Question

An electron jumps from the first excited state to the ground state of the hydrogen atom. What will be the percentage change in the speed of the electron

• A. $25\%$
• B. $50\%$
• C. $100\%$
• D. $200\%$

Answer 1

The correct option is C $100\%$

The orbital velocity of the electron in the $n^{th}$ bohr orbit is,

$v_n\propto \left(\frac{Z}{n}\right)$

As electron jumps from $(n_2=2)$ to $(n_1=1)$

$\Rightarrow \frac{v_1}{v_2}=\frac{2}{1}$

$\Rightarrow v_1=2\times v_2$

$\%\text{change in speed}=\frac{v_1-v_2}{v_2}\times 100$

$=\frac{2v_2-v_2}{v_2}\times 100=100\%$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question? Key concept: $\%$ change in the speed of the electron is, $\%\text{change in speed}=\frac{v_{final}-v_{initial}}{v_{initial}}\times 100$