A hydrogen‑like ion with nuclear charge $Z = 4$ is in the ground state. Using the Bohr formula $E_{n} = - 13.6, Z^{2} / n^{2} eV$ , what is the energy difference $Δ E$ required to excite the electron from $n = 1$ to $n = 2$ ?

easy MCQ 1 mark Class 12 · Physics · Atoms

Answer

163.2 eV

The correct answer is 163.2 eV — The excitation energy is the difference: $Δ E = E_{2} - E_{1} = (- 54.4) - (- 217.6) = 163.2 eV$ .

Step-by-step solution

Use the Bohr energy formula for the ground state: $E_{1} = - 13.6 \times 4^{2} / 1^{2} = - 217.6 eV$ .
Calculate the energy of the $n = 2$ level: $E_{2} = - 13.6 \times 4^{2} / 2^{2} = - 54.4 eV$ .
The excitation energy is the difference: $Δ E = E_{2} - E_{1} = (- 54.4) - (- 217.6) = 163.2 eV$ .

Common mistake: Using the absolute value of the energies instead of their signed values

Concept tested: Bohr model energy levels

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked