More practice questions

4. Q4. Find the wavelength of the light emitted when the electron of a hydrogen atom drops from n = 4 to n = 2 (NCERT Exercise 2.13, page 69).

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   |ΔE| = 2.18 × 10^⁻¹⁸ J × (1/2^² − 1/4^²) = 2.18 × 10^⁻¹⁸ × 0.1875 = 4.09 × 10^⁻¹⁹ J. λ = hc/|ΔE| = (6.626 × 10^⁻³⁴ × 3.0 × 10^⁸) ÷ (4.09 × 10^⁻¹⁹) = 4.86 × 10^⁻⁷ m = 486 nm, a blue-green Balmer line.

5. Q5. The ionisation energy of a hydrogen atom in the ground state is 2.18 × 10^⁻¹⁸ J. Find the energy needed for He^⁺(g) → He^²⁺(g) + e^⁻ (NCERT Exercise 2.34, page 71).

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   He^⁺ has one electron and Z = 2. Removing it from n = 1 takes 2.18 × 10^⁻¹⁸ J × 2^² = 8.72 × 10^⁻¹⁸ J per ion (about 5.25 × 10^³ kJ mol^⁻¹).

6. Q6. An electron in a hydrogen atom moves from an orbit of radius 1.3225 nm to one of radius 211.6 pm. Find the wavelength of the light emitted, the series, and the region of the spectrum (NCERT Exercise 2.56, page 72).

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   r_n = n^² × 52.9 pm: 1322.5 pm = 25 × 52.9 pm, so n = 5; 211.6 pm = 4 × 52.9 pm, so n = 2. |ΔE| = 2.18 × 10^⁻¹⁸ × (1/4 − 1/25) = 4.58 × 10^⁻¹⁹ J, so λ = (6.626 × 10^⁻³⁴ × 3.0 × 10^⁸) ÷ (4.58 × 10^⁻¹⁹) ≈ 434 nm. Balmer series, visible region.

7. Q7. The electron in the first Bohr orbit of hydrogen moves at about 2.19 × 10^⁶ m s^⁻¹, and the speed is proportional to Z/n. Find the speed in (a) the first orbit of Li^²⁺ and (b) the third orbit of hydrogen.

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   (a) Z = 3, n = 1: 3 × 2.19 × 10^⁶ ≈ 6.57 × 10^⁶ m s^⁻¹. (b) Z = 1, n = 3: 2.19 × 10^⁶/3 = 7.30 × 10^⁵ m s^⁻¹.

8. Q8. Find the longest and the shortest wavelengths of the Lyman series of hydrogen, using ν̄ = 1.09677 × 10^⁷ × (1/n₁² − 1/n₂²) m^⁻¹.

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   Longest: the smallest jump, 2 → 1. ν̄ = 1.09677 × 10^⁷ × (1 − 1/4) = 8.226 × 10^⁶ m^⁻¹, so λ = 1/ν̄ = 1.216 × 10^⁻⁷ m ≈ 122 nm. Shortest: the series limit, ∞ → 1. ν̄ = 1.09677 × 10^⁷ m^⁻¹, so λ = 9.12 × 10^⁻⁸ m = 91.2 nm. Both are in the ultraviolet.