NCERT Solutions Class 11 Chemistry (Part-I) Chapter-2: Structure of Atom (Free PDF)

In this unit, we have earlier provided you with a summary of notes on the structure of atoms that includes building blocks of matter, including their properties, various atomic theories, quantum theories of the atom, and the principles of electronic configuration. However, without practicing with unit-based questions, you may lack clarity and a deeper understanding. This blog provides some exercises along with their solutions to help you understand the concepts more easily.

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Chapter 1

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Chapter 4

Chapter 5

NCERT Solutions Class 11 Chemistry (Part-I) Chapter 2: Structure of Atom

Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part I) Chapter 2: Structure of Atom.

Exercises

1. (i) Calculate the number of electrons that will weigh one gram together. (ii) Calculate the mass and charge of one mole of electrons.
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.
   (iii) Find
    (a) The total number and
    (b) The total mass of protons in 34 mg of NH₃ at STP.
3. How many neutrons and protons are there in the following nuclei:

• ₆¹³C
• ₈¹⁶O
• ₁₂²⁴Mg
• ₂₆⁵⁶Fe
• ₃₈⁸⁸Sr

4. Will the answer change if the temperature and pressure are changed?

Write the complete symbol for the atom with the given atomic number (Z) and mass number (A):
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9

5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ν̅) of the yellow light.
6. Find the energy of each of the photons which:
   (i) Correspond to light of frequency 3 × 10¹⁵ Hz
   (ii) Have a wavelength of 0.50 Å
7. Calculate the wavelength, frequency, and wavenumber of a light wave whose period is 2.0 × 10⁻¹⁰ s.
8. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
9. A photon of wavelength 4 × 10⁻⁷ m strikes a metal surface. The work function of the metal is 2.13 eV. Calculate:
   (i) Energy of the photon (eV)
   (ii) Kinetic energy of the emission
   (iii) Velocity of the photoelectron
   (1 eV = 1.6020 × 10⁻¹⁹ J)
10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionization energy of sodium in kJ mol⁻¹.
11. A 25-watt bulb emits monochromatic yellow light of wavelength 0.57 µm. Calculate the rate of emission of quanta per second.
12. Electrons are emitted with zero velocity from a metal surface when exposed to radiation of wavelength 6800 Å. Calculate:

• Threshold frequency (ν₀)
• Work function (W₀) of the metal.

