Question

Among the given options choose the correct energy of transition:

• A. $H_{2\to1}$ $(6.8\text{ eV})$
• B. $Li^{2+}_{2\to1}$ $(13.6\text{ eV})$
• C. $He^{+}_{2\to1}$ $(40.8\text{ eV})$
• D. $Be^{3+}_{2\to1}$ $(13.6\text{ eV})$

Hint:

Always check whether the atom/ion is hydrogen-like (single electron). Use $Z^2$ scaling directly to compare transition energies quickly.

Answer 1

The Correct Option is C

Concept: For hydrogen-like species (single-electron systems), the energy of an electron in the $n^{\text{th}}$ orbit is given by: \[ E_n = -13.6\,\frac{Z^2}{n^2} \text{ eV} \] Energy of transition from $n_2$ to $n_1$ is: \[ \Delta E = 13.6\,Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]
Step 1: Transition considered is $2 \to 1$, so: \[ \Delta E = 13.6\,Z^2\left(1 - \frac{1}{4}\right) = 13.6\,Z^2\left(\frac{3}{4}\right) \]
Step 2: Check each option. Option (1): $H$ has $Z=1$ \[ \Delta E = 13.6 \times \frac{3}{4} = 10.2 \text{ eV} \neq 6.8 \text{ eV} \] Incorrect. Option (2): $Li^{2+}$ has $Z=3$ \[ \Delta E = 13.6 \times 9 \times \frac{3}{4} = 91.8 \text{ eV} \neq 13.6 \text{ eV} \] Incorrect. Option (3): $He^{+}$ has $Z=2$ \[ \Delta E = 13.6 \times 4 \times \frac{3}{4} = 40.8 \text{ eV} \] Correct. Option (4): $Be^{3+}$ has $Z=4$ \[ \Delta E = 13.6 \times 16 \times \frac{3}{4} = 163.2 \text{ eV} \neq 13.6 \text{ eV} \] Incorrect.
Step 3: Hence, the correct energy of transition is: \[ \boxed{40.8\text{ eV}} \]