**Question:**

In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.)

**Solution:**

Radius of the orbit of the Earth around the Sun, *r* = 1.5 × 1011 m

Orbital speed of the Earth, *ν* = 3 × 104 m/s

Mass of the Earth, *m* = 6.0 × 1024 kg

According to Bohr’s model, angular momentum is quantized and given as:

$m v r=\frac{n h}{2 \pi}$

Where,

*h* = Planck’s constant = 6.62 × 10−34 Js

*n* = Quantum number

$\therefore n=\frac{m v r 2 \pi}{h}$

$=\frac{2 \pi \times 6 \times 10^{24} \times 3 \times 10^{4} \times 1.5 \times 10^{11}}{6.62 \times 10^{-34}}$

$=25.61 \times 10^{73}=2.6 \times 10^{74}$

Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × 1074.