A) 16 : 1
B) 1 : 16
C) 8 : 1
D) 1 : 8
Correct Answer: A
Solution :
\[r=\frac{0.529\,{{n}^{2}}}{Z}\] \[\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{(2)}^{2}}}{{{(1)}^{2}}}=\frac{4}{1}\] Area of orbital\[=\pi {{r}^{2}}\] \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{\pi r_{2}^{2}}{\pi r_{1}^{2}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}={{\left( \frac{4}{1} \right)}^{2}}=\frac{16}{1}\]
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