JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 22)
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
828
Explanation
Let the electron jumps to $n^{\text {th }}$ orbit so
$$ \begin{aligned} & 12.75=13.6\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}\right] \\\\ & \Rightarrow n=4 \\\\ & \text { So, Angular momentum } L=\frac{n h}{2 \pi}=\frac{2 h}{\pi} \end{aligned} $$
$$ \begin{aligned} \text { Angular momentum }=\frac{2}{\pi} & \times 4.14 \times 10^{-15} \\\\ & =\frac{828 \times 10^{-17}}{\pi} \mathrm{eVs} \end{aligned} $$
$$ \begin{aligned} & 12.75=13.6\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}\right] \\\\ & \Rightarrow n=4 \\\\ & \text { So, Angular momentum } L=\frac{n h}{2 \pi}=\frac{2 h}{\pi} \end{aligned} $$
$$ \begin{aligned} \text { Angular momentum }=\frac{2}{\pi} & \times 4.14 \times 10^{-15} \\\\ & =\frac{828 \times 10^{-17}}{\pi} \mathrm{eVs} \end{aligned} $$
Comments (0)
