NCERT Solutions for Class 11 Chemistry Chapter 2 help students understand the internal structure of atoms and the behaviour of subatomic particles. The chapter covers electrons, protons, neutrons, atomic models, electromagnetic radiation, hydrogen spectra, quantum numbers, orbitals, and electronic configurations. The solutions explain both theoretical and numerical questions in a clear, step-by-step manner. They can be used for completing NCERT exercises, revising important concepts, and preparing for school examinations, JEE, and NEET.
Class 11 Chemistry Chapter 2: Structure of Atom – Overview
The Structure of Atom chapter explains the discovery of electrons, protons, and neutrons and describes their masses, charges, and locations inside an atom. It discusses Thomson’s atomic model, Rutherford’s nuclear model, and Bohr’s model of the hydrogen atom. Students also learn about atomic number, mass number, isotopes, and isobars. The chapter introduces electromagnetic radiation, wavelength, frequency, wave number, and Planck’s quantum theory. It also explains the hydrogen spectrum and the movement of electrons between different energy levels.
The chapter further discusses the dual nature of matter, de Broglie’s equation, and the Heisenberg Uncertainty Principle. These ideas lead to the quantum mechanical model, which describes electrons in terms of probability instead of fixed circular paths. Students learn about the four quantum numbers and the shapes of s, p, d, and f orbitals. The Aufbau Principle, Pauli Exclusion Principle, and Hund’s Rule are used to write electronic configurations correctly. These concepts provide the foundation for chemical bonding, periodic properties, and molecular structure.
NCERT Solutions For Class 11 Chemistry Chapter 2 – Structure of Atom
Ques 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 one electron = 9.10939 × 10-31 kg
Number of electrons that weigh 9.10939 × 10-31 kg = 1
Number of electrons that will weigh 1 g (1 × 10-3 kg) = (1 × 10-3 kg) / (9.10939 × 10-31 kg) = 1.098 × 1027 electrons.
(ii) Mass of one electron = 9.10939 × 10-31 kg
Mass of one mole of electrons = (6.022 × 1023) × (9.10939 × 10-31 kg) = 5.48 × 10-7 kg
Charge on one electron = 1.6022 × 10-19 C
Charge on one mole of electrons = (1.6022 × 10-19 C) × (6.022 × 1023) = 9.65 × 104 C
Ques 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 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 STP. Will the answer change if the temperature and pressure are changed?
Solution:
(i) Number of electrons in 1 molecule of methane (CH4) = 6 + 4(1) = 10
Number of electrons in 1 mole (6.022 × 1023 molecules) of methane = 6.022 × 1023 × 10 = 6.022 × 1024 electrons.
(ii) (a) 1 atom of 14C contains 14 – 6 = 8 neutrons.
Number of atoms in 14 g of 14C = 6.022 × 1023
Number of neutrons in 7 mg (7 × 10-3 g) of 14C = (6.022 × 1023 × 8 × 7 × 10-3) / 14 = 2.4088 × 1021
(b) Mass of total neutrons in 7 mg of 14C = (2.4088 × 1021) × (1.67493 × 10-27 kg) = 4.0347 × 10-6 kg
(iii) (a) Molar mass of NH3 = 17 g/mol
Number of protons in 1 molecule of NH3 = 7 + 3(1) = 10
Number of protons in 34 mg (34 × 10-3 g) of NH3 = (6.022 × 1023 × 10 × 34 × 10-3) / 17 = 1.2044 × 1022
(b) Mass of total protons = (1.2044 × 1022) × (1.6726 × 10-27 kg) = 2.0145 × 10-5 kg
The number and mass of protons remain unchanged with temperature and pressure changes.
Ques 3. How many neutrons and protons are there in the following nuclei?
136C, 168O, 2412Mg, 5626Fe, 8838Sr
Solution:
- 136C: Protons = 6, Neutrons = 13 – 6 = 7
- 168O: Protons = 8, Neutrons = 16 – 8 = 8
- 2412Mg: Protons = 12, Neutrons = 24 – 12 = 12
- 5626Fe: Protons = 26, Neutrons = 56 – 26 = 30
- 8838Sr: Protons = 38, Neutrons = 88 – 38 = 50
Ques 4. Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A):
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9
Solution:
(i) 3517Cl
(ii) 23392U
(iii) 94Be
Ques 5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number (ν̄) of the yellow light.
