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A $10 k g$ satellite circles earth once every $2 h$ in an orbit having a radius of $8000 k m$ . Assuming that Bohr's angular momentum postulate applies to a satellite just as it does to an electron in the hydrogen atom, then the quantum number of the orbit of satellite is:

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$7.8 \times 10^{50}$

$5.3 \times 10^{40}$

$5.3 \times 10^{45}$

$7.8 \times 10^{48}$

answer is B.

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Detailed Solution

Here, $m = 10 k g, r_{n} = 8 \times 10^{6} m$

$T = 2 \times 60 \times 60 = 7200 s$

Velocity of $n^{t h}$ orbit, $v_{n} = \frac{2 {πr}_{n}}{T}$ and from $m v_{n} r_{n} = \frac{n h}{2 π}$

$n = \frac{2 π}{h} \times m \times \frac{2 {πr}_{n}}{T} \times r_{n} = {(2 {πr}_{n})}^{2} \times \frac{m}{T \times h} = \frac{{(2 π \times 8 \times 10^{6})}^{2} \times 10}{7200 \times 6.626 \times 10^{- 34}}$

$= 5.3 \times 10^{45}$

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