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 $h=6.6\times 10^{-34}Js$)

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

Answer 1

Answer: (B)

Here, $m=10kg,$
$r=8000km=8000\times 10^{3}m=8\times 10^{6}m$,
$T=2h=7200s$
According to Bohr’s second postulate
$L=m\upsilon r=\frac{nh}{2\pi };$ As $v=\frac{2\pi r}{T}$
$\therefore m\left(\frac{2\pi r}{T}\right)r=\frac{nh}{2\pi }$
$n=\frac{{\left(2\pi r\right)}^{2}m}{Th}=\frac{{\left(2\times 3.14\times 8\times 10^6\right)}^2\times 10}{7200\times 6.6\times 10^{-34}}=5.3\times 10^{45}$