13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition from n = 4 to n = 2?
14. How much energy is required to ionise a hydrogen atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of a hydrogen atom from n = 1 orbit.
15. What is the maximum number of emission lines when an excited electron of a hydrogen atom in n = 6 drops to the ground state?
16. (i) The energy associated with the first orbit in a hydrogen atom is –2.18 × 10⁻¹⁸ J atom⁻¹. What is the energy associated with the fifth orbit?
17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of hydrogen.
18. What is the energy (in J) required to shift the electron of hydrogen from the first Bohr orbit to the fifth? Also, find the wavelength of light emitted when it returns to the ground state.
    (Ground state energy = –2.18 × 10⁻¹¹ ergs)
19. Using Eₙ = –2.18 × 10⁻¹⁸ / n² J, calculate the energy required to remove an electron from the n = 2 orbit. What is the longest wavelength of light (in cm) that can cause this transition?
20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ m/s.
21. The mass of an electron is 9.1 × 10⁻³¹ kg. If its kinetic energy is 3.0 × 10⁻²⁵ J, calculate its wavelength.
22. (i) Which of the following are isoelectronic species?
    Na⁺, K⁺, Mg²⁺, Ca²⁺, S²⁻, Ar (ii). Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.
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 with outermost electrons represented by:
    (a) 3s¹ (b) 2p³ (c) 3p⁵ (iii) Which atoms are indicated by these configurations?
    (a) [He] 2s¹
    (b) [Ne] 3s² 3p³
    (c) [Ar] 4s² 3d¹
24. What is the lowest value of n that allows g-orbitals to exist?
25. An electron is in one of the 3d orbitals. Give the possible values of n, l and mₗ for this electron.
26. An atom of an element contains 29 electrons and 35 neutrons. Deduce:
    (i) Number of protons
    (ii) Electronic configuration of the element
27. (i) An atomic orbital has n = 3. What are the possible values of l and mₗ?
    (ii) List quantum numbers (l and mₗ) of electrons in the 3d orbital.
    (iii) Which of the following orbitals are possible: 1p, 2s, 2p, 3f?
28. Using s, p, d notation, describe orbitals 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
29. How many electrons in an atom may have the following quantum numbers?
    (a) n = 4, mₛ = –½
    (b) n = 3, l = 0
30. 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.
31. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of the He⁺ spectrum?
32. Calculate the energy required for the process:
    He⁺(g) → He²⁺(g) + e⁻
    (Ionisation energy for H atom in the ground state is 2.18 × 10⁻¹⁸ J atom⁻¹)
33. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms that can be placed side by side in a straight line across a 20 cm long scale.
34. 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.
35. The diameter of a zinc atom is 2.6 Å. Calculate:
    (a) Radius of the zinc atom in pm
    (b) Number of atoms present in a length of 1.6 cm if zinc atoms are arranged side by side lengthwise
36. A certain particle carries 2.5 × 10⁻¹⁶ C of static electric charge. Calculate the number of electrons present in it.
37. In Millikan’s experiment, the static electric charge on an oil drop was –1.282 × 10⁻¹⁸ C. Calculate the number of electrons present on it.
38. In Rutherford’s experiment, thin foils of heavy atoms like gold or platinum were bombarded with α-particles. What difference in results would be observed if a foil of light atoms like aluminum were used instead?
39. Write the correct symbolic representation for:
    79/35 Br and explain why 79Br or 35Br alone are not acceptable.
40. An element with mass number 81 contains 31.7% more neutrons than protons. Assign the atomic symbol.
41. An ion with mass number 37 has one unit of negative charge and contains 11.1% more neutrons than electrons. Find the symbol of the ion.
42. An ion with mass number 56 has a 3+ charge and contains 30.4% more neutrons than electrons. Assign the symbol to this ion.
43. Arrange the following types of radiation in increasing order of frequency:
    (a) Radiation from a microwave oven
    (b) Amber light from the traffic signal
    (c) Radiation from FM radio
    (d) Cosmic rays from outer space
    (e) X-rays
44. A nitrogen laser produces radiation at a wavelength of 337.1 nm. If it emits 5.6 × 10²⁴ photons, calculate the power of this laser.
45. Neon gas is used in signboards. If it emits strongly at 616 nm, calculate:
    (a) Frequency of emission
    (b) Distance travelled by this radiation in 30 s
    (c) Energy of one quantum
    (d) Number of quanta if 2 J of energy is produced
46. In astronomical observations, signals from distant stars are weak. If the photon detector receives 3.15 × 10⁻¹⁸ J from 600 nm radiation, calculate the number of photons received.
47. A pulsed radiation source of 2 ns emits 2.5 × 10¹⁵ photons. Calculate the energy of the source.
48. The longest wavelength doublet absorption transition is observed at 589 nm and 589.6 nm. Calculate:

• Frequency of each transition
• Energy difference between the two excited states

49. The work function for the cesium atom is 1.9 eV. Calculate:
    (a) Threshold wavelength
    (b) Threshold frequency
    Also, if irradiated with 500 nm light, calculate the kinetic energy and velocity of the ejected photoelectron.
50. Sodium is irradiated with light of wavelengths 500 nm, 450 nm, and 400 nm. Corresponding photoelectron velocities (×10⁵ cm/s): 2.55, 4.35, 5.35.
    Calculate:
    (a) Threshold wavelength
    (b) Planck’s constant
51. In a photoelectric experiment, silver ejects electrons stopped by 0.35 V when irradiated with 256.7 nm radiation. Calculate the work function of silver.
52. A photon of 150 pm ejects an inner electron with velocity 1.5 × 10⁷ m/s. Calculate the binding energy of this electron to the nucleus.
53. Paschen series transitions end at n = 3 and are given by:
    ν = 3.29 × 10¹⁵ Hz × [1/3² – 1/n²]
    Calculate the value of n for the 1285 nm transition and the region of the spectrum.
54. A transition starts from an orbit of radius 1.3225 nm and ends at 211.6 pm.
    Calculate:

• Wavelength of emission
• Name of the series
• Region of the spectrum

55. An electron microscope uses electrons at a velocity of 1.6 × 10⁶ m/s. Calculate the de Broglie wavelength of the electron.
56. Neutron diffraction uses 800 pm wavelength. Calculate the characteristic velocity of the neutron.
57. If the velocity of an electron in Bohr’s first orbit is 2.19 × 10⁶ m/s, calculate the associated de Broglie wavelength.
58. A proton moving at 4.37 × 10⁵ m/s (from 1000 V potential). A hockey ball of mass 0.1 kg moves at the same speed. Calculate the wavelength associated with this velocity for both.
59. If the position of an electron is known within ±0.002 nm, calculate the uncertainty in momentum. Also, check consistency with momentum defined as h / (4πm × 0.05 nm).
60. Given the quantum numbers of six electrons:

