The energy of an electron in first Bohr orbit of \(\mathrm{H}\)-atom is \(-13.6 \mathrm{eV}\). The possible…
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 first excited state of \(\mathrm{Li}^{2+}\) is
\(-122.4 \mathrm{eV}\)
\(30.6 \mathrm{eV}\)
\(-30.6 \mathrm{eV}\)
\(13.6 \mathrm{eV}\)
Solution
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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