Question 441: Atoms & Nuclei

Question

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?

Accepted Answer

Velocity of electron in the $$\mathrm{n}^{	ext {th }}$$ 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,
$$	herefore \quad \frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{1}{2} \Rightarrow \mathrm{v}_2=\frac{\mathrm{v}_1}{2}$$
Change in velocity,
$$\begin{aligned}
\Delta \mathrm{v} & =\left|\mathrm{v}_2-\mathrm{v}_1ight| \
& =\left|\frac{\mathrm{v}_1}{2}-\mathrm{v}_1ight|=0.5 \mathrm{v}_1
\end{aligned}$$
Therefore, change in percentage is $$50 \%$$

Answer

25%

Answer

50%

Answer

75%

Answer

100%

Atoms & Nuclei

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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