Solution:
Frequency (ν) = c / λ = (3 × 108 m/s) / (580 × 10-9 m) = 5.17 × 1014 s-1
Wave number (ν̄) = 1 / λ = 1 / (580 × 10-9 m) = 1.72 × 106 m-1
Ques 6. Find energy of each of the photons which:
(i) correspond to light of frequency 3 × 1015 Hz.
(ii) have wavelength of 0.50 Å.
Solution:
(i) E = hν = (6.626 × 10-34 J·s) × (3 × 1015 s-1) = 1.988 × 10-18 J
(ii) E = hc / λ = [(6.626 × 10-34 J·s) × (3 × 108 m/s)] / (0.50 × 10-10 m) = 3.98 × 10-15 J
Ques 7. Calculate the wavelength, frequency, and wave number of a light wave whose period is 2.0 × 10-10 s.
Solution:
Frequency (ν) = 1 / T = 1 / (2.0 × 10-10 s) = 5.0 × 109 s-1
Wavelength (λ) = c / ν = (3.0 × 108 m/s) / (5.0 × 109 s-1) = 6.0 × 10-2 m
Wave number (ν̄) = 1 / λ = 1 / (6.0 × 10-2 m) = 16.66 m-1
Ques 8. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Solution:
E = (n · h · c) / λ ⇒ n = (E · λ) / (h · c)
n = [1 J × (4000 × 10-12 m)] / [(6.626 × 10-34 J·s) × (3 × 108 m/s)] = 2.012 × 1016 photons.
Ques 9. A photon of wavelength 4 × 10-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10-19 J).
Solution:
(i) E = hc / λ = [(6.626 × 10-34 J·s) × (3 × 108 m/s)] / (4 × 10-7 m) = 4.97 × 10-19 J = (4.97 × 10-19) / (1.602 × 10-19) = 3.10 eV
(ii) Ek = E – W0 = 3.10 eV – 2.13 eV = 0.97 eV = 1.55 × 10-19 J
(iii) v = √[(2Ek) / m] = √[(2 × 1.55 × 10-19 J) / (9.11 × 10-31 kg)] = 5.84 × 105 m/s
Ques 10. 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:
E = (NA · h · c) / λ = [(6.022 × 1023) × (6.626 × 10-34) × (3 × 108)] / (242 × 10-9) = 4.947 × 105 J/mol = 494.7 kJ/mol
Ques 11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm. Calculate the rate of emission of quanta per second.
Solution:
Energy per photon E = hc / λ = [(6.626 × 10-34) × (3 × 108)] / (0.57 × 10-6 m) = 3.487 × 10-19 J
Rate of emission = Power / E = (25 J/s) / (3.487 × 10-19 J) = 7.17 × 1019 quanta/s
Ques 12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0) and work function (W0) of the metal.
Solution:
Threshold frequency (ν0) = c / λ0 = (3 × 108 m/s) / (6800 × 10-10 m) = 4.41 × 1014 s-1
Work function (W0) = hν0 = (6.626 × 10-34 J·s) × (4.41 × 1014 s-1) = 2.922 × 10-19 J
Ques 13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?
Solution:
ΔE = 2.18 × 10-18 (1/42 – 1/22) = -4.0875 × 10-19 J
λ = hc / ΔE = [(6.626 × 10-34) × (3 × 108)] / (4.0875 × 10-19) = 4.86 × 10-7 m = 486 nm
Ques 14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).
Solution:
E5 = (-2.18 × 10-18) / 52 = -8.72 × 10-20 J
Ionization energy from n = 5 is ΔE5 = E∞ – E5 = 0 – (-8.72 × 10-20) = 8.72 × 10-20 J
Ionization energy from n = 1 is ΔE1 = 2.18 × 10-18 J
Comparison: ΔE1 / ΔE5 = (2.18 × 10-18) / (8.72 × 10-20) = 25. Removing an electron from n = 5 requires 25 times less energy than from n = 1.
Ques 15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Solution:
Number of spectral lines = [n(n – 1)] / 2 = (6 × 5) / 2 = 15 lines.
Ques 16. (i) The energy associated with the first orbit in the hydrogen atom is -2.18 × 10-18 J/atom. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for a hydrogen atom.
