A) 1.08 A
B) 2.12 A
C) 5.24 A
D) 6.36 A
Correct Answer: B
Solution :
We know that Bohr's radium \[=\frac{{{\text{(Number}\,\text{of}\,\text{excityed}\,\text{state)}}^{\text{2}}}}{\text{Atomic}\,\text{number}\,\text{of}\,\text{hydrogen}\,\text{atome}}\,\times \,\text{Radius}\,\text{of}\,\text{hydrogen}\,\text{atom}\] \[=\frac{{{(2)}^{2}}}{1\times 0.530}=4\times 0.530=2.120\,\overset{\text{o}}{\mathop{\text{A}}}\,.\] Hence, the correct option is (b).
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