A) 2:1
B) 4:1
C) 8:1
D) 16:1
Correct Answer: D
Solution :
[d] First excited state \[{{n}_{1}}=2\] Ground state \[{{n}_{2}}=1\] \[{{r}_{n}}=\frac{{{n}^{2}}}{Z}{{r}_{{}^\circ }}\]general expression \[{{r}_{{{n}_{1}}}}=\frac{{{(2)}^{2}}}{1}4{{r}_{{}^\circ }}Are{{a}_{{{n}_{1}}}}=\pi \,{{(4{{r}_{{}^\circ }})}^{2}}=16\pi r_{{}^\circ }^{2}\] \[{{r}_{{{n}_{1}}}}=\frac{{{(1)}^{2}}{{r}_{{}^\circ }}}{1}{{r}_{{}^\circ }}\] \[Are{{a}_{{{n}_{2}}}}=\pi r_{{}^\circ }^{2}=\pi r_{{}^\circ }^{2}\] \[\frac{{{A}_{({{n}_{1}})}}}{{{A}_{({{n}_{2}})}}}=\frac{16}{1}\]
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