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WBJEE · Chemistry · Structure of Atom

The energy of an electron in first Bohr orbit of \(\mathrm{H}\)-atom is \(-13.6 \mathrm{eV}\). The possible energy value of electron in the excited state of \(\mathrm{Li}^{2+}\) is

  1. A \(-122.4 \mathrm{eV}\)
  2. B \(-30.6 \mathrm{eV}\)
  3. C \(-30.6 \mathrm{eV}\)
  4. D \(13.6 \mathrm{eV}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(-30.6 \mathrm{eV}\)

Step-by-step Solution

Detailed explanation

Hints: \(E_n=\frac{E_1}{n^2} \times z^2\) \(=\frac{-13.6}{4} \times 9=-30.6 \mathrm{eV}\) For the excited state, \(\mathrm{n}=2\) and for \(\mathrm{Li}^{++}\)ion, \(\mathrm{z}=3\)