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 the satellite, then the quantum number of the orbit of the satellite is (Take $h = 6.6 × 10^{-34}\, J\, s$)

• A. $5.3 × 10^{47}$
• B. $5.3 × 10^{45}$
• C. $5.3 × 10^{43}$
• D. $5.3 × 10^{41}$

Answer 1

Answer: (B)

Here, $m = 10\, kg,$
$r = 8000 \,km = 8000 × 10^3\, m = 8 × 10^6\, m$,
$T = 2 \,h = 7200\, s$
According to Bohr’s second postulate
$L = m\upsilon r = \frac{nh}{2\pi};$ As $v = \frac{2\pi r}{T}$
$\therefore \quad m\left(\frac{2\pi r}{T}\right)r = \frac{nh}{2\pi }$
$n = \frac{\left(2\pi r\right)^{2} m}{Th} = \frac{\left(2\times3.14\times 8\times 10^{6}\right)^{2}\times 10}{7200\times 6.6\times 10^{-34}} = 5.3\times 10^{45}$