Solution:
(i) E5 = (-2.18 × 10-18) / 52 = -8.72 × 10-20 J
(ii) r5 = 0.529 × 52 Å = 13.225 Å = 1.3225 nm
Ques 17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
Solution:
For Balmer series, n1 = 2. For longest wavelength, n2 = 3.
ν̄ = RH (1/22 – 1/32) = (1.097 × 107 m-1) × (5/36) = 1.523 × 106 m-1
Ques 18. What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit, and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 × 10-11 ergs.
Solution:
Ground state energy = -2.18 × 10-18 J
ΔE = E5 – E1 = 2.18 × 10-18 (1 – 1/25) = 2.09 × 10-18 J
Wavelength (λ) = hc / ΔE = [(6.626 × 10-34) × (3 × 108)] / (2.09 × 10-18) = 9.51 × 10-8 m = 951 Å
Ques 19. The electron energy in a hydrogen atom is given by En = (-2.18 × 10-18) / n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Solution:
ΔE = E∞ – E2 = 0 – [(-2.18 × 10-18) / 22] = 5.45 × 10-19 J
λ = hc / ΔE = [(6.626 × 10-34) × (3 × 108)] / (5.45 × 10-19) = 3.674 × 10-7 m = 3.674 × 10-5 cm
Ques 20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m/s.
Solution:
λ = h / (m · v) = (6.626 × 10-34 kg·m2·s-1) / [(9.11 × 10-31 kg) × (2.05 × 107 m/s)] = 3.55 × 10-11 m
Ques 21. The mass of an electron is 9.11 × 10-31 kg. If its K.E. is 3.0 × 10-25 J, calculate its wavelength.
Solution:
v = √[(2 · K.E.) / m] = √[(2 × 3.0 × 10-25) / (9.11 × 10-31)] = 812 m/s
λ = h / (m · v) = (6.626 × 10-34) / [(9.11 × 10-31) × 812] = 8.967 × 10-7 m
Ques 22. Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na+, K+, Mg2+, Ca2+, S2-, Ar
Solution:
- Number of electrons in Na+ = 11 – 1 = 10
- Number of electrons in K+ = 19 – 1 = 18
- Number of electrons in Mg2+ = 12 – 2 = 10
- Number of electrons in Ca2+ = 20 – 2 = 18
- Number of electrons in S2- = 16 + 2 = 18
- Number of electrons in Ar = 18
Isoelectronic Groups:
1. Na+ and Mg2+ (10 electrons each)
2. K+, Ca2+, S2-, and Ar (18 electrons each)
Ques 23. (i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2- (d) F–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5?
(iii) Which atoms are indicated by the following configurations? (a) [He]2s1 (b) [Ne]3s23p3 (c) [Ar]4s23d1.
Solution:
(i) (a) H–: 1s2
(b) Na+: 1s2 2s2 2p6
(c) O2-: 1s2 2s2 2p6
(d) F–: 1s2 2s2 2p6
(ii) (a) 3s1 ⇒ 1s2 2s2 2p6 3s1 ⇒ Atomic number = 11
(b) 2p3 ⇒ 1s2 2s2 2p3 ⇒ Atomic number = 7
(c) 3p5 ⇒ 1s2 2s2 2p6 3s2 3p5 ⇒ Atomic number = 17
(iii) (a) [He]2s1 ⇒ Lithium (Li)
(b) [Ne]3s23p3 ⇒ Phosphorus (P)
(c) [Ar]4s23d1 ⇒ Scandium (Sc)
Ques 24. What is the lowest value of n that allows g orbitals to exist?
Solution:
For g orbitals, l = 4. Since l can take values up to n – 1, the minimum value of n required is n = l + 1 = 4 + 1 = 5.
Ques 25. An electron is in one of the 3d orbitals. Give the possible values of n, l, and ml for this electron.
Solution:
For 3d orbital: n = 3, l = 2, and ml = -2, -1, 0, +1, +2.
Ques 26. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Solution:
(i) For a neutral atom, Protons = Electrons = 29.
(ii) Electronic configuration (Z = 29, Copper): 1s2 2s2 2p6 3s2 3p6 3d10 4s1
Ques 27. Give the number of electrons in the species H2+, H2 and O2+.