• n = 4, l = 2, mₗ = –2, mₛ = –½
• n = 3, l = 2, mₗ = 1, mₛ = +½
• n = 4, l = 1, mₗ = 0, mₛ = +½
• n = 3, l = 2, mₗ = –2, mₛ = –½
• n = 3, l = 1, mₗ = –1, mₛ = +½
• n = 4, l = 1, mₗ = 0, mₛ = +½

Arrange in order of increasing energy and identify any degenerate states.

61. Br (Z = 35) has:

• 6 electrons in 2p
• 6 in 3p
• 5 in 4p

Which electron experiences the lowest effective nuclear charge?

62. Among the following pairs, which orbital experiences a larger effective nuclear charge?
    (i) 2s and 3s
    (ii) 4d and 4f
    (iii) 3d and 3p
63. Unpaired electrons in Al and Si are in 3p orbitals. Which experiences a more effective nuclear charge?
64. Indicate the number of unpaired electrons in:
    (a) P (b) Si (c) Cr (d) Fe (e) Kr
65. (a) How many subshells are associated with n = 4?
    (b) How many electrons will be present in the subshells having mₛ = –½ for n = 4?

Also Read: NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 12: Water (Oceans) (Free PDF)

Solutions

1. Mass of 1 electron = 9.109 × 10⁻³¹ kg
   1 g = 10⁻³ kg
   Number of electrons = (10⁻³) / (9.109 × 10⁻³¹) = 1.098 × 10²⁷ electrons
2. Mass of 1 mole of electrons = 9.109 × 10⁻³¹ × 6.022 × 10²³ = 5.485 × 10⁻⁴ kg
   Charge of 1 mole of electrons = 1.602 × 10⁻¹⁹ × 6.022 × 10²³ = 9.648 × 10⁴ C
3. Electrons in 1 CH₄ molecule = 6 (C) + 4 (H) = 10
   1 mole of methane has 6.022 × 10²³ molecules
   Total electrons = 10 × 6.022 × 10²³ = 6.022 × 10²⁴ electrons
4. Neutrons in one atom of ¹⁴C = 14 – 6 = 8
   Mass of ¹⁴C = 14 g/mol
   Moles in 7 mg = (7 × 10⁻³) / 14 = 0.5 × 10⁻³ mol
   Total atoms = 0.5 × 10⁻³ × 6.022 × 10²³ = 3.011 × 10²⁰
   Total neutrons = 8 × 3.011 × 10²⁰ = 2.41 × 10²¹
   Mass of 1 neutron = 1.675 × 10⁻²⁷ kg
   Total mass = 2.41 × 10²¹ × 1.675 × 10⁻²⁷ = 4.04 × 10⁻⁶ kg
5. Molar mass of NH₃ = 17 g
   Moles in 34 mg = (34 × 10⁻³) / 17 = 2 × 10⁻³ mol
   Protons in 1 NH₃ = 7 (N) + 3 (H) = 10
   Total protons = 2 × 10⁻³ × 6.022 × 10²³ × 10 = 1.204 × 10²²
   Mass of 1 proton = 1.6726 × 10⁻²⁷ kg
   Total mass = 1.204 × 10²² × 1.6726 × 10⁻²⁷ = 2.01 × 10⁻⁵ kg
6. ₆¹³C → Protons = 6, Neutrons = 13 − 6 = 7
   ₈¹⁶O → Protons = 8, Neutrons = 16 − 8 = 8
   ₁₂²⁴Mg → Protons = 12, Neutrons = 24 − 12 = 12
   ₂₆⁵⁶Fe → Protons = 26, Neutrons = 56 − 26 = 30
   ₃₈⁸⁸Sr → Protons = 38, Neutrons = 88 − 38 = 50
   Changing temperature and pressure has no effect on the number of protons and neutrons.
7. (i) Z = 17, A = 35 → ₁₇³⁵Cl
   (ii) Z = 92, A = 233 → ₉₂²³³U
   (iii) Z = 4, A = 9 → ₄⁹Be
8. Wavelength λ = 580 nm = 580 × 10⁻⁹ m
