodal plane.

Shape of d – orbital : For d-orbital l=2, thus five various values of magnetic quan
tum numbers are -2,-1, 0, +1, +2 which are represented as ,d_xy d_yz ,d_zx,d_{x² – y²},d_z² . Their sizes are different but energy is same, Size of three orbitals d_xy,d_xz, d_yz are same but their four lobes represent high electron density. They are of double dumb-bell shape. d_{x² – y²}
orbital is like d_xy, but the lobes are on X-and Y-axis. Lobes of d_z² orbital are on Z-axis. It has a ring representing high electron density which lies at XY-plane.

Question 7.
Explain Quantum Mechanical model of an atom.
Answer:
Concept of Quantum Mechanical Model of Atom :
Louis de Broglie’s hypothesis and Heisenberg’s uncertainty principle led to the development of new model of atom called quantum mechanical model.
de-Broglie, on the basis of similarity expressed the behavior of an electron like light i. e., wave and particle nature. He gave a mathematical expression for the wave obtained by minute particles like electrons and protons according to which
λ = \(\frac{h}{p}\)
or
λ = \(\frac{h}{m v}\)
Where λ = wavelength of wave obtained from particle, p = momentum of particle,
m mass of particle, ν = velocity of particle.

Heisenberg’s uncertainty principle :
Heisenberg pointed out uncertainty in deter-mination of two conjugate properties simultaneously. According to this principle, it is im-possible to determine charge momentum and charge position of electron simultaneously.
Mathematical form of uncertainty principle,
Δx × Δp ≥\(\frac{h}{4 \pi}\)
Where, Δx = Uncertainty in position, Δp = Uncertainty in momentmh, h = Planck’s constant.
On the basis of wave nature and uncertainty principle of an electron structure of atom can be explained.

Probability picture of an Electron:
The concept of well-defined orbits suggested by Bohr is ruled out because according to the uncertainty principle the simultaneous determination of position and momentum of electron is not possible. The quantum mechanical model, on the other hand, considers the probability of presence of an electron with a particular energy in certain region of space. The probability of finding electron at a particular part in an atom is given by square of wave function i.e., Ψ² corresponding to that location.

The probability of finding electron may differ from one location to the other and even at a large distance from the nucleus it does not become zero, though it may become negligible. Due to this it is not possible to draw boundary enclosing the region of 100% probability. However, for simplicity we draw arbitrary boundary enclosing the region of maximum probability of 90-95% called orbital. Thus, orbital may be defined as “Region of space around the nucleus where the probability of finding on electron is maximum. ”

Schrodinger Equation :
In this model behaviour of the electron in an atom is de-scribed by an equation called Schrodinger wave equation.

Where, m is mass of the electron,
h is the Planck’s constant,
E is the total energy,
V is the potential energy of the electron,
x, y and z are the three space coordinate.
Ψ (Greek letter psi) is amplitude of the electron wave called wave function.
By the solution of Schrodinger’s equation three quantum numbers n, l and m are obtained which determine the size and shape of electrons.
Value of l may be from 0 to (n-1) i.e.
For n = 1 value of l = 0 subshell is one s
For n = 2 value of l = 0,1, subshell are two s and p.

Question 8.
How does Schrodinger Equation represent the possibility of the presence of electron near the nucleus ? What do you understand by orbital?
Answer:
Possibility of Presence of Electron : Schrodinger Equation keeps into account the particle and wave nature of de-Broglie. Both the principles were given a mathematical form known as Schrodinger equation.

Where x,y, z axis, m= mass of electron, h = Planck’s constant, E = Energy, V = Potential energy, Φ = wave function.
Solution of equation, gives the knowledge of various energy levels, radiated frequency for hydrogen atom which is represented by the symbol θ.

Utility of Wave-function :
In an atom, orbitals of electron, wave function or rihas no physical significance on itself. Square of wave function in a point represents the probability of density of electron at that point. In Schrodinger equation energy level is represented by the square of wave function. Higher the value of square of wave function higher is the possibility of the presence of electron. If the value of square of wave function (θ²) of a point is zero, then the probability of finding the electron at that place is also zero.
θ² is also known as probability density and it is positive. In an atom at various points by the value of θ² can predict the maximum probability of the presence of electron at a place.

By this equation energy of electrons in an atom is quantized i. e., it has certain specific values. Corresponding wave functions related to every energy level is also found. Corresponding wave functions are characterized by a group of three quantum numbers. These are Principal, Azimuthal and Magnetic Quantum numbers. These quantum numbers are obtained by the solution of Schrodinger wave equation. According to Heisenberg’s uncertainty principle actual position and actual momentum of an electron cannot be determined simultaneously.

Path of electron is not accurate but only possible.. This led to the concept of atomic orbital. Atomic orbit is denoted by Φ. For every electron, such types of various wave functions are possible. Thus, their corresponding atomic orbital are also possible. Energy of electron in every orbit is definite. These orbitals cannot have more than 2 electrons. Region around the nucleus where the probability of presence of an electron is maximum 90-95% is called orbital.

In the stage of H- atom, there is le^– with charge -e and nuclear charge +e, its potential energy can be given as follows :
V = \(\frac{-e^{2}}{r}\)
∴ On the basis of above facts Schrodinger wave equation can be written as follows :

Structure of Atom Class 11 Important Numerical Questions

Question 1.
(i) Calculate the number of electrons which will together weigh one gram,
(ii) Calculate the mass and charge of one mole of electrons.
Solution:
(i) ∵ Mass of 1 electron = 9.11 × 10^-28 g (or 9.11 × 10^-31 kg )
No. of electrons in mass of 1 gm = \(\frac{1}{9 \cdot 11 \times 10^{-28}}\) = 1.0976 × 10²⁷ electrons

(ii) ∵ Mass of 1 electron = 9.11 × 10^-31 kg
Mass of 1 mole electron = 9.11 × 10^-31 × 6.022 × 10²³
= 54.86 × 10<sup-8kg = 5.486 × 10^-7kg
Charge on 1 electron = 1.602 × 10^-19 C
Charge on 1 mole electron = 1.602 × 10^-19 × 6.022 × 10²³
= 9.647 × 10⁴C

Question 2.
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C.
[Assume that mass of a neutron = 1.675 × 10^-27 kg ]
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at S.T.P.
Will the answer change if the temperature and pressure are changed?
Solution:
(i) No. of electrons in one molecule of methane (CH₄)
= 6 + 4= 10
(6 from carbon and 1 from each hydrogen)
∴ No. of electrons present in 1 mole methane
= 6.022 × 10²³ × 10
= 6.022 × 10²⁴

(ii)
(a) 1 mole ¹⁴C = 14g = 6.022 × 10²³carbonatom.
∵ No. of neutrons in 1 carbon atom = Mass number – Atomic number
= 14-6 = 8 neutrons.
∵ No. of neutrons in 6.022 × 10²³ carbon atoms
= 6.022 x10²³ × 8
Or, No. of neutrons in 14 g of Carbon-14 = 6.022 × 10²³ × 8

∴ No of neutrons in 7 mg (or 7 × 10^-3 g) = \(\frac{7 \times 10^{-3} \times 6 \cdot 022 \times 10^{23} \times 8}{14}\)
= 24.088 × 10²⁰
= 2.4088 × 10²¹ Neutrons.

(b) ∵ Mass of 1 neutron = 1.675 × 10²⁷ kg
∴ Massof 2.4088 × 10²¹ neutrons = 2.4088 × 10²¹ × 1.675 × 10^-27kg
= 4.0347 × 10^-6 kg.

