Class 12 Physics
CBSE and UP Board
Atoms
Chapter-12 Exercise 12.10
NCERT Solutions for Class 12 Physics Chapter 12 Question-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 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6.0 × 10²⁴ kg.)
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Rdius of the orbit of the Earth around the Sun, r= 1.5 x 1011 m
Orbital speed of the Earth, v = 3 x 104 m/s
Mass of the Earth, m = 6.0 x 10²4 kg
According to Bohr’s model, angular momentum is quantized and given as:
mvr = nh/2π
Where,
h = Planck’s constant = 6.62 x 10⁻34 Js
n = Quantum number
Therefore , n = mvr2π/h
= 2π x (6.0 x 10²4 ) x (3 x 104 ) x (1.5 x 1011) /(6.62 x 10⁻34)
= 25.61 x 10⁷³= 2.6 x 10⁷⁴
Hence, the quanta number that characterizes the Earth’ revolution is 2.6 x 1074.