Solution:
• H2+ = 2 – 1 = 1 electron
• H2 = 1 + 1 = 2 electrons
• O2+ = 16 – 1 = 15 electrons
Ques 28. (i) An atomic orbital has n = 3. What are the possible values of l and ml?
(ii) List the quantum numbers (ml and l) of electrons for the 3d orbital.
(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f.
Solution:
(i) For n = 3: l = 0, 1, 2.
• l = 0 ⇒ ml = 0
• l = 1 ⇒ ml = -1, 0, +1
• l = 2 ⇒ ml = -2, -1, 0, +1, +2
(ii) For 3d orbital: l = 2, ml = -2, -1, 0, +1, +2.
(iii)
• 1p: Not possible (l=1 not allowed when n=1).
• 2s: Possible.
• 2p: Possible.
• 3f: Not possible (l=3 not allowed when n=3).
Ques 29. Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n=1, l=0; (b) n=3, l=1; (c) n=4, l=2; (d) n=4, l=3.
Solution:
(a) 1s
(b) 3p
(c) 4d
(d) 4f
Ques 30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) n = 0, l = 0, ml = 0, ms = +1/2
(b) n = 1, l = 0, ml = 0, ms = -1/2
(c) n = 1, l = 1, ml = 0, ms = +1/2
(d) n = 2, l = 1, ml = 0, ms = -1/2
(e) n = 3, l = 3, ml = -3, ms = +1/2
(f) n = 3, l = 1, ml = 0, ms = +1/2
Solution:
(a) Not possible because n cannot be 0.
(b) Possible.
(c) Not possible because l must be strictly less than n (when n=1, l can only be 0).
(d) Possible.
(e) Not possible because l cannot be equal to n (when n=3, l can only be 0, 1, or 2).
(f) Possible.
Ques 31. How many electrons in an atom may have the following quantum numbers?
(a) n = 4, ms = -1/2 (b) n = 3, l = 0
Solution:
(a) Total electrons in n = 4 is 2n2 = 2(4)2 = 32. Half of them have ms = -1/2, so 16 electrons.
(b) n=3, l=0 corresponds to the 3s orbital, which holds a maximum of 2 electrons.
Ques 32. Show that the circumference 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’s postulate of angular momentum:
mvr = (n · h) / (2π) ⇒ 2πr = (n · h) / (m · v)
According to de Broglie equation: λ = h / (m · v)
Substituting λ:
2πr = nλ
Thus, the circumference of the Bohr orbit (2πr) is an integral multiple (n) of the de Broglie wavelength (λ).
Ques 33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of the He+ spectrum?
Solution:
For He+ (Z = 2):
ν̄ = RH · Z2 (1/22 – 1/42) = RH · 4 · (3/16) = 3RH / 4
For Hydrogen (Z = 1):
ν̄ = RH (1/n12 – 1/n22) = 3RH / 4 ⇒ (1/n12 – 1/n22) = 3/4
This condition is satisfied when n1 = 1 and n2 = 2. Thus, the transition is from n = 2 to n = 1.
Ques 34. Calculate the energy required for the process: He+(g) → He2+(g) + e–. The ionization energy for the H atom in the ground state is 2.18 × 10-18 J/atom.
Solution:
En ∝ Z2. For He+, Z = 2.
I.E.(He+) = I.E.(H) × Z2 = (2.18 × 10-18 J) × 22 = 8.72 × 10-18 J
Ques 35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
Solution:
Diameter = 0.15 nm = 1.5 × 10-10 m
Total length = 20 cm = 0.2 m
Number of atoms = 0.2 m / (1.5 × 10-10 m) = 1.33 × 109 atoms
Ques 36. 2 × 108 atoms of carbon are arranged side by side. Calculate the radius of a carbon atom if the length of this arrangement is 2.4 cm.
Solution:
Diameter = 2.4 cm / (2 × 108) = 1.2 × 10-8 cm = 1.2 × 10-10 m
Radius = Diameter / 2 = 0.6 × 10-10 m = 0.06 nm
Ques 37. 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 arranged side by side lengthwise.
Solution:
(a) Radius = 2.6 Å / 2 = 1.3 Å = 130 pm
(b) Number of atoms = (1.6 × 10-2 m) / (2.6 × 10-10 m) = 6.154 × 107 atoms
Ques 38. A certain particle carries 2.5 × 10-16 C of static electric charge. Calculate the number of electrons present in it.
