JEE Physics Atoms MCQs Set 14

Physics Objective Questions and Answers: Atoms

Access targeted multiple-choice questions for Atoms designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: Rydberg constant R is equal to
(a) \( \frac{me^2}{8\epsilon_0^2ch^3} \)
(b) \( \frac{me^4}{8\epsilon_0^2ch^3} \)
(c) \( \frac{m^2e^4}{8\epsilon_0^2ch^3} \)
(d) \( \frac{m^4e^4}{8\epsilon_0^2ch^3} \)
Show Answer & Explanation

Answer: (b) \( \frac{me^4}{8\epsilon_0^2ch^3} \)

Explanation:
1. R = me⁴/(8ε₀²ch³).

Teacher's Note:
1. Recall the basic fact.

Question 2: In a hypothetical Bohr hydrogen atom, the mass of the electron is doubled. The energy E'₀ of the first orbit will be (r₀ is the Bohr radius)
(a) −11.2 eV
(b) −6.8 eV
(c) −13.6 eV
(d) −27.2 eV
Show Answer & Explanation

Answer: (d) −27.2 eV

Explanation:
1. E ∝ m, so doubling the mass doubles the energy: −27.2 eV.

Teacher's Note:
1. Recall the basic fact.

Question 3: Energy required for the electron excitation in Li⁺⁺ from the first to the third Bohr orbit is
(a) 36.3 eV
(b) 108.8 eV
(c) 122.4 eV
(d) 12.1 eV
Show Answer & Explanation

Answer: (b) 108.8 eV

Explanation:
1. ΔE = 13.6 × Z²(1 − 1/9) = 13.6 × 9 × 8/9 = 108.8 eV.

Teacher's Note:
1. Recall the basic fact.

Question 4: The shortest wavelength in Balmer's series for hydrogen atom is ___A___ and this is obtained by substituting ___B___ in Balmer's formula. Here A and B refer to
(a) 656.3 nm, n = 3
(b) 486.1 nm, n = 4
(c) 410.2 nm, n = 5
(d) 364.6 nm, n = ∞
Show Answer & Explanation

Answer: (d) 364.6 nm, n = ∞

Explanation:
1. The shortest wavelength is the series limit: 1/λ = R/4 for n = ∞, giving 364.6 nm.

Teacher's Note:
1. Recall the basic fact.

Question 5: Hydrogen, deuterium, singly ionised helium and doubly ionised lithium all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ₁, λ₂, λ₃ and λ₄ respectively, then approximately which one of the following is correct?
(a) 4λ₁ = 2λ₂ = 2λ₃ = λ₄
(b) λ₁ = 2λ₂ = 2λ₃ = λ₄
(c) λ₁ = λ₂ = 4λ₃ = 9λ₄
(d) λ₁ = 2λ₂ = 3λ₃ = 4λ₄
Show Answer & Explanation

Answer: (c) λ₁ = λ₂ = 4λ₃ = 9λ₄

Explanation:
1. λ ∝ 1/Z² (ignoring nuclear mass differences).
2. H and D have Z = 1, He⁺ has Z = 2 and Li⁺⁺ has Z = 3.
3. So λ₁ = λ₂ = 4λ₃ = 9λ₄.

Teacher's Note:
1. Recall the basic fact.

Question 6: If ν₁ is the frequency of the series limit of Lyman, ν₂ is the frequency of the first line of Lyman series and ν₃ is the frequency of the series limit of the Balmer series then
(a) \( \nu_1-\nu_2=\nu_3 \)
(b) \( \nu_1=\nu_2-\nu_3 \)
(c) \( \frac{1}{\nu_2}=\frac{1}{\nu_1}+\frac{1}{\nu_3} \)
(d) \( \frac{1}{\nu_1}=\frac{1}{\nu_2}+\frac{1}{\nu_3} \)
Show Answer & Explanation

Answer: (a) \( \nu_1-\nu_2=\nu_3 \)

Explanation:
1. ν₁ = Rc, ν₂ = 3Rc/4, ν₃ = Rc/4.
2. ν₁ − ν₂ = Rc/4 = ν₃.

Teacher's Note:
1. Recall the basic fact.

Question 7: Bohr's atom model is the modification of Rutherford's atom model by the application of
(a) Newton's theory
(b) Huygens' theory
(c) Maxwell's theory
(d) Planck's quantum theory
Show Answer & Explanation

Answer: (d) Planck's quantum theory

Explanation:
1. Bohr applied Planck's quantum idea to Rutherford's model.

Teacher's Note:
1. Recall the basic fact.