(iii) No. of protons in 1 mole NH₃ = 7 + 3 = 10 mole proton,
(7 in N and 1 in each H)
No. of 10 mole protons = 6.022 × 10²³ × 10
= 6.022 × 10²⁴ protons.

(a) ∵ No. of protons in 17 g NH₃ = 6.022 × 10²⁴ protons
∴ No. of protons in 34 mg (or 34 × 10^-3g) NH₃
\(=\frac{34 \times 10^{-3} \times 6 \cdot 022 \times 10^{24}}{17}\)
= 12.044 × 10²¹ protons.
= 1.2044 × 10²² protons.

(b) ∵ Mass of 1 proton = 1.6726 × 10^-27kg
∴ Mass of 1.2044 × 10²²protons = 1.2044 × 10²² × 1.6726 × 10^-27kg
= 2.01447 × 10^-5 kg
There is no effect of pressure and temperature on the number of basic particles. Thus, answer will be same.

Question 3.
Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (υ) and wave number (\(\text { (\overline{U) } }\)) of the yellow light.
Solution:
Frequency (υ) = \(\frac{c}{\lambda}\) ,(1nm = 10m^-9m)
= 580 nm = 580 × 10^-9m = 580 × 10^-7cm
υ = \(\frac{c}{\lambda}=\frac{3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{580 \times 10^{-9} \mathrm{~m}}\) = 5.17 × 10¹⁴s^-1 (∵ Velocity of light, c = 3 × 10⁸ms^-1)
Wave number, \(\bar{v}=\frac{1}{\lambda}=\frac{1}{580 \times 10^{-7} \mathrm{~cm}}\) = 1.724 10⁴cm^-1

Question 4.
Calculate the frequency of a light wave whose time period is 2.0 × 10^-10s.
Solution:
Frequency υ = \(\frac{1}{\text { Time period }}\) = \(\frac{1}{2 \cdot 0 \times 10^{-10} \mathrm{~s}}\) = 5 × 10^-9 s^-1.

Question 5.
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Solution:
Energy of 1 photon E = \(\frac{h c}{\lambda}\) = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{Js} \times 3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{4000 \times 10^{-12} \mathrm{~m}}\)
(∵ 1pm = 10^-12m
= 4.9695 × 10^-17J
No. of photons N = \(\frac{1 \mathrm{~J}}{4 \cdot 9695 \times 10^{-17} \mathrm{~J}}\) = 2.0122 × 10¹⁶.

Question 6.
A photon of wavelength 4 × 10^-7 m strikes on metal surface, the work function of the metal being 2-13eV. Calculate
(i) The energy of photon (eV)
(ii) The kinetic energy of emission and
(iii) The velocity of the photoelectron. (leV = 1.6020 × 10^-19J)
Solution:
(i) Energy of a photon E = \(\frac{h c}{\lambda}\) = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{Js} \times 3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{4 \times 10^{-7} \mathrm{~m}}\)
E = 4.969 × 10^-19.J (1.602 × 10^-19J = 1eV)
or
E = \(\frac{4 \cdot 969 \times 10^{-19}}{1 \cdot 602 \times 10^{-19}}\) = 3.10eV

(ii) Kinetic Energy of emitted electron, K.E. = hυ – hυ₀
hυ = 3.10e V (Energy of striking photon)
hυ₀ = W₀ =2.13eV (Work function of metal)
K.E. = 3.10 – 2.13 = 0.97eV

(iii) K.E. = \(\frac{1}{2}\)mν² = 0.97eV (∵ 1eV = 1.602 × 10^-19J)
or
\(\frac{1}{2}\)mν² = 0.97 × 1.602 × 10^-19 J
or
= \(\frac{1}{2}\) × 9.1 1 × 10^-31kg × v² = 0.97 × 1.602 × 10^-19 J
∵(Mass of1e = 9.11 × 10^-31kg)
or
ν² = \(\frac{0.97 \times 1.602 \times 10^{-19} \mathrm{~J} \times 2}{9.11 \times 10^{-31} \mathrm{~kg}}\) = 0.341 × 10¹²
or
ν = \(\sqrt{0 \cdot 341 \times 10^{12}}\)
or
ν = 5.84 10⁵ms^-1

Question 7.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.
Solution:
Wavelength λ = 242 × 10^-9 m, (∵ 1 nm = 10^-9 m)
Energy E = hυ = \(\frac{h c}{\lambda}\) = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{Js} \times 3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{242 \times 10^{-9} \mathrm{~m}}\)
E = 0.0821 × 10^-17 J/atom
Above energy is sufficient for the ionisation of 1 Na atom. Energy for 1 mole Na is the ionisation energy of Na.
Energy for 1 mole Na (Ionisation energy)
E = 6.02 × 10²³ × 0.0821 × 10^-17 J/mol
= 4.945 × 10⁵ J/mol
\(=\frac{4 \cdot 945 \times 10^{5}}{1000}\)KJ/mol ( ∵1KJ = 1000J)
= 4.945 × 10² kJ/mol

Question 8.
A 25 watt bulb emits monochromatic yellow light of wavelength 0.57µm. Calculate the rate of emission of quanta per second.
Solution:
Wavelength λ = 0.57 µpm = 0.57 × 10^-6m (∵ 1µm = 10^-6 m)
Energy of photon E hυ = \(\frac{h c}{\lambda}\) = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{Js} \times 3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{0.57 \times 10^{-6} \mathrm{~m}}\)
or
E = 34.84 × 10^-20J
25 watt = 25Js^-1 ( ∵1watt = 1Js^-1)
Number of photons emitted per second = \(\frac{\text { Total Energy per second }}{\text { Energy of } 1 \text { photon }}\)
\(=\frac{25 \mathrm{Js}^{-1}}{34 \cdot 87 \times 10^{-20} \mathrm{~J}}\)
= 0.7169 × 10²⁰ photon/second
= 7.169 × 10¹⁹ photon/second.

Question 9.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (υ₀) and work function (W₀) of the metal.
Solution:
Threshold wavelength, λ₀ =6800Å = 6800 × 10^-10m (∵ 1Å= 10^-10m)
Threshold Frequency υ₀ = \(\frac{c}{\lambda_{0}}\) = \(\frac{3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{6800 \times 10^{-10} \mathrm{~m}}\) = 4.41 × 10¹⁴ s^-1
Work function W₀ = hυ₀
= 6.626 × 10^-34Js × 4.41 × 10¹⁴s^-1
= 29.22 × 10^-20J=2.922 × 10^-19J.

Question 10.
What is the wavelength of Light emitted when the electron in a hydrogen
atom undergoes transition from an energý level with n = 4 to an energy level with
n = 2?
Solution:
Wavenumber, \(\vec{v}=\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)
(Where R Rydberg constant = 109677cm^-1)

= 486 × 10^-9m = 486nm
Colour of this wavelength of light is blue.

Question 11.
How much energy is required to ionise a H-atom if the electron occupies n = 5 orbit. Compare your answer with the ionisation enthalpy of H-atom (energy required to remove an electron from n = 1 orbit is ionisation Enthalpy)
Solution:
Energy change ΔE = E_f – E_i
ΔE = 2.18 ×10^-18\(\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right)\) J
When n_i, = 5 and n_f = ∝Change in Energy
ΔE = 2.18 ×10^-18 \(\left(\frac{1}{c^{2}}-\frac{1}{\infty}\right)\) J=0.0872 × 10^-18J
When n_i = 1 and n_f = ∝, Energy change
ΔE1 = 2.18 × 10^-18\(\left(\frac{1}{1^{2}}-\frac{1}{\infty}\right)\)J = 2.18 10^-18J
∴ \(\frac{\Delta \mathrm{E}^{1}}{\Delta \mathrm{E}}\) = \(\frac{2 \cdot 18 \times 10^{-18}}{0 \cdot 0872 \times 10^{-18}}\) = 25
Thus, energy required to emit an electron from the first orbit is 25 times the energy required to emit an electron from the fifth orbit.