Solution:
Number of electrons = Q / e = (2.5 × 10-16 C) / (1.6022 × 10-19 C) = 1560 electrons
Ques 39. In Millikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is -1.282 × 10-18 C, calculate the number of electrons present on it.
Solution:
Number of electrons = (1.282 × 10-18 C) / (1.6022 × 10-19 C) = 8 electrons
Ques 40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?
Solution:
Heavy nuclei contain a large amount of positive charge, causing significant electrostatic repulsion that deflects α-particles at large angles. Light nuclei contain much less positive charge, so the deflections would be very small, and almost all α-particles would pass straight through without noticeable scattering.
Ques 41. Symbols 7935Br and 79Br can be written, whereas symbols 3579Br and 35Br are not acceptable. Answer briefly.
Solution:
The atomic number must always be written as a subscript and mass number as a superscript, making 3579Br incorrect. Furthermore, 35Br is invalid because atomic number is a constant property of the element, whereas mass number varies across isotopes and must be explicitly stated to identify the specific nuclide.
Ques 42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Solution:
Let protons = x. Then neutrons = x + 0.317x = 1.317x.
Mass number = Protons + Neutrons = x + 1.317x = 81
2.317x = 81 ⇒ x = 35
Atomic number 35 corresponds to Bromine (Br). Symbol: 8135Br.
Ques 43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than electrons, find the symbol of the ion.
Solution:
Let electrons in ion = x. Neutrons = x + 0.111x = 1.111x.
Protons = x – 1 (since it has a -1 charge).
Mass number = (x – 1) + 1.111x = 37 ⇒ 2.111x = 38 ⇒ x = 18.
Protons = 18 – 1 = 17 (Chlorine). Symbol: 3717Cl–.
Ques 44. 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 electrons in ion = x. Neutrons = x + 0.304x = 1.304x.
Protons = x + 3 (since it has a +3 charge).
Mass number = (x + 3) + 1.304x = 56 ⇒ 2.304x = 53 ⇒ x = 23.
Protons = 23 + 3 = 26 (Iron). Symbol: 5626Fe3+.
Ques 45. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
Solution:
FM radio < Microwave < Amber light < X-rays < Cosmic rays.
Ques 46. Nitrogen laser produces radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser.
Solution:
E = (N · h · c) / λ = [(5.6 × 1024) × (6.626 × 10-34) × (3 × 108)] / (337.1 × 10-9 m) = 3.3 × 106 J
Ques 47. Neon gas is generally used in the signboards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s, (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
Solution:
(a) ν = c / λ = (3 × 108) / (616 × 10-9) = 4.87 × 1014 s-1
(b) Distance = c × t = (3 × 108 m/s) × 30 s = 9.0 × 109 m
(c) E = hν = (6.626 × 10-34) × (4.87 × 1014) = 3.227 × 10-19 J
(d) Number of quanta = 2 J / (3.227 × 10-19 J) = 6.2 × 1018
Ques 48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.
Solution:
Ephoton = hc / λ = [(6.626 × 10-34) × (3 × 108)] / (600 × 10-9) = 3.313 × 10-19 J
Number of photons = (3.15 × 10-18) / (3.313 × 10-19) ≈ 10 photons.
Ques 49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.
Solution:
Frequency (ν) = 1 / (2 × 10-9 s) = 0.5 × 109 s-1
Energy = N · h · ν = (2.5 × 1015) × (6.626 × 10-34) × (0.5 × 109) = 8.28 × 10-10 J
Ques 50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Solution:
ν1 = (3 × 108) / (589 × 10-9) = 5.093 × 1014 s-1
ν2 = (3 × 108) / (589.6 × 10-9) = 5.088 × 1014 s-1
ΔE = h(ν1 – ν2) = (6.626 × 10-34) × (0.005 × 1014) = 3.31 × 10-22 J
Ques 51. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Solution:
(a) ν0 = W0 / h = (1.9 × 1.602 × 10-19 J) / (6.626 × 10-34 J·s) = 4.59 × 1014 s-1
(b) λ0 = c / ν0 = (3 × 108) / (4.59 × 1014) = 654 nm
(c) Eincident = hc / λ = [(6.626 × 10-34) × (3 × 108)] / (500 × 10-9) = 3.975 × 10-19 J
W0 = 3.04 × 10-19 J
K.E. = E – W0 = 9.35 × 10-20 J
v = √[(2 · K.E.) / m] = √[(2 × 9.35 × 10-20) / (9.11 × 10-31)] = 4.53 × 105 m/s
Ques 52. Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.
