In accordance with the Bohr’s mode | Class 12 Physics Chapter Atoms, Atoms NCERT Solutions

Welcome to the NCERT Solutions for Class 12 Physics - Chapter Atoms. This page offers a step-by-step solution to the specific question from Exercise 1, Question 10: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

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.)

Answer:

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....