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.)
Radius of the orbit of the Earth around the Sun, r = 1.5 × 1011 m
Orbital speed of the Earth, ν = 3 × 104 m/s
Mass of the Earth, m = 6.0 × 1024 kg
According to Bohr’s model, angular momentum is quantized and given as:
mvr = nh/2π
Where,
h = Planck’s constant = 6.62 × 10−34 Js
n = Quantum number
∴ n = mvr2π/h
= (2πx6x1024x3x104x1.5x1011)/(6.62x10-34)
= 25.61x1073 = 2.6 x 1074
Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × 1074 .
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 10: In accordance with the Bohr’s model, find the quantum number that characterises the earth&rsqu....
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