| λ (nm) | 500 | 450 | 400 |
|---|---|---|---|
| v × 10-5 (cm/s) | 2.55 | 4.35 | 5.20 |
Solution:
Using hc(1/λ – 1/λ0) = ½mv2, we solve for average threshold wavelength λ0 ≈ 543 nm.
Substituting λ0 into the equation yields Planck’s constant h ≈ 6.626 × 10-34 J·s.
Ques 53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying a voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Solution:
Ephoton = hc / λ = [(6.626 × 10-34) × (3 × 108)] / (256.7 × 10-9) = 7.74 × 10-19 J = 4.83 eV
K.E. = e × Vstop = 0.35 eV
W0 = E – K.E. = 4.83 eV – 0.35 eV = 4.48 eV
Ques 54. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 m/s, calculate the energy with which it is bound to the nucleus.
Solution:
Ephoton = hc / λ = [(6.626 × 10-34) × (3 × 108)] / (150 × 10-12) = 13.25 × 10-16 J
K.E. = ½mv2 = ½(9.11 × 10-31)(1.5 × 107)2 = 1.025 × 10-16 J
Binding Energy = Ephoton – K.E. = 12.225 × 10-16 J = 7.63 × 103 eV
Ques 55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as ν = 3.29 × 1015 (1/32 – 1/n2) Hz. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Solution:
ν = c / λ = (3 × 108) / (1285 × 10-9) = 2.33 × 1014 Hz
2.33 × 1014 = 3.29 × 1015 (1/9 – 1/n2) ⇒ n = 5
1285 nm lies in the Infrared region.
Ques 56. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Solution:
r ∝ n2 ⇒ r1 / r2 = n12 / n22 ⇒ 1322.5 / 211.6 = 6.25 ⇒ n1 / n2 = 2.5
For integer values, n2 = 2 and n1 = 5.
Since n2 = 2, it belongs to the Balmer series.
1 / λ = RH (1/22 – 1/52) ⇒ λ = 434 nm (Visible region).
Ques 57. 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 × 106 m/s, calculate de Broglie wavelength associated with this electron.
Solution:
λ = h / (m · v) = (6.626 × 10-34) / [(9.11 × 10-31) × (1.6 × 106)] = 4.55 × 10-10 m = 455 pm
Ques 58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Solution:
v = h / (m · λ) = (6.626 × 10-34) / [(1.675 × 10-27 kg) × (800 × 10-12 m)] = 4.94 × 104 m/s
Ques 59. If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 m/s, calculate the de Broglie wavelength associated with it.
Solution:
λ = h / (m · v) = (6.626 × 10-34) / [(9.11 × 10-31) × (2.19 × 106)] = 3.32 × 10-10 m = 332 pm
Ques 60. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 m/s. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Solution:
λ = h / (m · v) = (6.626 × 10-34) / [0.1 kg × (4.37 × 105 m/s)] = 1.516 × 10-38 m
Ques 61. If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h / (4π × 0.05 nm), is there any problem in defining this value?
Solution:
Δp = h / (4π · Δx) = (6.626 × 10-34) / [4π × (2 × 10-12 m)] = 2.638 × 10-23 kg·m/s
Actual Momentum = h / [4π × (5 × 10-11 m)] = 1.055 × 10-24 kg·m/s
Since the uncertainty in momentum (Δp) is larger than the actual momentum, this value cannot be defined precisely.
Ques 62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, ml = -2, ms = -1/2
2. n = 3, l = 2, ml = 1, ms = +1/2
3. n = 4, l = 1, ml = 0, ms = +1/2
4. n = 3, l = 2, ml = -2, ms = -1/2
5. n = 3, l = 1, ml = -1, ms = +1/2
6. n = 4, l = 1, ml = 0, ms = +1/2
Solution:
Occupied orbitals: (1) 4d, (2) 3d, (3) 4p, (4) 3d, (5) 3p, (6) 4p
Increasing energy order: (5) < (2) = (4) < (3) = (6) < (1)
Ques 63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?
