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The radius of electron’s second stationary orbit in Bohr’s atom is R. the radius of 3rd orbit will be

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a

3 R

b

9R

c

2.25 R

d

$\frac{R}{3}$

answer is C.

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Detailed Solution

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Bohr’s atom = hydrogen atom
$\begin{array}{l} r_{n} = \frac{n^{2} h^{2} \in_{0}}{{πmze}^{2}} \\ z = 1 \Rightarrow r_{n} \propto n^{2} \\ \frac{r_{2}}{r_{3}} = \frac{(2)^{2}}{(3)^{2}} \\ \frac{R}{r_{3}} = \frac{4}{9} \Rightarrow r_{3} = \frac{9 R}{4} = 2.25 R \end{array}$

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