Class 12 Physics Chapter 12: Atoms worksheet with answer key

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.

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.

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.

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.

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.

E = hc/λ = 1240/434 ≈ 2.86 eV. The levels are E₂ = −3.40 eV and E₅ = −13.6/25 = −0.544 eV, and E₅ − E₂ = 2.86 eV, so the transition is n = 5 to n = 2. A line ending on n = 2 belongs to the Balmer series (visible).

• photon energy 2.86 eV (1)
• matches E₅ − E₂: transition n = 5 to n = 2, Balmer series (1)

In Thomson's model the positive charge is spread uniformly through the atom, so the field it produces is far too small to turn an α-particle back; yet about 1 in 8000 α-particles is deflected by more than 90°. Rutherford's model is unstable because its orbiting electrons are accelerated and, by classical electromagnetic theory, radiate energy continuously, so they would spiral into the nucleus.

• uniform charge gives far too small a field, yet about 1 in 8000 α-particles is deflected by more than 90° (1)
• Rutherford: accelerated orbiting electron radiates and spirals into the nucleus (1)

An emission line spectrum is obtained when the gas, excited at low pressure (for example by an electric current), emits only certain wavelengths: bright lines on a dark background. An absorption spectrum is obtained when white light passes through the gas: dark lines appear at the same wavelengths as the emission lines. Each element has its own characteristic set of lines, so the spectrum acts as a fingerprint of the gas.

• emission: bright lines on dark background from excited gas; absorption: dark lines in white light passed through gas, at the same wavelengths (1)
• lines are characteristic of each element, so they identify the gas (1)

(a) The number of lines is n(n − 1)/2 = 4 × 3/2 = 6. Lyman (ending on n = 1): 4→1, 3→1, 2→1, three lines. Balmer (ending on n = 2): 4→2, 3→2, two lines. Paschen (ending on n = 3): 4→3, one line. (b) The longest wavelength is the smallest energy gap, 4→3: ΔE = 13.6 × (1/9 − 1/16) = 0.661 eV, λ = 1240/0.661 ≈ 1876 nm (infrared).

• six lines (1)
• Lyman 3, Balmer 2, Paschen 1 (1)
• longest wavelength from 4→3, about 1876 nm (1)

(a) As with waves on a string, only standing waves survive: a wave that does not fit a whole number of times round the orbit interferes with itself on going round, and its amplitude quickly drops to zero. So 2πr_n = nλ. (b) Three wavelengths means n = 3, so r₃ = 9a₀ = 4.77 Å, and λ = 2πr₃/3 = 2π × 4.77/3 ≈ 9.99 Å, about 10 Å.

• only resonant standing waves persist; other waves cancel themselves (1)
• n = 3, r = 4.77 Å (1)
• λ ≈ 10 Å (1)

For a stable orbit the Coulomb force gives the centripetal force: (1/4πε₀)e²/r² = mv²/r, so mv²r = e²/4πε₀. Bohr's postulate: mvr = nh/2π, so v = nh/(2πmr). Substituting, m(nh/2πmr)²r = e²/4πε₀, which gives r_n = (n²/m)(h/2π)²(4πε₀/e²), proportional to n². So r_n/r₂ = n²/4 = 9 gives n² = 36 and n = 6.

• force balance (1/4πε₀)e²/r² = mv²/r (1)
• uses mvr = nh/2π to eliminate v, r_n ∝ n² (1)
• n = 6 (1)