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If the radius of the $3^{r d}$ Bohr's orbit of hydrogen atom is $r_{3}$ and the radius of $4^{t h}$ Bohr's orbit is $r_{4}$ . Then :

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$r_{4} = \frac{9}{16} r_{3}$

$r_{4} = \frac{16}{9} r_{3}$

$r_{4} = \frac{3}{4} r_{3}$

$r_{4} = \frac{4}{3} r_{3}$

answer is B.

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Detailed Solution

$r = 0.529 \times \frac{n^{2}}{z}$

$r_{3} = 0.529 \times \frac{3^{2}}{1}$

$r_{4} = 0.529 \times \frac{4^{2}}{1}$

$\frac{r_{4}}{r_{3}} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$

$r_{4} = \frac{16 r_{3}}{9}$

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