Question 8: In the ground state, in ___A___ electrons are in stable equilibrium while in ___B___ electrons always experience a net force. Here A and B refer to
(a) Dalton's theory, Rutherford model
(b) Rutherford's model, Bohr's model
(c) Thomson's model, Rutherford's model
(d) Rutherford's model, Thomson's model
Show Answer & Explanation

Answer: (c) Thomson's model, Rutherford's model

Explanation:
1. In Thomson's model electrons sit in stable equilibrium. In Rutherford's model there is always a net centripetal force.

Teacher's Note:
1. Recall the basic fact.

Question 9: An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be (R is the Rydberg constant, m is the mass of the atom)
(a) \( \frac{24hR}{25m} \)
(b) \( \frac{25hR}{24m} \)
(c) \( \frac{25m}{24hR} \)
(d) \( \frac{24m}{25hR} \)
Show Answer & Explanation

Answer: (a) \( \frac{24hR}{25m} \)

Explanation:
1. Photon wave number = R(1 − 1/25) = 24R/25, so p = h/λ = 24hR/25.
2. Momentum conservation gives atom velocity v = p/m = 24hR/(25m).

Teacher's Note:
1. Recall the basic fact.

Question 10:
Assertion (A): Electrons in the atom are held by Coulomb forces.
Reason (R): The atom is stable only because the centripetal force due to Coulomb's law is balanced by the centrifugal force.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Coulomb attraction does hold the electrons.
2. Stability is due to the quantisation of angular momentum and non-radiating orbits, not a balance with a centrifugal force, so the reason is incorrect.

Teacher's Note:
1. Stability needs the quantum postulates.

Question 11: The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is
(a) 3
(b) 4
(c) 1
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. Lyman first line: 1/λ = 3R/4.
2. Ion second Balmer line: 1/λ = Z² R × 3/16.
3. Equating: Z² = 4, so Z = 2.

Teacher's Note:
1. Recall the basic fact.

Question 12: The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10⁻⁴ T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to
(a) 1.8 eV
(b) 1.1 eV
(c) 0.8 eV
(d) 1.6 eV
Show Answer & Explanation

Answer: (b) 1.1 eV

Explanation:
1. Photon energy = 1.89 eV.
2. Kmax = (eBr)²/2m = (1.6 × 10⁻¹⁹ × 3 × 10⁻⁴ × 0.01)²/(2 × 9.1 × 10⁻³¹) = 1.27 × 10⁻¹⁹ J = 0.79 eV.
3. φ = 1.89 − 0.79 ≈ 1.1 eV.

Teacher's Note:
1. Recall the basic fact.

Question 13: According to classical theory, the circular path of an electron in Rutherford atom model is
(a) spiral
(b) circular
(c) parabolic
(d) straight line
Show Answer & Explanation

Answer: (a) spiral

Explanation:
1. The orbiting electron would radiate energy and spiral into the nucleus.

Teacher's Note:
1. Recall the basic fact.

Question 14: According to the Bohr theory of H-atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number n, respectively, as
(a) \( \frac{1}{n},n^2,\frac{1}{n^2} \)
(b) \( n,\frac{1}{n^2},n^2 \)
(c) \( n,\frac{1}{n^2},\frac{1}{n} \)
(d) \( \frac{1}{n},\frac{1}{n^2},n^2 \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{n},\frac{1}{n^2},n^2 \)

Explanation:
1. v ∝ 1/n, E ∝ 1/n² (in magnitude) and r ∝ n².

Teacher's Note:
1. Recall the basic fact.

Question 15: The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionised helium atom is
(a) 1215 Å
(b) 1640 Å
(c) 2430 Å
(d) 4687 Å
Show Answer & Explanation

Answer: (a) 1215 Å

Explanation:
1. H first line: 1/λ = 5R/36. He⁺ second Balmer line: 1/λ = 4R × 3/16 = 3R/4.
2. λ = 6561 × (5/36)/(3/4) = 1215 Å.

Teacher's Note:
1. Recall the basic fact.

Question 16: The extreme wavelengths of Paschen series are
(a) 0.365 μm and 0.565 μm
(b) 0.818 μm and 1.89 μm
(c) 1.45 μm and 4.04 μm
(d) 2.27 μm and 7.43 μm
Show Answer & Explanation

Answer: (b) 0.818 μm and 1.89 μm

Explanation:
1. Shortest (n = ∞): 9/R = 0.82 μm.
2. Longest (4 → 3): 144/7R = 1.875 μm ≈ 1.89 μm.

Teacher's Note:
1. Recall the basic fact.

Question 17: The radius of the nth Bohr's orbit of H atom is given by
(a) \( \frac{\epsilon_0 n^2h^2}{\pi m e^2} \)
(b) \( \frac{n^2h^2}{\epsilon_0\pi m e^2} \)
(c) \( \frac{\pi m e^2}{\epsilon_0 n^2h^2} \)
(d) \( n^2h^2 \)
Show Answer & Explanation

Answer: (a) \( \frac{\epsilon_0 n^2h^2}{\pi m e^2} \)

Explanation:
1. rₙ = ε₀n²h²/(πme²).

