Question

Find out the correct energy for the ground state or energy transition. (Symbols have usual meaning and \(n \to m\) gives the transition)

• A. \( \mathrm{H}\;(-6.8\ \text{eV}) \)
• B. \( \mathrm{Li^{2+}}\;(-13.6\ \text{eV}) \)
• C. \( \mathrm{He^+}\;(2 \to 1)\;(40.8\ \text{eV}) \)
• D. \( \mathrm{Be^{3+}}\;(2 \to 1)\;(+13.6\ \text{eV}) \)

Hint:

For hydrogen-like ions, energy scales as \(Z^2\). Always check both magnitude and sign for ground-state energies and transitions.

Answer 1

The Correct Option is C

Concept:
For hydrogen-like species, the energy of the electron in the \(n^{\text{th}}\) orbit is: \[ E_n = -13.6\,\frac{Z^2}{n^2}\ \text{eV} \] Energy released or absorbed during a transition: \[ \Delta E = 13.6\,Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \]
Step 1: Check each option. Option (A): Hydrogen ground state \[ E_1 = -13.6\ \text{eV} \] Given value is \(-6.8\ \text{eV}\) ❌ (incorrect) Option (B): \( \mathrm{Li^{2+}} \) ground state (\(Z=3\)) \[ E_1 = -13.6 \times 3^2 = -122.4\ \text{eV} \] Given value is \(-13.6\ \text{eV}\) ❌ Option (C): \( \mathrm{He^+} \), transition \(2 \to 1\) (\(Z=2\)) \[ \Delta E = 13.6 \times 2^2\left(1 - \frac{1}{4}\right) \] \[ = 13.6 \times 4 \times \frac{3}{4} = 40.8\ \text{eV} \] ✔️ Correct Option (D): \( \mathrm{Be^{3+}} \), transition \(2 \to 1\) (\(Z=4\)) \[ \Delta E = 13.6 \times 16 \times \frac{3}{4} = 163.2\ \text{eV} \] Also energy released should be negative, not positive ❌ Conclusion: Only option (3) matches the correct calculated value.