Question 12.
Dual behaviour of matter proposed by de-Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is
1.6 × 10⁶ms Calculate de-Broglie wavelength associated with this electron.
Solution:
de-Broglie wavelength, λ = \(\frac{h}{m v}\)
λ = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{Js}}{9 \cdot 11 \times 10^{-31} \mathrm{~kg} \times 1 \cdot 6 \times 10^{6} \mathrm{~ms}^{-1}}\) (1J = 1kgm²s^-2)
= 0.455 × 10^-9m = 4.55 × 10^-10m
= 455 × 10^-12m = 455pm.

Question 13.
Similar to electron diffraction, neutron diffraction microscope often used for the highly magnified images of biological molecules and other type of material. If the wavelength used here is 800pm, calculate the characteristic velocity associated
with the neutron. (Mass of Neutron = 1.675 × 10^-27kg)
Solution:
λ = \(\frac{h}{m v}\)
λ = 800pm = 800 × 10<sup-12m, m = 1.675 × 10^-27kg
∴ ν = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}}{1 \cdot 675 \times 10^{-27} \mathrm{~kg} \times 800 \times 10^{-12} \mathrm{~m}}\)
= 0.494 × 10³ ms^-1
= 4.94 × 10²ms^-1

Question 14.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225nm and ends at 211.6pm. Name the series to which this transition belongs and the region of the spectrum.
Solution:
Radius of n^th orbit of H like species :

If n₁ = 5 and n₂ = 2, then emission transition occurs from fifth orbit to second orbit. Thus, it represents Balmer series :

Question 15.
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as υ = 3.29 x 10¹⁵ \(\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right)\) Hz. Calculate the value of
n if the transition is observed at 1285 nm. Find the region of spectrum.
Solution:


0.0709 = 0.1111 – \(\frac{1}{n^{2}}\)
or
\(\frac{1}{n^{2}}\) = 0.1111-0.0709 = 0.0402 = 0.04 = \(\frac{1}{25}\)
or
n² = 25 or n = 5
∴ Electron jumps from n = 5 to n = 3. i.e., transition occurs in Paschen series and is present in infrared region.

Question 16.
An ion with mass number 56 contains 3 units of positive charge and 30*4% more neutrons than electrons. Assign the symbol to this ion.
Solution:
Let, No. of electrons in ion = x .
∴ No. of neutrons = x + \(\frac{30 \cdot 4}{100}\)x = 1.304x
(∵ No. of neutrons is 30.4% more than No. of electrons)
No. of electrons in neutral atom = x + 3 (∵ + 3 charge is on the ion)
Thus, No. of protons = x + 3
We know that, Mass no. = n + p = 1.304x + x + 3 = 56
2.304x = 53
or
x = \(\frac{53}{2 \cdot 304}\) = 23.003 ≈ 23
No. of protons = 23 + 3 = 26 = Atomic No.
Thus, symbol of ion is \({ }_{26}^{56} \mathrm{Fe}^{+3}\).

Question 17.
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons. Find the symbol of the ion.
Solution:
Mass no. of ion = 37
Thus, p + n = 37
Unit negative charge is one on the ion.
Let no. of Protons = x ?
No. of Electrons = x + 1
No. of Neutrons = 37- x
According to the equation,
x + 1 + x + 1 ×\(\frac{11 \cdot 1}{100}\) = 37 – x
or
(x + 1) \(\left[1+\frac{1 \cdot 1}{100}\right]\) = 37 – x
111.1(x+1) = 3700 – 100x
or
211.1x = 3700 -111.1 = 3588 .9
x = \(\frac{3588 \cdot 9}{211 \cdot 1}\) =17
No. of Protons = 17
No. of Electrons = 17+1 18
No. ofNeutrons =37 – 17=20
Thus, symbol of the ion will be \({ }_{17}^{37} \mathrm{Cl}^{-1}\)

Question 18.
Neon gas is generally used in sign boards. If it emits strongly at 616 nm.
Calculate
(a) The frequency of emission,
(b) Distance travelled by this radiation in 30s,
(c) Energy of quanta,
(d) No. of quanta present ¡fit produces 2J of energy.
Solution:
(a) Frequency u = \(\frac{c}{\lambda}\)
λ = 616nm = 616 × 10^-9 m
υ = \(\frac{3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{616 \times 10^{-9} \mathrm{~m}}\)
= 4.870 × 10¹⁴s^-1

(b) Distance travelled Speed × Time
= 3.0 × 10⁸ms^-1× 30s=9.0 × 10 ⁹m.

(c) Energy of quanta (or photon)
E = hυ = 6.626 × 10^-34Js × 4.870 × 10¹⁴s^-1
= 32.268 × 10^-20J= 3.2268 × 10^-19J.

(d) No. of quanta present = \(\frac{\text { Total energy produced }}{\text { Energy of one photon }}\)
= \(\frac{2 \mathrm{~J}}{3 \cdot 232 \times 10^{-19} \mathrm{~J}}\)
= 6.691 × 10¹⁸

Question 19.
The work function of sodium atom is 1.9 eV. Calculate
(a) The threshold wavelength of emitted radiation.
(b) Threshold frequency.
(c) If sodium element is irradiated with a wavelength of 500 nm, then calculate the kinetic energy of photoelectrons.
(d) Velocity of photoelectrons.
Solution:
(a) Work function W₀ = hυ₀
W₀ = 1.9eV = 1.9 × 1.602 × 10J
υ₀ = \(\frac{\mathrm{W}_{\mathrm{o}}}{h}\) = \(\frac{1 \cdot 9 \times 1 \cdot 602 \times 10^{-19} \mathrm{~J}}{6 \cdot 626 \times 10^{-34} \mathrm{Js}}\)
υ₀ = 4.59 × 10¹⁴S^-1

(b) Threshold wavelength λ₀ = \(\frac{c}{v_{0}}\) = \(\frac{3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{4 \cdot 59 \times 10^{14} \mathrm{~s}^{-1}}\) = 6.536 × 10^-7 m
= 653.6 × 10^-9m = 653.6nm.

(c) Kinetic energy of emitted photoelectrons, K.E. = h(υ – υ₀)
Wavelength of particle after striking λ = 500nm = 500 × 10^-9m
υ = \(\frac{c}{\lambda}\) = \(\frac{3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{500 \times 10^{-9} \mathrm{~m}}\) = 6.0 × 10¹⁴s^-1
∴ K.E. = 6.626 × 10^-34Js(6.0 × 10¹⁴s^-1-4.59 × 10¹⁴s^-1)
or
K.E. = 9.34 × 10^-20 J.

(d) Calculation of velocity :
Kinetic energy = \(\frac { 1 }{ 2 }\)mν²
9.34 × 10^-20 J = \(\frac { 1 }{ 2 }\) × 9.1 1 × 10^-31 kg × ν²
(Mass of electron =9.11 × 10^-31kg)
or
ν² = 2.050 × 10¹¹m²s^-2 (1J = 1kgm²s^-2
ν = \(\sqrt{20 \cdot 50 \times 10^{10} \mathrm{~m}^{2} \mathrm{~s}^{-2}}\)
ν = 4.527 × 10⁵ms^-1

Question 20.
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) Threshold wavelength and (b) Planck’s constant.

