A) \[25\]%
B) \[50\]%
C) \[100\,\]%
D) \[200\,\]%
Correct Answer: B
Solution :
Velocity of \[{{e}^{-}}\] in \[{{n}^{th}}\] orbit \[{{v}_{n}}\propto \frac{1}{n}\], First excited state corresponds to \[\operatorname{n}= 2\] and ground state corresponds to\[n=1\]. \[\frac{{{v}_{2}}}{{{v}_{1}}}=\frac{1}{2}\,\,\,\Rightarrow \,\,\,\,{{v}_{2}}=\frac{{{v}_{1}}}{2}\] \[\therefore \,\, Change in velocity\,\,\Delta v={{v}_{1}}-{{v}_{2}}=\frac{{{v}_{1}}}{2}\]i.e., \[50\,.\]%,
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