A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :
Solution
Change in energy
<br/><br/>
$$
\begin{aligned}
&\Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{2} \Rightarrow 12.09=\mathrm{E}_{\mathrm{f}}-(-13.6) \Rightarrow \mathrm{E}_{\mathrm{f}}=-1.51 \mathrm{eV} \\\\
&\Rightarrow \frac{-13.6}{n^{2}}=-1.51 \Rightarrow \mathrm{n}^{2}=9 \Rightarrow \mathrm{n}=3 \\\\
&\text { So, } \Delta \mathrm{L}=\mathrm{L}_{\mathrm{f}}-\mathrm{L}_{\mathrm{i}} \\\\
&=\frac{h}{2 \pi}(3-1)=\frac{2 h}{2 \pi}=\frac{h}{\pi}=\frac{6.63 \times 10^{-34}}{3.14}=2.11 \times 10^{-34} \mathrm{Js}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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