Solution:
Kinetic Energy (K.E.) = h(υ – υ₀) = \(\frac { 1 }{ 2 }\) mν²
Substituting the values obtained from the three experiments in the equation,

(i) 1^st Experiment:
λ = 500nm = 500 × 10^-9m



∴ h = 7.043 × 10^-34JS. (∵ 1J = 1 kg m²s^-2)

Question 21.
The ejection of photoelectrons from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35V when the radiation of 256.7 nm is used. Calculate the work function for silver metal.
Solution:
Energy of radiation used hυ = hυ₀ + K.E.
E = hυ = \(\frac{h c}{\lambda}\) = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{Js} \times 3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{256 \cdot 7 \times 10^{-9} \mathrm{~m}}\)
= 7.74 × 10^-19J = \(\frac{7.74 \times 10^{-19} \mathrm{~J}}{1.602 \times 10^{-19}}\) = 4.8eV
Kinetic energy is obtained by the potential applied
K.E. = eV₀
= 1.6 × 10^-19× 0.35 = \(\frac{0.56 \times 10^{19} \mathrm{~J}}{1 \cdot 6 \times 10^{-19}}\) = eV = 0.35eV
K.E. = 0.35eV
Work function, W₀ = hυ₀ = hυ – K.E.
= 4.83eV-0.35eV = 4.48eV

Question 22.
If the photon of wavelength 150 pm strikes an atom and one of this inner bound electrons is ejected out with a velocity of 1.5 × 10⁷ ms^-1. Calculate the energy with which it is bound to the nucleus.
Solution:
Energy of reflected radiation hυ = hυ₀ + \(\frac { 1 }{ 2 }\) mν²
E = hυ =\(\frac{h c}{\lambda}\) = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{Js} \times 3 \cdot 0 \times 10^{8} \mathrm{~ms}^{-1}}{150 \times 10^{-12} \mathrm{~m}}\)
(∵ 1pm = 10^-12m)
E = 13.25 × 10^-16J

Kinetic Energy of ejected electron K.E. = \(\frac { 1 }{ 2 }\) mν²
= \(\frac { 1 }{ 2 }\) × 9.11 × 10^-31kg × (1.5 × 10⁷ms^-1)²
K.E. = 1.025 × 10^-16J
W₀ = hυ₀ = hυ -K.E.
= 13.25 × 10^-16 J-1.025 × 10^-16 J= 12.225 × ^-16 J
= \(\frac{12 \cdot 225 \times 10^{-16}}{1 \cdot 602 \times 10^{-19}}\)eV =
7.63 × 10³eV

Question 23.
The velocity related to a proton moving with a potential difference of 1000V is 4.37 × 10⁵ ms^-1. If a hockey bail of mass 0.1 kg is moving with this velocity, then calculate the wavelength related to this.
Solution:
Wavelength related to hockey ball,
λ = \(\frac{h}{m v}\) = \(\frac{6 \cdot 626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}}{0 \cdot 1 \mathrm{~kg} \times 4 \cdot 37 \times 10^{5} \mathrm{~ms}^{-1}}\)
= 15.16 × 10^-39m = 1.516 × 10^-38m.

Question 24.
Nitrogen laser produces a radiation at wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 10²⁴. Calculate the power of this laser.
Solution:
If n photons are emitted from the laser, then total energy of the ejected photons will be equal to the capacity of the laser.
Energy of L photon, = \(\frac{h c}{\lambda}\) =
(∵ 1nm = 10^-9m

= 0-.05896 × 10^-17uJ
-Energy of 5.6 × 10<sup24photons = 0.05896 × 10^-17 × 5.6 × 10²⁴J
= 0.3302 × 10⁷J
= 3.302 × 10⁶J.

Question 25.
What transition in the hydrogen spectrum would have the same wave-length as Balmer transition n = 4 to n = 2 of He⁺ spectrum?

On solving, n₁ = 1 and n₂ = 2
Thus, wavelength of transition from n = 2 to n₁ = 1 in H spectrum will be equal to transition of n = 3 to n = 2 in Balmer series in He⁺.

Question 26.
Calculate the energy required for the process
He_(g)⁺ → He_(g)²⁺ + e^–
The ionisation energy for the H atom in the ground state is 2.18 10^-18 J atom^-1.
Solution:
Energy of electron in atomic orbital

E_n = \(\frac{-2 \pi^{2} m Z^{2} e^{4}}{n^{2} h^{2}}\)
For H-atom, ionisation energy (IE) =E_∝ – E₁
IE = 0 – \(\left(-\frac{2 \pi^{2} 1^{2} e^{4}}{1^{2} h^{2}}\right)\)
(∵ For H-atom Z = 1 and n = 1)
IE = 2.18 × 10^-18J atom^-1

For He⁺, IE = E_∝ – E₁, E₁ = 0 \(\left(\frac{-2 \pi^{2} 2^{2} e^{4}}{1^{2} h^{2}}\right)=4 \times \frac{2 \pi^{2} m e^{4}}{h^{2}}\)
=4 × 2.18 × 10^-18J atom^-1
= 8.72 × 10^-18J atom^-1
Thus, Energy required for He⁺ → He²⁺ +e^–
= 8.72 × 10^-18J atom^-1.

Question 27.
In Milikan’s experiment, statk electric charge on the oil drops has been
obtained by shining X-rays. If the static electric charge on the oil drop ¡s-1.282 × 10^-18C. Calculate the number of electrons present on it.
Solution:
No. of electrons = \(\frac{\text { Total charge on oil drop }}{\text { Charge on } 1 \text { electron }}\)

= \(\frac{-1 \cdot 282 \times 10^{-18} \mathrm{C}}{-1 \cdot 6022 \times 10^{-19} \mathrm{C}}\) = 0.800 × 10=8.0.

Question 28.
A certain particle carries 25 × 10^-16C of static electric charge. Calculate
the number of electrons present in it.
Solution:
Charge on 1 electron = 1.6022 × 10^-19C
Total charge = 2.5 × 10^-16 C
Number of electrons present in particles
\(=\frac{2 \cdot 5 \times 10^{-16} \mathrm{C}}{1 \cdot 6022 \times 10^{-19} \mathrm{C}}\) = 1.5603 × 10³
= 1560.

Question 29.
29.2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Solution:
Total length = 2.4 cm.
No. of carbon atoms arranged in line = 2 × 10⁸
∴ Diameter of 1 carbon atom = \(\frac{2 \cdot 4 \mathrm{~cm}}{2 \times 10^{8}}\) = 1.2 × 10^-8 cm
Radius of 1 carbon atom \(=\frac{\text { Diameter }}{2}=\frac{1 \cdot 2 \times 10^{-8}}{2}\)
= 0.60 × 10^-8cm
= 0.060 × 10^-7cm = 0.060 × 10^-9m = 0.060nm.

