A) 8 : 1
B) 4 : 1
C) 2 : 1
D) 1 : 2
Correct Answer: C
Solution :
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 \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{r}_{1}}}{{{r}_{2}}}}\] \[=\sqrt{\frac{4}{1}}=2:1\]You need to login to perform this action.
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