An electron in the hydrogen atom jumps from the first excited state'to the ground state. What will be the…
An electron in the hydrogen atom jumps from the first excited state'to the ground state. What will be the percentage change in the speed of electron?
- $25 \%$
- $50 \%$
- $75 \%$
- $100 \%$
Solution
Velocity of electron in the $\mathrm{n}^{\text {th }}$ orbit is
$\begin{aligned}
& \mathrm{v}_{\mathrm{n}}=\frac{\mathrm{e}^2}{2 \varepsilon_0 \mathrm{nh}} \\
& \Rightarrow \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}}
\end{aligned}$
Taking the ratio,
$\frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{1}{2} \Rightarrow \mathrm{v}_2=\frac{\mathrm{v}_1}{2}$
Change in velocity,
$\begin{aligned}
\Delta \mathrm{v} & =\left|\mathrm{v}_2-\mathrm{v}_1\right| \\
& =\left|\frac{\mathrm{v}_1}{2}-\mathrm{v}_1\right|=0.5 \mathrm{v}_1
\end{aligned}$
Therefore, change in percentage is $50 \%$
Asked in: MHT CET 2023 (11 May Shift 1)
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