Question 30.
The diameter of Zinc atom is 2.6 Å. Calculate (a) Radius of Zinc atom in pm and (b) Number of atoms present in a length of 1.6 cm if the Zinc atoms are ar-ranged side by side lengthwise.
Solution:
(a) Diameter of Zn atom = 2.6Å = 2.6 × 10^-10 m
Radius of Zn atom = \(\frac{2 \cdot 6 \times 10^{-10} \mathrm{~m}}{2}\) = 1.3 × 10^-10m
= 130 × 10^-12m = 130pm

(b) Given, Length = 1.6 cm = 1.6 × 10^-2 m
Number of Zn atoms in 1.6 × 10^-2 m long line
\(=\frac{1 \cdot 6 \times 10^{-2} \mathrm{~m}}{2 \cdot 6 \times 10^{-10} \mathrm{~m}}\) = 0.6154 × 10⁸ = 6.154 × 10⁷

Structure of Atom Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Atomic number of an element is 27, and number of neutrons in its atom are 14, number of electrons will be :
(a) 14
(b) 27
(c) 13
(d) 41.
Answer:
(c) 13

Question 2.
Protons and Neutrons are collectively known as :
(a) Deutron
(b) Positron
(c) Meson
(d) Nucleon.
Answer:
(d) Nucleon.

Question 3.
Similarity found in Na⁺ ion and Ne element is :
(a) In number of protons
(b) In number of Neutrons
(c) In number of Electrons
(d) In Atomic mass.
Answer:
(c) In number of Electrons

Question 4.
According to Hund’s Rule Electronic configuration (1s²,2s²,2p^x1, 2p^y1 2p^z1,) is of which element:
(a) Of Oxygen
(b) Of Nitrogen
(c) Of Fluorine
(d) Of Boron.
Answer:
(b) Of Nitrogen

Question 5.
Value of all four quantum numbers of seventh electron of nitrogen is :
(a) n = 2,l = 1, m = +1, s = –\(\frac{1}{2}\)
(b) n = 2,l = 2, m = -1, s = +\(\frac{1}{2}\)
(c) n = 2, l =1, m = +1, s = +\(\frac{1}{2}\)
(d) n = 2,l = 2, m = +2,s = +\(\frac{1}{2}\)
Answer:
(c) n = 2, l =1, m = +1, s = +\(\frac{1}{2}\)

Question 6.
Total number of orbitals in Azimuthal quantum number of l subshell are :
(a) 2l + 1
(b) 3l + 1
(c) 4l + 1
(d)2(l+l).
Answer:
(a) 2l + 1

Question 7.
In Isotopes number of this is same:
(a) Proton
(b) Neutron
(c) Proton and Neutron
(d) Nucleon.
Answer:
(a) Proton

Question 8.
Which of the following expression represents de-Broglie relation:
(a) \(\frac{h}{m v}\) = p
(b) λ = \(\frac{h}{m v}\)
(c) λ = \(\frac{h}{m p}\)
(d) λm = \(\frac{h}{p}\)
Answer:
(b) λ = \(\frac{h}{m v}\)

Question 9.
In Rutherford’s experiment a-particles are bombarded on which metal:
(a) Aluminium
(b) Silver
(c) Iron
(d) Gold.
Answer:
(d) Gold.

Question 10.
What will be the number of electrons in a molecule of CO₂
(a) 22
(b) 44
(c) 11
(d) 38.
Answer:
(a) 22

Question 11.
A neutral atom of an element contain two K, nine M and two N electrons, then
total number of d electrons will be:
(a) 2
(b) 3
(c) 1
(d) 4.
Answer:
(c) 1

Question 12.
According to Heisenberg’s uncertainty principle :
(a) Δx.Δp ≥ \(\frac{h}{4 \pi}\)
(b) Δx.Δυ≥ \(\frac{h}{4 \pi}\)
(c) Δx\(\frac{c}{\lambda} \geq \frac{h}{4 \pi}\)
(d) Δx.Δm≥\(\frac{h}{4 p}\)
Answer:
(a) Δx.Δp ≥ \(\frac{h}{4 \pi}\)

Question 13
For an atom n = 3, / = 2 then value of m will be :
(a) -2, – 1,0, + 1, + 2
(b) -1, 0, +1
(c) -3, -1, 0, + 1, + 3
(d) -2, 0+1.
Answer:
(a) -2, – 1,0, + 1, + 2

Question 14.
By the removal of a neutron from the nucleus of an atom :
(a) Positive charge of atom increases
(b) Atomic mass decreases
(c) α and β-particles are ejected
(d) None of these.
Answer:
(b) Atomic mass decreases

Question 15.
Number of Electrons in Cl^– ion will be :
(a) 17
(b) 18
(c) 35
(d) 37.
Answer:
(b) 18

Question 16.
Which of the following configuration is not according to Aufbau’s Principle :
(a) 1s²,2s²p¹
(b) [Kr]4d¹⁰,5s²
(c) [Ar]3d¹⁰,4s¹
(d) [Ar]3d⁴,4s².
Answer:
(c) [Ar]3d¹⁰,4s¹

Question 17.
Number of orbitals in f sub-shell is :
(a) 3
(b) 4
(c) 5
(d) 7.
Answer:
(d) 7.

Question 18.
Lightest particle of an atom is :
(a) Proton
(b) Electron
(c) Neutron
(d) α -particle.
Answer:
(b) Electron

Question 19.
In Rutherford’s Experiment on which metal are α particles bombarded :
(a) Al
(b) Ag
(c) Fe
(d) Au.
Answer:
(d) Au.

Question 20.
For n = 3 values of l will be :
(a) 1, 2, 3
(b) 0, 1, 2
(c) 0 , 1, 3
(d) None of these.
Answer:
(b) 0, 1, 2

Question 21.
Which shell is nearest to the nucleus :
(a) M
(b) N
(c) K
(d) L.
Answer:
(c) K

Question 22.
Bohr’s Atomic model can explain :
(a) Spectrum of only hydrogen atom
(b) Spectrum of an atom or ion containing only one electron
(c) Spectrum of hydrogen molecule
(d) Spectrum of sun.
Answer:
(b) Spectrum of an atom or ion containing only one electron

Question 23.
Rutherford’s a-particle scattering experiment reaches the conclusion :
(a) Mass and energy are interrelated
(b) Electron resides in the region around the nucleus
(c) Neutron is deeply inserted in the nucleus
(d) Position of a point in a substance can be determined accurately.
Answer:
(b) Electron resides in the region around the nucleus

Question 24.
Order of mass of Neutron is :
(a) 10^-23 kg
(b) 10^-24 kg
(c) 10^-26 kg
(d) 10^-27 kg.
Answer:
(d) 10^-27 kg.

Question 25.
Two electrons of K shell will differ:
(a) In principle quantum number
(b) In azimuthal quantum number
(c) In magnetic quantum number
(d) In spin quantum number.
Answer:
(d) In spin quantum number.

Question 26.
Magnetic quantum number is related to :
(a) Size
(b) Shape
(c) Orientation
(d) Spin.
Answer:
(c) Orientation

Question 27.
Number of unpaired electrons in Ni2+ ion (For Ni, Z = 28):
(a) 0
(b) 2
(c) 4
(d) 8.
Answer:
(b) 2

Question 28.
In hydrogen spectrum lines of Lyman series are present in :
(a) UV
(b) IR
(c) Visible
(d) Remote.
Answer:
(a) UV

Question 29.
Electronic configuration (outermost) of Mn²⁺ ion Atomic no. of manganese = 25 in minimum energy state is :
(a) 3d⁵
(b) 2d⁴ 4s²
(c) 3d<sup3 4s²
(d) 3d²4s²4p².
Answer:
(a) 3d⁵

Question 30.
If nucleus of Electron, Hydrogen, Helium and Neon are moving with velocity of light, then order of wavelength connected to these particles will be :
(a) Electron > Hydrogen > Helium > Neon
(b) Electron > Helium > Hydrogen > Neon
(c) Electron < Hydrogen < Helium < Neon (d) Neon > Hydrogen > Helium > Electron.
Answer:
(a) Electron > Hydrogen > Helium > Neon

