A) 0.13
B) 1.06
C) 4.77
D) 2.12
Correct Answer: D
Solution :
\[r\propto \,n_{1}^{2}\,{{Z}^{2}}\] where n = number of orbit Z = Atomic number \[\because \] \[{{r}_{1}}\propto n_{1}^{2}\] \[{{r}_{2}}\propto \,n_{2}^{2}\,(Z=1\,for\,H-atom)\] \[So,\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{n_{1}^{2}}{n_{2}^{2}}\] \[\frac{0.530}{{{r}_{2}}}=\frac{{{1}^{2}}}{{{2}^{2}}}\] \[\therefore {{r}_{2}}=0.530\times 4=2.120\,{\AA}\]
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