A hydrogen‑like ion has nuclear charge $Z = 2$ (e.g., Δ $^{+}$ ). Using the Bohr formula $E_{n} = - 13.6, Z^{2} / n^{2} eV$ , calculate the energy difference $Δ E$ needed to excite an electron from the ground state ( $n = 1$ ) to the $n = 3$ orbit.

medium 2 marks Class 12 · Physics · Atoms

Answer

48.36 eV

The correct answer is 48.36 eV — Round to two decimal places to get $48.36 eV$ .

Step-by-step solution

Calculate ground‑state energy: $E_{1} = - 13.6 \times 2^{2} / 1^{2} = - 54.4 eV$ .
Calculate $n = 3$ energy: $E_{3} = - 13.6 \times 2^{2} / 3^{2} = - 6.0444 eV$ .
Energy required for excitation is the absolute difference: $Δ E = ∣ E_{1} - E_{3} ∣ = ∣ - 54.4 - (- 6.0444) ∣ = 48.3556 eV$ .
Round to two decimal places to get $48.36 eV$ .

Common mistake: Subtracting

E_{3}

from

E_{1}

instead of taking absolute difference

Concept tested: Bohr energy levels

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked