12.10 In accordance with the Bohr’s model, find the quantum number that characterizes 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.)

Hint: In Bohr's model of atom, the angular momentum is quantized.
Step 1: Find the relation between the angular momentum and the quantum number.
The radius of the orbit of the Earth around the Sun, r=1.5x${10}^{11}$m
The orbital speed of the Earth, v = 3 x 104 m/s
Mass of the Earth, m=6.0 x ${10}^{24}$ kg
According to Bohr's model, angular momentum is quantized and given as:
$\mathrm{mvr}=\frac{\mathrm{nh}}{2\mathrm{\pi }}$
where,
h = Planck's constant =
n = Quantum number
Step 2: Find the quantum number corresponding to the earth revolution around the sun.
$\therefore \mathrm{n}=\frac{\mathrm{\left(mvr\right)}2\mathrm{\pi }}{\mathrm{h}}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{\pi }×6×{10}^{24}×3×{10}^{4}×1.5×{10}^{11}}{6.62×{10}^{-34}}\phantom{\rule{0ex}{0ex}}=25.61×{10}^{73}=2.6×{10}^{74}$
Hence, the quantum number that characterizes the Earth's revolution is $2.6×{10}^{74}$.