A) \[{{10}^{-8}}\,\,m\]
B) \[{{10}^{-9}}\,\,m\]
C) \[{{10}^{-11}}\,\,m\]
D) \[{{10}^{-10}}\,\,m\]
Correct Answer: D
Solution :
Radius of \[{{n}^{th}}\] orbit \[{{r}_{n}}=0.53\frac{{{n}^{2}}}{Z}\overset{o}{\mathop{A}}\,\] Here, \[n=1\] and for hydrogen atom \[Z=1\] \[\therefore \,\,{{r}_{1}}=0.53\frac{{{(1)}^{2}}}{1}\overset{o}{\mathop{A}}\,=0.53\times {{10}^{-10}}m=53\,pm\]
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