If the radius of the second Bohr orbit of hydrogen atom is $r_{2}$ , the radius of the third Bohr orbit will be

$\frac{4}{9} r_{2}$

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$4 r_{2}$

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$\frac{9}{4} r_{2}$

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$9 r_{2}$

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Solution

The correct option is C
$\frac{9}{4} r_{2}$

According to Bohr's postulates,

$r$ = $\frac{n^{2} h^{2}}{4 π^{2} m Z e^{2}}$

Now, $r_{2}$ = $r_{1} \times n_{2} = r_{1} \times (2)^{2} = 4 r_{1}$

$r_{3} = r_{1} \times n_{2} = r_{1} \times (3)^{2} = 9 r_{1}$

& here, $r_{2} = \frac{(2)^{2} h^{2}}{4 π^{2} m Z e^{2}}$

& $r_{3} = \frac{(3)^{2} h^{2}}{4 π^{2} m Z e^{2}}$

Therefore,

$\frac{r_{2}}{r_{3}} = \frac{(2)^{2}}{(3)^{2}}$

$\Rightarrow \frac{r_{2}}{r_{3}} = \frac{4}{9}$

$\Rightarrow r_{3} = \frac{9}{4} r_{2}$

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