Question 31.
Atomic number of Ca is 20 and atomic mass is 40, which of the following statement is incorrect for Ca atom :
(a) Number of electrons is equal to number of neutrons
(b) Number of nucleons is twice the number of electrons
(c) Number of protons are half the number of neutrons
(d) Number of nucleons is twice the atomic number.
Answer:
(c) Number of protons are half the number of neutrons

Question 32.
Four quantum numbers of outermost electrons of K (Atomic No. = 19) :
(a) n = 2,l = 0,m = 0,s = +\(\frac{1}{2}\)
(b) n = 4,l = 0,m 0,s = +\(\frac{1}{2}\)
(c) n = 3,l = 1 ,m = 1,s = +\(\frac{1}{2}\)
(d) n = 4,l = 2, m = -1, s = +\(\frac{1}{2}\)
Answer:
(b) n = 4,l = 0,m 0,s = +\(\frac{1}{2}\)

Question 33.
How many mole electrons are in one kilogram :
(a) 6.023 × 10²⁵
(b) \(\frac{1}{9 \cdot 108}\) × 10³¹
(c) \(\frac{6 \cdot 02}{9 \cdot 108}\) × 10⁵⁴
(d) \(\frac{1}{9 \cdot 108 \times 6 \cdot 02}\) × 10⁸
Answer:
(d) \(\frac{1}{9 \cdot 108 \times 6 \cdot 02}\) × 10⁸

Question 34.
Total number of orbitals for principal quantum number n =:
(a) 2n
(b) n²
(c) 2n²
(d) n + l.
Answer:
(b) n²

Question 35.
An outermost electron represents the following quantum numbers n = 3, l= 2, m = + 2 and s = +1/2. Atomic number will be :
(a) 13
(b) 21
(c) 29
(d) 39.
Answer:
(c) 29

Question 36.
An ion isoelectronic with carbon monoxide is represented as :
(a) O₂^–
(b) N₂⁺
(c) CN^–
(d) O₂⁺
Answer:
(c) CN^–

2. Fill in the blanks:

1. Atomic -number of an atom is 20 and atomic mass is 40. Number of neutrons in its
nucleus will be ………………….
Answer:
20

2. Atomic number of chlorine gas is 17. Number of electrons in its outermost shell will
be ………………………
Answer:
7

3. If the sum of total magnetic quantum numbers is 7, then value of azimuthal quantum
number will be …………………….
Answer:
3

4. Nucleus of an atom contain ………………….. and ………………….
Answer:
Proton, Neutron

5. Neutron was discovered by ………………..
Answer:
Chadwick

6. Magnitude of mass of neutron is ………………….
Answer:
10^-27 kg

7. Lines of Lymann series in hydrogen spectrum are present in ………………….
Answer:
Ultraviolet rays

8. Radiations of highest wavelength are …………………
Answer:
Radiowaves

9. If value of azimuthal quantum number is one, then shape of orbital will be ………………..
Answer:
Dumb-bell

10. Region in space around the nucleus where there is maximum probability of presence
of electron is called ……………………..
Answer:
Orbital

11. Atomic number of an element is 30 and its mass number is 66. In its atom number of
protons is ………………… and number of neutrons is ………………….
Answer:
30, 36

12. Magnetic quantum number is related to …………………
Answer:
Orientation

13. In Isotopes number of ……………….. is same.
Answer:
Proton

3. Match the following:

‘A’	‘B’
1. Goldstein	(a) Atomic number
2. J.J. Thomson	(b) Molarity
3. Moseley	(c) Anode rays
4. Rutherford	(d) Structure of atom
5. Rydberg	(e) Cathode rays
6. Soddy	(f) Uncertainty principle
7. Heisenberg	(g) Large elliptical orbit
8. Sommer Field	(h) Isotopes.

Answer:
1. (c) Anode rays
2. (e) Cathode rays
3. (a) Atomic number
4. (d) Structure of atom
5. (b) Molarity
6. (h) Isotopes.
7. (f) Uncertainty principle
8. (g) Large elliptical orbit

4. Answer in one word/sentence:

1. Who discovered electron?
Answer:
J. J. Thomson

2. What is de-Broglie equation?
Answer:
λ = \(\frac{h}{m v}=\frac{h}{p}\)

3. Azimuthal quantum number is represented by.
Answer:
l

4. Formula of Heisenberg’s uncertainty principle is.
Answer:
Δx.Δp ≥ \(\frac{h}{4 \pi}\)

5. Who discovered neutron?
Answer:
Chadwick

6. Maximum number of electrons in first and second shell is.
Answer:
2, 8

These MP Board Class 11th Biology Notes for Chapter 13 Photosynthesis in Higher Plants help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 13 Photosynthesis in Higher Plants

→ Photosynthesis is a basic most significant process on which the life on earth entirely depends.

→ It is the formation of carbon containing compounds from CO₂ and H₂O by illuminated green cells.

→ There are two phases in the process of photosynthesis, photochemical (light dependent) and chemical (light independent).

→ Two kinds of photosystems control the entire light phase. The pigment system-I (PS-I) and pigment system-II (PS-II).

→ Two kinds of photophosphorylation are there : cyclic and non-cyclic.

→ The dark or chemical phase is called Calvin cycle. This cycle includes carboxylative phase, reductive phase and regenerative phase. Many enzymes control the process.

→ The first CO₂ acceptor is ribulose diphosphate and the first stable product is phosphoglycer- aldehyde.

→ Hatch and Slack cycle also called C₄ cycle is reported in many grasses and some dicots. The CO₂ acceptor is phosphoenol pyruvate (PEP) and the first stable product is oxaloacetate.

→ Blackman and his associates (1905-1915) discovered the light and dark reactions of the pho-tosynthesis.

→ The law of minimum was proposed by Liebig.

→ Law of limiting factors was proposed by Blackman in 1905.

→ Melvin, Calvin and Andrews Benson (1940-1950) have described the C₃ cycle of photosyn-thesis.

→ Daniel Arnon has shown that the light reaction of photosynthesis takes place in grana, whereas dark reaction takes place within stroma.

→ Rouhani and his associates (1973) have discovered the CAM cycle of photosynthesis in Sedum (a succulent plant).

→ Robert Emerson discovered the red drop and Emerson enhancement effect. Two pigment system was also discovered by them.

→ Chl-a 700 (P 700) was discovered by Clayton which is the reaction centre of PS-I.

→ Photochemical breakdown of water always takes place during PS-II only because of the presence of a strong oxidant NADP.

→ The end-products of light reaction are ATP, O₂ and NADPH₂.

→ The acceptor of CO₂ molecule in C₃ cycle is Ribulose-1, 6-Diphosphate.

→ RuDP carboxylase enzyme catalyze the carboxylation reaction of C₃ cycle and is active only in presence of higher CO₂ concentration.

→ The first stable product of C₃ cycle is PGA.

→ The CO₂ acceptor molecule of C₄ cycle is PEP.

→ PEP carboxylase enzyme catalyzes the carboxylation reaction in C₄ plants and is active in presence of lower CO₂ concentration too.

→ The leaves of C₄ plants exhibit Kranz type of leaf anatomy.

→ In C₄ plants PEP-C enzyme is found within the chloroplast of mesophyll cells, whereas RuDP-C enzyme is found within the chloroplast of bundle sheath cells.

→ In C₄ plants carbohydrates are synthesized within bundle sheath cells.

→ The amount of chlorophyll in plants is measured with the help of colorimeter.

→ H₂O is the hydrogen donor of green plants, whereas H₂S functioning as hydrogen donor in bacterial photosynthesis.

