In accordance with the Bohr's model, the quantum number that characterize the earth's revolution around the sun in an orbit of radius $1.5 \times 10^{11} m$ with orbital speed $3 \times 10^{4} m s^{- 1}$ is: (Mass of earth $= 6 \times 10^{24} k g$ )

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$5.98 \times 10^{86}$

$2.57 \times 10^{74}$

$2.57 \times 10^{38}$

$8.57 \times 10^{64}$

answer is D.

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Detailed Solution

Here, $r = 1.5 \times 10^{11} m; v = 3 \times 10^{4} m s^{- 1}$

$m = 6.0 \times 10^{24} k g$

According to Bohr's model

$m v r = \frac{n h}{2 π} \Rightarrow n = \frac{2 πmvr}{h}$

$\Rightarrow n = \frac{2 \times 22}{7} \times \frac{6 \times 10^{24} \times 3 \times 10^{4} \times 1.5 \times 10^{11}}{6.6 \times 10^{- 34}} = 2.57 \times 10^{74} .$

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