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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?
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Verified Answer
The correct answer is:
$50 \%$
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 $50 \%$
$$
\begin{aligned}
& \mathrm{v}_{\mathrm{n}}=\frac{\mathrm{e}^2}{2 \varepsilon_0 \mathrm{nh}} \\
& \Rightarrow \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}}
\end{aligned}
$$
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 $50 \%$
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