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Question : 10 of 18
Marks: +1, -0
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 × 101110^{11} m with orbital speed 3 × 104 m s−110^4\,\text{m s}^{-1}.
(Mass of earth = 6.0 × 102410^{24} kg)
Solution:  
According to Bohr’s quantization condition of angular momentum,
Angular momentum of the earth around the sun,
mvr = nh2π\frac{nh}{2\pi}
∴ n = 2πmrh\frac{2\pi mr}{h}
=
2×3.14×6.0×1024×1.5×1011×3×1046.6×10−34\frac{2\times3.14\times6.0\times10^{24}\times1.5\times10^{11}\times3\times10^4}{6.6\times10^{-34}}
= 2.57 × 107410^{74}
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