A light of energy $12.75 ~\mathrm{eV}$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{\pi} \times 10^{-17} ~\mathrm{eVs}$. The value of $x$ is ___________ (use $h=4.14 \times 10^{-15} ~\mathrm{eVs}, c=3 \times 10^{8} \mathrm{~ms}^{-1}$ ).
Answer (integer)
828
Solution
Let the electron jumps to $n^{\text {th }}$ orbit so
<br/><br/>$$
\begin{aligned}
& 12.75=13.6\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}\right] \\\\
& \Rightarrow n=4 \\\\
& \text { So, Angular momentum } L=\frac{n h}{2 \pi}=\frac{2 h}{\pi}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\text { Angular momentum }=\frac{2}{\pi} & \times 4.14 \times 10^{-15} \\\\
& =\frac{828 \times 10^{-17}}{\pi} \mathrm{eVs}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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