Physics · Atoms and nuclei · chapter 12 of 14

Class 12 Physics Chapter 12: Atoms MCQs with answers

10 multiple-choice questions on Atoms, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
In the model of the atom proposed by J. J. Thomson,
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Answer: (b) the positive charge is spread uniformly through the volume of the atom and the electrons are embedded in it

Thomson pictured the positive charge spread uniformly throughout the atom with the electrons embedded in it like seeds in a watermelon (the plum pudding model). Option a is Rutherford's nuclear model and option c is Bohr's quantum condition; option d contradicts the neutrality of the atom.

Question 2
Bohr's model can be applied directly to which one of the following?
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Answer: (c) A doubly ionised lithium ion, Li²⁺

Bohr's model works for hydrogenic atoms: a nucleus of charge +Ze with a single electron. Lithium has three electrons, so Li²⁺ has one left. Neutral helium has two electrons, Li⁺ has two and neutral lithium three, so each has electron-electron interactions the model leaves out.

Question 3
In the alpha-particle scattering experiment, most α-particles pass straight through the gold foil. This shows that
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Answer: (c) most of the volume of an atom is empty space

The nucleus is 10⁻¹⁵ m to 10⁻¹⁴ m across while the atom is about 10⁻¹⁰ m, so most of an atom is empty space and most α-particles meet no nucleus on their way. A few are deflected through large angles, so the atom does carry concentrated positive charge (a, b wrong); a uniform spread (d) could not turn any α-particle back.

Question 4
An α-particle approaches a gold nucleus with impact parameter zero. Its angle of scattering is close to
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Answer: (d) 180°

Zero impact parameter means a head-on collision: the α-particle is stopped by the repulsion and rebounds along its path, θ ≅ π = 180°. A very large impact parameter gives θ ≅ 0°; 90° and 45° belong to intermediate impact parameters.

Question 5
An α-particle aimed head-on at a nucleus comes momentarily to rest at a distance d from its centre. If the kinetic energy of the α-particle is doubled, the distance of closest approach becomes how many times d?
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Answer: (b) 0.5

At the turning point K = (1/4πε₀)(2e)(Ze)/d, so d ∝ 1/K. Doubling K halves d: 0.5 d. 2 takes d proportional to K; 0.25 takes d ∝ 1/K²; 1.41 takes d ∝ √K.

Question 6
An α-particle and a proton, each of kinetic energy K, are in turn aimed head-on at the same nucleus. The ratio of the distance of closest approach of the α-particle to that of the proton is (charge of α-particle = 2e, charge of proton = e)
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Answer: (d) 2

For a projectile of charge q, K = (1/4πε₀)(q)(Ze)/d, so d = qZe/(4πε₀K) ∝ q at the same K and Z. dα/dp = 2e/e = 2. The mass of the projectile does not enter, since K is given; 4 squares the charge ratio and 1/2 inverts it.

Question 7
An α-particle fired head-on at a tin nucleus (Z = 50) comes momentarily to rest 24 fm from its centre. Its initial kinetic energy is (1/4πε₀ = 9 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C, 1 MeV = 1.6 × 10⁻¹³ J)
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Answer: (b) 6.0 MeV

All the kinetic energy is stored as potential energy at the turning point: K = (1/4πε₀)(2e)(Ze)/d = 9 × 10⁹ × 2 × 50 × (1.6 × 10⁻¹⁹)²/(24 × 10⁻¹⁵) J = 9.6 × 10⁻¹³ J = 6.0 MeV. 3.0 MeV takes the charge of the α-particle as e instead of 2e; 0.60 MeV and 60 MeV are slips of a power of ten in fm (10⁻¹⁵ m).

Question 8
α-particles of the same kinetic energy are aimed head-on, in turn, at a gold nucleus (Z = 79) and a silver nucleus (Z = 47). The ratio of the distance of closest approach for gold to that for silver is about
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Answer: (d) 1.68

d = 2Ze²/(4πε₀K), so at the same K, d ∝ Z: d(Au)/d(Ag) = 79/47 ≈ 1.68. 0.59 inverts the ratio; 1.30 takes d ∝ √Z; 2.83 takes d ∝ Z².

Question 9
According to classical electromagnetic theory, an electron revolving round the nucleus in Rutherford's model
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Answer: (b) radiates energy continuously and spirals into the nucleus

A charge moving in a circle is constantly accelerated, and an accelerated charge emits electromagnetic radiation. The electron's energy would fall continuously and it would spiral into the nucleus, so the atom could not be stable. Constant speed (d) does not mean zero acceleration; option c is Bohr's postulate, not classical theory.

Question 10
Bohr's first postulate states that an electron in an atom
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Answer: (a) can revolve in certain stable orbits without emitting radiant energy

The first postulate: the electron can revolve in certain stable orbits, the stationary states, each of definite total energy, without emitting radiant energy, contrary to electromagnetic theory. Option b is the classical expectation; option c contradicts the second postulate; emission happens only in a transition (third postulate), not while in an orbit.

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