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
- A \(-122.4 \mathrm{eV}\)
- B \(-30.6 \mathrm{eV}\)
- C \(-30.6 \mathrm{eV}\)
- D \(13.6 \mathrm{eV}\)
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, \(\mathrm{z}=3\)
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