The minimum excitation energy of an electron revolving in the first orbit of hydrogen is
- $3.4 \mathrm{eV}$
- $8.5 \mathrm{eV}$
- $10.2 \mathrm{eV}$
- $13.6 \mathrm{eV}$
Solution
$\mathrm{E}_{\mathrm{n}}=\frac{-13.6 \mathrm{eV}}{\mathrm{n}^2}$
Excitation energy of electron from $n=1$ to $n=2$
$\mathrm{E}=-13.6\left[\frac{1}{2^2}-\frac{1}{1}\right]=-13.6\left[\frac{1}{4}-\frac{1}{1}\right]=10^{-2} \mathrm{eV}$
So, Minimum excitation energy $=10.2 \mathrm{eV}$
Asked in: AP EAMCET 2023 (19 May Shift 1)
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