→ The number of oxygen atoms evolved from per quantum of light is called as quantum yield.

→ Red light shows highest rate of photosynthesis and blue light shows second highest rate of photosynthesis.

These MP Board Class 11th Biology Notes for Chapter 12 Mineral Nutrition help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 12 Mineral Nutrition

→ Mineral elements which are necessary for the growth of plants are called essential elements.

→ Mineral elements which are required in very small amounts are called trace elements.

→ Mineral elements which are usually radioactive and are used to trace out some metabolic processes are called tracer elements.

→ Mineral elements are absorbed in molecular form.

→ The necessity of a particular element can be tested by growing plants in soilless or solution culture (hydroponics).

→ Soilless culture or water culture of plants is known as hydroponics.

→ The deficiency of Mg results in the inhibition of chlorophyll synthesis.

→ Mn, B, Zn, Cu, Mo, and Cl are the micro or trace elements.

→ Premature leaf fall is caused due to deficiency of phosphorus.

→ Mg and Fe is necessary for chlorophyll synthesis.

→ Osmosis was first discovered by Nollet in 1748, using pigs bladder as a membrane filled with alcohol.

→ N, S, P, Mg, K are some very important macronutrients needed for the metabolic activities of plants, Zn, Cu, Fe, Mn, B, Mo are some of the micronutrients essential for the plant and are mainly catalytic in function.

→ Active absorption of molecules and ions requires metabolic energy and also a carrier molecule.

→ Root hair absorb only capillary water of the soil.

→ The molybdenum (Mo) is a metal activator of the enzyme nitrate reductase.

→ Deficiency of molybdenum causes whiptail disease in Cauliflower and also inhibits the flow-ering.

→ Coenzyme A, vitamin, thiamine and biotin are sulphur containing compounds.

→ Leghaemoglobin is a pink coloured pigment found in the root nodules of leguminous plants.

→ Bacterium Rhizobium is found in the root nodules of leguminous plants. It can fix atmospheric nitrogen in presence of leghaemoglobin.

→ Dionaea is an insectivorous plant.

→ Ringing or girdling experiment was done for the first time by Curtis (1925).

→ Organic compounds are translocated through phloem while inorganic compounds and water molecules are translocated through lumen of xylem.

→ Electroosmotic theory of translocation of organic solutes was proposed by Fenson (1957) and Spanner (1958).

→ Mass flow hypothesis for translocation of organic solutes was proposed by Munch (1930).

→ Protoplasmic streaming or cytoplasmic streaming theory of translocation of organic solutes was proposed by de Vries (1885).

→ The organic food formed by leaves is transported in the form of sucrose.

These MP Board Class 11th Biology Notes for Chapter 11 Transport in Plants help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 11 Transport in Plants

→ Loss of water in the form of vapour from aerial part of the plants through stomata, lenticel or cuticle is called as transpiration.

→ Loss of water in the form of vapour from any surface of the plant body is called as evapora¬tion.

→ In stomatal transpiration guard cells control the opening and closing of stomata.

→ Loftfield has classified the plants into three groups on the basis of daily stomatal movement: (i) Alfalfa type, (ii) Potato type, (iii) Barley type.

→ Guard cells contain a large number of chloroplasts.

→ The inner wall of guard cells is thick while outer wall is thin. Both walls are elastic.

→ According to Curtis, ‘Transpiration is a necessary evil’.

→ The opening and closing of stomata is being controlled by turgidity and flaccidity of guard cells respectively.

→ Stomatal pores open when guard cells are fully turgid.

→ During daytime stomata opens because guard cells photosynthesize and produce osmotically active substance.

→ Osmotic theory of stomatal opening and closing was proposed by Von Mohl (1856).

→ Starch ⇌ sugar hypothesis of stomatal opening and closing was proposed by Lloyd.

→ According to Steward, stomata is open during daytime when starch is converted into sugar at high pH and closed when sugars are converted into starch at low pH.

→ Phosphorylase enzyme was discovered by Hanes. It converts starch into glucose-1 -phosphate.

→ According to Steward, the OP of guard cells is not effected unless glucose-1 -phosphate is not converted into glucose and inorganic phosphate.

→ The rate of transpiration is measured with the help of potometer.

→ K⁺ ion transport method for stomatal opening and closing was proposed by Health, Zelitsch and Ruscka.

→ Proton transport method for the opening and closing of stomata was proposed by Levitt in 1974.

→ According to Levitt, when K⁺ ions enter the guard cell, stomata opens.

→ Loss of water in the form of water droplets from the margins of leaves is called as guttation.

→ Guttation takes place through hydathodes.

→ The cavity and loose tissue present below the hydathodes is called epithem.

→ In succulent plants stomata are open during daytime and closed during night-time. It is called as photoactive opening of stomata and plants are called photoactive.

→ The term guttation was proposed by Burgrstein, for exudation of water in the form of liquid from the margins of the leaves.

→ Xerophytic plants have higher rate of transpiration than mesophytes.

→ Percentage of water in the soil at permanent wilting point is called wilting coefficient.

→ Abscisic acid (ABA) results in the closure of stomata. It appears to block the active excretion of H+ ions from guard cells.

→ Plants can utilize only capillary water of the soil.

→ Difference between the diffusion pressure of a solution and its solvent is called as diffusion pressure deficit (DPD).
DPD = OP – TP or WP.

These MP Board Class 11th Biology Notes for Chapter 10 Cell Cycle and Cell Division help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 10 Cell Cycle and Cell Division

→ Multicellular organism grows and develop by mitosis cell division, whereas some unicellular organisms reproduce by this division.

→ Body of some lower category of organisms are made up of haploid cells. Zygote of these organ-isms divides by meiosis division to form haploid individual.

→ Meiosis cell division helps in the maintenance of chromosome number in the species.

→ Spindles regulate movement of chromosomes during cell division.

→ Mitosis help to maintain genetic stability.

→ Nerve cells never divide.

→ Generally in normal condition cells complete cell cycle in 20-25 hours.

→ Duration of cell cycle depends upon type of cell and some external factors such as temperature, nutrition, availability of oxygen, etc.

→ Actual cell division occurs in M-phase of the cell cycle.

→ Duration of cell cycle in bacteria is 20 minutes, in epithelial cell is 8-10 hours and in root cells of onion is 20 hours.

→ Cell division is the basis of continuity of life.

→ The main difference between the cell division of plant cell and animal cell are in centriole mechanism and cytokinesis process.

These MP Board Class 11th Biology Notes for Chapter 9 Biomolecules help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 9 Biomolecules

→ Body of the living organisms is made up of many mineral elements.

→ Elements are found either in free form or in the form of compound in the living body.

→ Collection of various types of molecules in a cell is called as cellular pool.

→ Smaller molecules found in the cells are called as micromolecules and larger molecules found in the cells are called as macromolecules.

→ Sugar, amino acids, nucleotides, minerals and water are the important micromolecules found in the cells.

→ Carbohydrates are formed by the union of carbon, hydrogen and oxygen in the ratio of 1: 2 :1.

→ Fat and fat like substances are called as lipids.

→ Carbohydrates and fats provide energy to the body.

→ Some nucleotides and their derivatives act as coenzymes.

→ General formula of carbohydrates is C_x(H₂O)_y.

→ C, H and O forms 93% part of the body.

→ Approximately 22 types of amino acids are found in the cells of living organisms.

→ Proteins are formed by the polymerization of amino acids.

→ Nucleotides are the basic units of nucleic acids.

→ Minerals regulate metabolic activities of the body.

