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(B) The velocity of an electron in the $n^{	ext{th}}$ orbit is given by $v_n = \frac{e^2}{2 \varepsilon_0 nh}$.This implies $v_n \propto \frac{1}{n}$

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$50$

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(B) The velocity of an electron in the $n^{	ext{th}}$ orbit is given by $v_n = \frac{e^2}{2 \varepsilon_0 nh}$.This implies $v_n \propto \frac{1}{n}$.For the ground state,$n_1 = 1$,and for the first excited state,$n_2 = 2$.Thus,the ratio of velocities is $\frac{v_2}{v_1} = \frac{n_1}{n_2} = \frac{1}{2}$,which means $v_2 = 0.5 v_1$.The change in velocity is $\Delta v = |v_2 - v_1| = |0.5 v_1 - v_1| = 0.5 v_1$.The percentage change is $\frac{\Delta v}{v_1} 	imes 100\% = \frac{0.5 v_1}{v_1} 	imes 100\% = 50\%$.

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 the electron (in $\%$)?

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