   Speed of light c = 3.0 × 10⁸ m/s
   Frequency ν = c / λ = 3.0 × 10⁸ / 580 × 10⁻⁹ = 5.17 × 10¹⁴ Hz
   Wavenumber ν̅ = 1 / λ (in cm) = 1 / (580 × 10⁻⁷) = 1.72 × 10⁴ cm⁻¹
9. (i) E = hν = 6.626 × 10⁻³⁴ × 3 × 10¹⁵ = 1.9878 × 10⁻¹⁸ J
   (ii) λ = 0.50 Å = 0.50 × 10⁻¹⁰ m
   E = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (0.50 × 10⁻¹⁰) = 3.976 × 10⁻¹⁵ J
10. T = 2.0 × 10⁻¹⁰ s
    Frequency ν = 1 / T = 5.0 × 10⁹ Hz
    λ = c / ν = 3.0 × 10⁸ / 5.0 × 10⁹ = 0.06 m
    Wavenumber ν̅ = 1 / λ (in cm) = 1 / (6.0 × 10⁻¹) = 1.67 cm⁻¹
11. Energy of one photon (E) = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (4000 × 10⁻¹²) = 4.97 × 10⁻²⁰ J
    Number of photons = 1 / 4.97 × 10⁻²⁰ = 2.01 × 10¹⁹
12. (i) E = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (4 × 10⁻⁷) = 4.97 × 10⁻¹⁹ J
    Convert to eV: E = 4.97 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ = 3.10 eV
    (ii) KE = 3.10 − 2.13 = 0.97 eV
    (iii) KE (in J) = 0.97 × 1.602 × 10⁻¹⁹ = 1.554 × 10⁻¹⁹ J
    v = √(2 × KE / m) = √(2 × 1.554 × 10⁻¹⁹ / 9.109 × 10⁻³¹) = 5.86 × 10⁵ m/s
13. λ = 242 nm = 242 × 10⁻⁹ m
    E (per photon) = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (242 × 10⁻⁹) = 8.21 × 10⁻¹⁹ J
    E (per mole) = 8.21 × 10⁻¹⁹ × 6.022 × 10²³ = 494.5 kJ/mol
14. P = 25 W = 25 J/s
    λ = 0.57 µm = 5.7 × 10⁻⁷ m
    E per photon = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (5.7 × 10⁻⁷) = 3.49 × 10⁻¹⁹ J
    Number of photons/sec = 25 / 3.49 × 10⁻¹⁹ = 7.16 × 10¹⁹
15. λ = 6800 Å = 6800 × 10⁻¹⁰ m
    ν = c / λ = 3 × 10⁸ / 6.8 × 10⁻⁷ = 4.41 × 10¹⁴ Hz
    ν₀ = 4.41 × 10¹⁴ Hz
    W₀ = hν₀ = 6.626 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹ J
16. For transition n = 4 to n = 2 in hydrogen:
    1 / λ = R (1/2² − 1/4²) = 1.097 × 10⁷ × (1/4 − 1/16) = 1.097 × 10⁷ × 3/16 = 2.056 × 10⁶ m⁻¹
    λ = 1 / 2.056 × 10⁶ = 486.1 nm
17. Energy at n = 5 = −2.18 × 10⁻¹⁸ / 5² = −8.72 × 10⁻²⁰ J
    Energy required to ionise = 0 − (−8.72 × 10⁻²⁰) = 8.72 × 10⁻²⁰ J
    Ionisation enthalpy from n = 1 = 2.18 × 10⁻¹⁸ J
    Energy required from n = 5 is much smaller.
18. Maximum number of emission lines from n = 6 to n = 1 = n(n − 1)/2 = 6(6 − 1)/2 = 15
19. Energy of first orbit = −2.18 × 10⁻¹⁸ J
    Energy of fifth orbit = −2.18 × 10⁻¹⁸ / 5² = −8.72 × 10⁻²⁰ 
20. The longest wavelength in the Balmer series is for n = 3 to n = 2
    1 / λ = R (1/2² − 1/3²) = 1.097 × 10⁷ × (5/36) = 1
21. λ = h / (mv) = (6.626 × 10⁻³⁴) / (9.1 × 10⁻³¹ × 2.05 × 10⁷) = 3.55 × 10⁻¹¹ m
22. K.E. = 3.0 × 10⁻²⁵ J
    p = √(2mK.E.) = √(2 × 9.1 × 10⁻³¹ × 3.0 × 10⁻²⁵)
