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?

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 \(50 \%\)