A hydrogen‑like ion with nuclear charge $Z = 5$ (boron $^{4 +}$ ) is in its ground state ( $n = 1$ ).\

hard 4 marks Class 12 · Physics · Atoms

Answer

(i)

340 eV

; (ii)

λ \approx 26.3 nm

The correct answer is (i) $340 eV$ ; (ii) $λ \approx 26.3 nm$ — Photon wavelength from $λ = \frac{h c}{Δ E}$ : $h c = 1240 eV\cdotnm$ , so $λ = \frac{1240}{47.22} \approx 26.3 nm$ .

Step-by-step solution

(i) Ground‑state energy: $E_{1} = - 13.6 \frac{Z ^{2}}{1 ^{2}} = - 13.6 \times 5^{2} = - 13.6 \times 25 = - 340 eV$ .
Ionization energy is the energy needed to bring the electron to $E = 0$ , so $IE = ∣ E_{1} ∣ = 340 eV$ .
(ii) Energy of the $n = 3$ level: $E_{3} = - 13.6 \frac{5 ^{2}}{3 ^{2}} = - 13.6 \times \frac{25}{9} = - \frac{340}{9} \approx - 37.78 eV$ .
Energy of the $n = 2$ level: $E_{2} = - 13.6 \frac{5 ^{2}}{2 ^{2}} = - 13.6 \times \frac{25}{4} = - \frac{340}{4} = - 85 eV$ .
Energy released in the transition: $Δ E = E_{2} - E_{3} = (- 85) - (- 37.78) = 47.22 eV$ .
Photon wavelength from $λ = \frac{h c}{Δ E}$ : $h c = 1240 eV\cdotnm$ , so $λ = \frac{1240}{47.22} \approx 26.3 nm$ .

Common mistake: Using

Z

instead of

Z^{2}

in the Bohr energy formula, which gives a much smaller energy difference.

Concept tested: Energy levels of hydrogen‑like ions

Published 15 September 2026 · How this solution was prepared and checked

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A hydrogen‑like ion with nuclear charge Z=5 (boron4+) is in its ground state (n=1).\

Step-by-step solution

A hydrogen‑like ion with nuclear charge $Z = 5$ (boron $^{4 +}$ ) is in its ground state ( $n = 1$ ).\