    λ = h / p = 6.626 × 10⁻³⁴ / √(5.46 × 10⁻⁵⁵) = 9.66 × 10⁻¹⁰ m
23. Isoelectronic species (same electrons):
    Na⁺ (10), Mg²⁺ (10), F⁻ (10), O²⁻ (10), Ne (10),
    K⁺ (18), Ca²⁺ (18), S²⁻ (18), Ar (18)
    Bohr radius (rₙ) = 0.529 × n² = 0.529 × 25 = 13.23 Å = 1323 pm
24.  (a) H⁻: 1s²
    (b) Na⁺: 1s² 2s² 2p⁶
    (c) O²⁻: 1s² 2s² 2p⁶
    (d) F⁻: 1s² 2s² 2p⁶
    (ii) (a) Z = 11 (b) Z = 7 (c) Z = 17
    (iii) (a) Li (b) P (c) Sc
25.  g-orbitals: l = 4, so minimum n = 5
26.  For 3d orbital: n = 3, l = 2, mₗ = –2, –1, 0, 1, 2
27.  (i) 29 protons
    (ii) Electronic configuration: [Ar] 3d¹⁰ 4s¹
28.  (i) l = 0, 1, 2; mₗ = −l to +l
    (ii) 3d → l = 2, mₗ = −2, −1, 0, 1, 2
    (iii) 1p , 2s , 2p , 3f 
29.  (a) n=1, l=0 → 1s
    (b) n=3, l=1 → 3p
    (c) n=4, l=2 → 4d
    (d) n=4, l=3 → 4f
30.  (a) n=4, mₛ = −½ → 8 electrons
    (b) n=3, l=0 → 2 electrons
31.  2πr = nλ → Bohr’s orbit is an integral multiple of de Broglie wavelength.
32.  Transition n=4 to n=2 in He⁺ is the same as n=8 to n=4 in H.
33.  Energy required = Z² × 2.18 × 10⁻¹⁸ = 4 × 2.18 × 10⁻¹⁸ = 8.72 × 10⁻¹⁸ J
34.  No. of atoms = 0.2 m / 1.5 × 10⁻¹⁰ m = 1.33 × 10⁹
35.  Radius = 2.4 / (2 × 10⁸) = 1.2 × 10⁻⁸ m = 120 pm
36.  Radius = 1.3 Å = 130 pm
    Atoms = 1.6 × 10⁻² m / 2.6 × 10⁻¹⁰ m = 6.15 × 10⁷
37.  Electrons = Q / e = 2.5 × 10⁻¹⁶ / 1.6 × 10⁻¹⁹ = 1562.5 ≈ 1.56 × 10³
38.  Electrons = 1.282 × 10⁻¹⁸ / 1.6 × 10⁻¹⁹ = 8.01 ≈ 8 electrons
39.  Light atoms deflect α-particles more, with fewer large-angle deflections than with heavy atoms.
40.  Only ⁷⁹₃₅Br is correct. Just ⁷⁹Br or ³⁵Br are incomplete (lack atomic/mass number).
41.  Let protons = x → neutrons = x + 0.317x = 1.317x
    x + 1.317x = 81 → x = 34.9 → Z = 35, so symbol = ⁸¹₃₅Br
42.  Let electrons = x → neutrons = x + 0.111x = 1.111x
    x + 1.111x = 36, x = 17.05 → electrons = 17, protons = 18
    Symbol = ³⁷₁₈Ar⁻
43.  x + 1.304x = 56 → x = 24.2 → electrons = 24, protons = 27
    Symbol = ⁵⁶₂₇Co³⁺
44.  Frequency order (increasing):
    FM radio < microwave oven < amber light < X-rays < cosmic rays
45.  E per photon = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (337.1 × 10⁻⁹) = 5.9 × 10⁻¹⁹ J
    Total energy = 5.6 × 10²⁴ × 5.9 × 10⁻¹⁹ = 3.3 × 10⁶ J
    Power = E / t = 3.3 × 10⁶ W
46.  (a) ν = c / λ = 3 × 10⁸ / 616 × 10⁻⁹ = 4.87 × 10¹⁴ Hz
    (b) Distance = c × t = 3 × 10⁸ × 30 = 9 × 10⁹ m
    (c) E = hc / λ = 3.22 × 10⁻¹⁹ J
    (d) Quanta = 2 / 3.22 × 10⁻¹⁹ = 6.2 × 10¹⁸
47.  E = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (600 × 10⁻⁹) = 3.313 × 10⁻¹⁹
