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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)
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Solution
Radius of the orbit of the earth around the sun, r = 1.5 × 1011 m
Orbital speed of the earth, v = 3 × 104 m/s
Mass of the earth, m = 6.0 × 1024 kg
According to Bohr’s model, angular momentum is quantised and given as:
mvr = `(nh)/(2pi)`
Where,
h = Planck’s constant = 6.62 × 10−34 Js
n = Quantum number
∴ n = `(mvr 2pi)/h`
= `(2pi xx 6 xx 10^24xx 3 xx 10^4 xx1.5 xx 10^(11))/(6.62 xx 10^(-34))`
= 25.61 × 1073
= 2.6 × 1074
Hence, the quantum number that characterises the earth’s revolution is 2.6 × 1074.
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