Question

An electron in a hydrogen atom makes a transition $ {{n}_{1}}\to {{n}_{2}} $ where $ {{n}_{1}} $ and $ {{n}_{2}} $ are principal quantum numbers to be valid. The time period of the electron in initial state is eight times that in the final state. Then the ratio of $ {{n}_{1}} $ and $ {{n}_{2}} $ will be

• A. 8 : 1
• B. 4 : 1
• C. 2 : 1
• D. 1 : 2

Answer 1

Answer: (C)

Let T be the periodic time and r the radius of circular orbit, then $ {{T}^{2}}\propto {{r}^{3}} $ So, $ {{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}\propto {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}} $ or $ \frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2/3}} $ $ ={{(8)}^{2/3}}=4 $ If n is principal quantum number, then $ r=\frac{{{\varepsilon }_{0}}{{h}^{2}}{{n}^{2}}}{\pi m{{e}^{2}}}\propto {{n}^{2}} $ So, $ \frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{n}_{1}}}{{{n}_{2}}} \right)}^{2}} $ or $ \frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{r}_{1}}}{{{r}_{2}}}} $ $ =\sqrt{\frac{4}{1}}=2:1 $