Question 380: Atoms & Nuclei

Question

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 $$\frac{x}{\pi} 	imes 10^{-17} ~\mathrm{eVs}$$. The value of $$x$$ is ___________ (use $$h=4.14 	imes 10^{-15} ~\mathrm{eVs}, c=3 	imes 10^{8} \mathrm{~ms}^{-1}$$ ).

Accepted Answer

Let the electron jumps to $n^{	ext {th }}$ orbit so
$$
\begin{aligned}
& 12.75=13.6\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}ight] \\
& \Rightarrow n=4 \\
& 	ext { So, Angular momentum } L=\frac{n h}{2 \pi}=\frac{2 h}{\pi} 
\end{aligned}
$$
$$
\begin{aligned}
	ext { Angular momentum }=\frac{2}{\pi} & 	imes 4.14 	imes 10^{-15} \\
& =\frac{828 	imes 10^{-17}}{\pi} \mathrm{eVs}
\end{aligned}
$$

Atoms & Nuclei

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