A) 1:2
B) 1 : 16
C) 8: 1
D) 16: 1
Correct Answer: D
Solution :
According to Bohr \[{{r}_{n}}\propto \frac{{{n}^{2}}}{Z}\left[ r=\frac{0.529{{n}^{2}}}{Z}\overset{\text{o}}{\mathop{\text{A}}}\, \right]\] \[\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{(2)}^{2}}}{{{(1)}^{2}}}=\frac{4}{1}\] \[Area=\pi {{r}^{2}}\] \[\Rightarrow \]\[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{\pi r_{2}^{2}}{\pi t_{1}^{2}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}={{\left( \frac{4}{1} \right)}^{2}}=\frac{16}{1}\]
You need to login to perform this action.
You will be redirected in
3 sec
