# Q 12.10  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   $1.5 \times 10^{11}m$ m with orbital speed $3 \times 10^{4}m/s$  (Mass of earth = $6.0 \times 10^{24}kg$.)

As per the Bohr's model, the angular of the Earth will be quantized and will be a multiple of $\frac{h}{2 \pi}$

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

n = 2.56$\times$1074

Therefore the quantum number that characterises the earth’s revolution around the sun in an orbit of radius   $1.5 \times 10^{11}m$ m with an orbital speed $3 \times 10^{4}m/s$

is 2.56$\times$1074

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