Solution:
The 4p electrons experience the lowest effective nuclear charge as they are located furthest from the nucleus and shielded by inner shell electrons.
Ques 64. Among the following pairs of orbitals, which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
Solution:
(i) 2s (closer to the nucleus)
(ii) 4d (more penetrating than 4f)
(iii) 3p (more penetrating than 3d)
Ques 65. The unpaired electrons in Al and Si are present in the 3p orbital. Which electrons will experience a more effective nuclear charge from the nucleus?
Solution:
The unpaired 3p electron in Silicon (Si) experiences a larger effective nuclear charge because Silicon has a higher nuclear charge (Z = 14) compared to Aluminium (Z = 13).
Ques 66. Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe, and (e) Kr.
Solution:
(a) P (Z=15): 3 unpaired electrons (3p3)
(b) Si (Z=14): 2 unpaired electrons (3p2)
(c) Cr (Z=24): 6 unpaired electrons (3d5 4s1)
(d) Fe (Z=26): 4 unpaired electrons (3d6)
(e) Kr (Z=36): 0 unpaired electrons (Fully filled)
Ques 67. (a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having ms value of -1/2 for n = 4?
Solution:
(a) For n = 4, l = 0, 1, 2, 3 ⇒ 4 sub-shells (4s, 4p, 4d, 4f).
(b) Number of orbitals in n = 4 shell = n2 = 16. Each orbital holds one electron with spin ms = -1/2 ⇒ 16 electrons.
Why Class 11 Chemistry Chapter 2 Matters in NEET and JEE
Class 11 Chemistry Chapter 2 is important because it explains the arrangement and behaviour of electrons, protons and neutrons inside an atom. It helps students understand atomic models, electromagnetic radiation, hydrogen spectra, energy levels, orbitals, quantum numbers and electronic configurations. These concepts are used extensively in chemical bonding, periodic classification, molecular structure, coordination chemistry and spectroscopy. The chapter also strengthens numerical problem-solving skills through calculations involving wavelength, frequency, photon energy, electronic transitions and the de Broglie equation. Questions from atomic structure are regularly asked in school examinations, JEE and NEET. A clear understanding of this chapter makes advanced chemistry concepts easier to understand and apply.
Preparation Tips for Class 11 Chemistry Chapter 2
Class 11 Chemistry Chapter 2 is important because it explains the arrangement and behaviour of electrons, protons and neutrons inside an atom. It helps students understand atomic models, electromagnetic radiation, hydrogen spectra, energy levels, orbitals, quantum numbers and electronic configurations. These concepts are used extensively in chemical bonding, periodic classification, molecular structure, coordination chemistry and spectroscopy. The chapter also strengthens numerical problem-solving skills through calculations involving wavelength, frequency, photon energy, electronic transitions and the de Broglie equation. Questions from atomic structure are regularly asked in school examinations, JEE and NEET. A clear understanding of this chapter makes advanced chemistry concepts easier to understand and apply.
Class 11 Chemistry Chapter 2 FAQs
Q.1. Why should I study the Structure of Atom Class 11 NCERT Solutions?
Ans. Studying these NCERT solutions helps you understand the chapter easily. The answers are explained simply, so you can learn the basics clearly. It also helps in solving textbook questions, preparing for exams, and revising during exams.
Q.2. What is an atom?
Ans. An atom is the smallest unit of matter. Everything around us is made of atoms. It has a center called the nucleus, which includes protons and neutrons. Electrons move around the nucleus.
Q.3. Are Class 11 Chemistry Chapter 2 NCERT Solutions tough to understand?
Ans. These NCERT solutions for Chemistry class 11 chapter 2 are written in a simple and clear way. If you read the textbook and the solutions, you will find the chapter easy to understand.
Q.4. What does an atomic structure mean?
Ans. Atomic structure means how an atom is built. It explains where the protons, neutrons, and electrons are inside an atom. Learning atomic structure helps you understand the behavior of elements in Chemistry.
Q.5. What is an electron?
Ans. An electron is a tiny particle present in an atom. It has a negative charge and moves around the nucleus in circular paths called orbits or shells. Electrons play a big role in chemical reactions and bonding.