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.)
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.
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