    Photons = 3.15 × 10⁻¹⁸ / 3.313 × 10⁻¹⁹ = 9.51 photons
48.  E per photon = hc / λ = 5.91 × 10⁻¹⁹
    Total energy = 2.5 × 10¹⁵ × 5.91 × 10⁻¹⁹ = 1.48 × 10⁻³ J
49.  ν₁ = 3 × 10⁸ / 589 × 10⁻⁹ = 5.09 × 10¹⁴
    ν₂ = 3 × 10⁸ / 589.6 × 10⁻⁹ = 5.087 × 10¹⁴
    ΔE = h(ν₂ − ν₁) = 1.06 × 10⁻²⁰ J
50.  (a) λ = hc / E = 6.56 × 10⁻⁷ m
    (b) ν = c / λ = 4.57 × 10¹⁴ Hz
    E = hc / λ = 3.98 × 10⁻¹⁹
    KE = 3.98 − 1.9 = 2.08 eV
    v = √(2 × KE × e / m) = 8.52 × 10⁵ m/s
51.  v² = 2KE / m = gives threshold KE = 0 when λ = 688 nm
    Use E = hc / λ, find h = (KE × λ) / c → h = 6.626 × 10⁻³⁴ J·s
52.  KE = eV = 0.35 eV = 5.6 × 10⁻²⁰
    E = hc / λ = 7.74 × 10⁻¹⁹
    Work function = 7.74 × 10⁻¹⁹ − 5.6 × 10⁻²⁰ = 7.18 × 10⁻¹⁹ J
53.  E = hc / λ = 1.326 × 10⁻¹⁵
    KE = ½mv² = 1.025 × 10⁻¹⁹
    Binding energy = E − KE = 1.223 × 10⁻¹⁵ J
54.  1285 nm = 1.285 × 10⁻⁶ m
    ν = c / λ = 2.33 × 10¹⁴
    Solve: 3.29 × 10¹⁵ (1/9 − 1/n²) = 2.33 × 10¹⁴ → n = 6
    Region: Infrared
55.  R ∝ n²
    r₁ = 1.3225 nm, r₂ = 211.6 pm
    Find n₁ and n₂, use E = hc / λ
    Wavelength = 9.75 × 10⁻⁸ m
    Series = Lyman
    Region = UV
56.  λ = h / mv = 6.626 × 10⁻³⁴ / (9.1 × 10⁻³¹ × 1.6 × 10⁶) = 4.54 × 10⁻¹⁰ m
57.  λ = 800 pm = h / mv → v = h / (mλ) = 9.06 × 10² m/s
58.  λ = h / mv = 6.626 × 10⁻³⁴ / (9.1 × 10⁻³¹ × 2.19 × 10⁶) = 3.31 × 10⁻¹⁰ m
59.  m (proton) = 1.67 × 10⁻²⁷ kg
    λ = h / mv = 9.68 × 10⁻¹⁰ m
    For hockey ball: λ = 1.51 × 10⁻³³ m
60.  Δx = 0.002 nm = 2 × 10⁻¹² m
    Δp ≥ h / (4πΔx) = 2.63 × 10⁻²³ kg·m/s
    Given p = h / (4πm × 0.05 nm) = 1.16 × 10⁻²³, consistent
61.  Energy order (increasing):
    (5) < (2), (4) < (3), (6) = degenerate
62.  Least Z_eff felt by 4p electron (furthest from nucleus)
63.  (i) 2s > 3s → 2s feels more Z_eff
    (ii) 4d > 4f → 4d feels more
    (iii) 3p > 3d → 3p feels more
64.  Si 3p electron feels more Z_eff due to the higher atomic number
65. (a) P = 3, Si = 2, Cr = 6, Fe = 4, Kr = 0
    (b) Subshells for n=4: 4s, 4p, 4d, 4f → 4 subshells
    Each orbital holds 2 electrons, half with mₛ = −½: 16 electrons

Also Read: NCERT Solutions Class 11 Geography Fundamentals of Physical Geography Chapter 11: World Climate and Climate Change (Free PDF)

Download NCERT Solutions Class 11 Chemistry (Part-I) Chapter 2: Structure of Atom

Download PDF of NCERT Solutions Class 11 Chemistry (Part-1) Chapter 2: Structure of Atom

Download the Solutions of Other Chapters of Class 11 Psychology

Chapter 1

Chapter 3

Chapter 4

Chapter 5

Chapter 6

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