An electron revolving around H-nucleus in the ground state absorbs 10.2 eV energy. Its angular momen

Question

An electron revolving around H-nucleus in the ground state absorbs 10.2 eV energy. Its angular momentum increases by

Accepted Answer

$\Delta E = 13.6z^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)\text{ eV}$

$\therefore 10.2 = 13.6 \times 1^{2}\left( \frac{1}{1^{2}} - \frac{1}{n_{2}^{2}} \right) \Rightarrow n_{2} = 2$

Now, change in angular momentum $= \frac{2 \times h}{2\pi} - \frac{1 \times h}{2\pi} = \frac{h}{2\pi}$

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