Question

A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. 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

• A. $5.5 \times 10^{40}$
• B. $5.3 \times 10^{45}$
• C. $7.8 \times 10^{48}$
• D. $7.8 \times 10^{50}$

Answer 1

Answer: (B)

$5.3 \times 10^{45}$

Answer 2

Here m = 10 kg $r_{n} = 8 \times 10^{6}m$

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

Velocity of nth orbit $v_{n} = \frac{2\pi r_{n}}{T}$ and from $mv_{n}r_{n} = \frac{nh}{2\pi}$

$n = \frac{2\pi}{h} \times m \times \frac{2\pi r_{n}}{T} \times r_{n}$

$= (2\pi r_{n})^{2} \times \frac{m}{T \times h} = \frac{(2\pi \times 8 \times 10^{6})^{2} \times 10}{7200 \times 6.64 \times 10^{- 34}}$

$= 5.3 \times 10^{45}$