→ Fructose is the sweetest natural sugar.

→ Mammary glands synthesize lactose (milk sugar) from glucose and galactose.

→ Water soluble vitamins are vitamin C and vitamin B-complex.

→ Fat soluble vitamins are A, D, E and K.

→ Deficiency of iodine causes the enlargement of thyroid gland known as goitre. It can be pre-vented by taking iodized salt.

→ Deficiency of iron causes anaemia.

→ Proteins are the polymers of amino acids.

→ Ecdysone is a steroid hormone which promotes the transformation of larva to the pupal stage during insect metamorphosis.

→ Glycosidic bond is between alcoholic group of one and aldose / ketose group of another sugar.

→ Extracellular fluid : Most abundant is Na, followed by Cl, carbonate, Ca, K, P04 and Mg.

→ Intracellular pool: Most abundant is K followed by P04, Mg, C03, Na, Cl and Ca.

→ One molecule of cellulose has about 6,000 glucose residues.

→ Cellulose can be hydrolyzed into soluble sugars.

→ Artificial fibre, rayon is manufactured by dissolving cellulosic materials in alkali.

→ The vitreous humour of the eye and the synovial fluid also contains mucopolysaccharides (mucilages).

→ Polysaccharides are the high molecular weight polymers of monosaccharides.

→ In diabetic persons, excess amount of sugar is not converted into glycogen.

→ Cellulose is not digested in our body. Ruminants like cow and buffalo can digest it.

→ Chitin forms the exoskeleton in arthropods its periodical pealing off is known as moulting or ecdysis.

→ Nucleic acids are composed of four types of nucleotides.

→ DNA is known as the ‘master molecule of the body’.

→ DNA is the hereditary material known as gene. Genes are found in the nucleus located on chromosomes.

→ Proteins are the ‘building blocks’ of the living organism.

→ Tertiary and quaternary proteins are always globular in nature.

→ Antibody is a protein produced in our body which neutralizes antigen and protects our body.

→ In meat, percentage of proteins is higher than carbohydrate but in rice it is reverse.

→ Two types of pentose sugar are found in nucleic acids. Ribose (in RNA) and deoxyribose (in DNA). Deoxyribose has one oxygen atom less than ribose sugar.

→ Keratin is a fibrous, insoluble protein which is called as scleroprotein and found in ectoder¬mal cells, hair, nails and horns of animals.

→ DNA produces RNA by a process called transcription (heterocatalytic function).

→ Monellin is a protein which is the sweetest chemical.

→ All enzymes are proteins, but all proteins are not enzymes.

→ Enzymes are the biological catalysts.

→ Enzymes are manufactured in the living cells but they are non-living.

→ DNA of all cells act as carrier of message for synthesis of enzyme.

→ Enzyme substrate complex is short lived.

→ Enzymes are very specific in their mode of action.

→ Pepsin is secreted as proenzyme, in an inactive form known as pepsinogen.

→ Inactive pepsinogen is hydrolysed to active pepsin in the stomach by gastric hydrochloric acid or by the pre-existing pepsin.

→ Enzymes have two sites : One is called as active site and another is called as allosteric site. By active site enzymes combine with the substrate, whereas by allosteric site it combines with other substances such as activator or inhibitor.

→ Due to presence of three dimensional structure, enzymes are more active than catalysts.

→ High temperature causes unfolding of protein structure which is also called as denaturation.

→ Many vitamins form prosthetic group of important enzymes. Some of the such vitamins are thymine, nicotinic acid, riboflavin and pyridoxin of vitamin B-complex.

→ Enzymology : Study of enzymes, their actions and functions.

→ Allozymes : Similar enzymes formed by different genes.

→ Allosteric enzyme : Structure was studied by Monod et al (1965). The enzymes do not obey Michaelis-Menten or Km constant.

→ AIDS is tested by ELISA (Enzyme Linked Immunesor bent Assay).

These MP Board Class 11th Biology Notes for Chapter 8 Cell: The Unit of Life help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 8 Cell: The Unit of Life

→ The blue-green algae resembles bacteria in many details but most of them form multicellular colonies.

→ The smallest cells observed so far are of Pleuropneumonia like organism (PPLO) namely Mycoplasma gallisepticum.

→ Mature nerve cells never divide but may reach length of 90 cm.

→ In our body many red blood cells and outer cells of the epidermis of the skin are destroyed daily.

→ The Ostrich egg cell is considered as the largest cell, its diameter is about 175 mm.

→ Liver cells and muscle cells retain mitotic ability, but rarely divide.

→ Multicellular organisms are adapted for longer life span.

→ Tissue culture is a technique by which cells are cultured on an artificial culture medium. The technique was developed by White (1932).

→ Embryo developed outside the cell other than zygote is called embryoids.

→ Term ‘cell’ was used for the first time by Robert Hooke.

→ Rudolf Virchow states that cells are originated from pre-existing ones.

→ Viruses do not have cellular structure hence they are acellular bodies.

→ Centriole and ribosomes are non-membranous organelles.

→ Plant cell contains cell wall of cellulose which is absent in animal cells.

→ Prokaryotic cell: A cell which has an incipient nucleus and lacks membrane-bound organelles, is called as prokaryotic cell.

→ Eukaryotic cell: A cell which has a well defined nucleus and membrane-bound organelles, is called as eukaryotic cell.

→ Mesosomes : Mesosomes are extensions of plasma membrane into the cell in the form of vesicles, tubules and lamellae, usually in prokaryotic cells.

→ Lysozyme : It is an enzyme present in body secretions like tear, saliva and sweat; it lyses the cell wall of bacteria.

→ Transmembrane proteins : The membrane proteins which extend through the phospholipid bimolecular layer as a single helix are called transmembrane proteins.

→ Plasmodesmata : The cytoplasmic bridges in the primary cell wall between adjacent cells are called plasmodesmata.

→ Symplasm : The cytoplasm (living matter) present in the plasmodesmata and cells is called symplasm.

→ Apoplasm : The non-living substances present in the intercellular spaces (outside the cell) constitute apoplasm.

→ Pits : The unthickened areas in the secondary cell wall are called pits.

→ Tonoplast: The single semipermeable membrane boundary of vacuoles in plant cells is called tonoplast.

→ Dictyosomes : In plant cells, the Golgi apparatus is present in the form of unconnected units, called dictyosomes.

→ GERL (Golgi-Endoplasmic Reticulum-Lysosome): It is a region of golgi bodies, which is thought to be involved in the formation of lysosomes and the secretory granules.

→ Diffusion : It is the phenomenon of movement of ions, atoms or molecules of any substance (gas, liquid or solid) from a region of higher concentration to a region of lower concentration, until equilibrium is reached.

→ Osmosis : It is the phenomenon of movement of solvent/water molecules from a region of higher concentration or weaker solution to a region of lower concentration or stronger solution through a semi-permeable membrane, until equilibrium is reached.

→ Endocytosis : It is a process of active cellular intake of such materials (in bulk) that cannot pass through the unbroken plasma membrane.

→ Exocytosis : It is a process of active cellular extrusion/expulsion of such materials (in bulk) which cannot pass through the unbroken plasma membrane.

→ Pinocytosis : It is the process of active cellular intake of droplets of fluid along with submicroscopic particles.

→ Phagocytosis : It is the process of active cellular intake (ingestion) of solid particles.

→ Osmotic pressure: The hydrostatic pressure which balances and prevents any osmotic entry of water into the concerned solution, is called osmotic pressure.

→ Electrical gradient: The difference of electrical charges between two sides of a membrane is called as electrical gradient.