A) 1:1
B) 1:16
C) 8:1
D) 16:1
Correct Answer: D
Solution :
[d] \[16:1\] \[{{r}_{n}}\propto \,{{n}^{2}}\] \[\therefore \]area covered \[(An)\propto {{n}^{4}}(\therefore \,Area\,=\pi {{r}^{2}})\] \[\therefore \frac{{{A}_{2}}}{{{A}_{1}}}=\frac{{{2}^{4}}}{{{1}^{4}}}=\frac{(In\,sec\,ond\,orbital)}{(In\,first\,orbital)}=\frac{16}{1}\]You need to login to perform this action.
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