Teacher's Note:
1. Recall the basic fact.

Question 18: As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion
(a) kinetic energy decreases, potential energy increases but total energy remains same
(b) kinetic energy and total energy decrease but potential energy increases
(c) its kinetic energy increases but potential energy and total energy decrease
(d) kinetic energy, potential energy and total energy decrease
Show Answer & Explanation

Answer: (c) its kinetic energy increases but potential energy and total energy decrease

Explanation:
1. KE = −E increases, while PE = 2E and total E decrease as n falls.

Teacher's Note:
1. Recall the basic fact.

Question 19: Balmer series lies in which spectrum?
(a) visible
(b) ultraviolet
(c) infrared
(d) partially visible, partially infrared
Show Answer & Explanation

Answer: (a) visible

Explanation:
1. The Balmer lines (656 nm to 365 nm) lie in the visible region.

Teacher's Note:
1. Recall the basic fact.

Question 20: The observations of Geiger-Marsden experiment are I. many of α particles pass straight through the gold foil II. only 0.14% of α-particles scatter by more than 1° III. about 1 in 8000 α-particles is deflected more than 90° IV. very few α-particles are reflected back
(a) I, II and IV
(b) I, II and III
(c) II, III and IV
(d) I, II, III and IV
Show Answer & Explanation

Answer: (d) I, II, III and IV

Explanation:
1. All four statements describe the observations.

Teacher's Note:
1. Recall the basic fact.

Question 21: Doubly ionised helium and hydrogen ions are accelerated, from rest, through the same potential difference. The ratio of final velocities of helium and of hydrogen is
(a) \( 1:\sqrt2 \)
(b) \( \sqrt2:1 \)
(c) \( 1:2 \)
(d) \( 2:1 \)
Show Answer & Explanation

Answer: (a) \( 1:\sqrt2 \)

Explanation:
1. v = √(2qV/m).
2. For He²⁺, q/m = 2/4 = 1/2 (in units of e/mp), for H⁺ it is 1.
3. So vHe/vH = √(1/2) = 1 : √2.

Teacher's Note:
1. Recall the basic fact.

Question 22: Match the quantities in Column I with the expressions in Column II for the nth orbit.
Column IColumn II
(A) Radius of nth orbit(1) 2πkze²/(nh)
(B) Velocity of electron in nth orbit(2) −kze²/(rₙ)
(C) Potential energy in nth orbit(3) kze²/(2rₙ)
(D) Kinetic energy in nth orbit(4) n²h²/(4π²kze²m)

(a) A-4, B-3, C-2, D-1
(b) A-4, B-1, C-2, D-3
(c) A-4, B-2, C-1, D-3
(d) A-4, B-1, C-3, D-2
Show Answer & Explanation

Answer: (b) A-4, B-1, C-2, D-3

Explanation:
1. r = n²h²/(4π²kze²m).
2. v = 2πkze²/(nh).
3. PE = −kze²/r.
4. KE = kze²/2r.

Teacher's Note:
1. Recall the Bohr formulae.

Question 23: Maximum energy is evolved during which of the following transitions?
(a) n = 1 to n = 2
(b) n = 2 to n = 1
(c) n = 2 to n = 6
(d) n = 6 to n = 2
Show Answer & Explanation

Answer: (b) n = 2 to n = 1

Explanation:
1. Energy is emitted in a downward transition, and 2 → 1 gives the largest release among 6 → 2 and 2 → 1.

Teacher's Note:
1. Recall the basic fact.

Question 24: The longest wavelength of the Balmer series is 6563 Å. Rydberg's constant is
(a) 1.09 × 10⁵ m⁻¹
(b) 1.09 × 10⁶ m⁻¹
(c) 1.09 × 10⁷ m⁻¹
(d) 1.09 × 10⁸ m⁻¹
Show Answer & Explanation

Answer: (c) 1.09 × 10⁷ m⁻¹

Explanation:
1. 1/λ = 5R/36, so R = 36/(5 × 6.563 × 10⁻⁷) = 1.097 × 10⁷ m⁻¹.

Teacher's Note:
1. Recall the basic fact.

Question 25: The Lyman transitions involve
(a) largest changes of energy
(b) smallest changes of energy
(c) largest changes of potential energy
(d) smallest changes of potential energy
Show Answer & Explanation

Answer: (a) largest changes of energy

Explanation:
1. Transitions to n = 1 have the largest energy gaps.

Teacher's Note:
1. Recall the basic fact.

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Multiple Choice Questions (MCQs) for JEE Physics Atoms

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