When an electron in the hydrogen atom in ground state absorbs a photon of energy $12 \cdot 1 eV$. Calculate the change in angular momentum.

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**SOLUTION :** After absorbing a photon of energy $12 \cdot 1 eV$ electron jumps from ground state $(n=1)$ to sccond excited state $(n=3)$. Therefore, change in angular momentum.

$\begin{aligned}\Delta L &=L_{3}-L_{1} \quad\left(L=n \frac{h}{2 \pi}\right) \\&=3\left(\frac{h}{2 \pi}\right)-\left(\frac{h}{2 \pi}\right)=\frac{h}{\pi} \\&=\frac{6 \cdot 626 \times 10^{-34}}{3 \cdot 14} J . S \\&= 2 \cdot 11 \times 10^{-34} J . s .\end{aligned}$

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Calculate the wave number, wavelength and frequency of light with energy of $2 \times 10^{-12} erg$.

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