Class 12 Physics

CBSE and UP Board

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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 10

^{11}mOrbital speed of the Earth, v = 3 x 10

^{4}m/sMass of the Earth, m = 6.0 x 10²

^{4}kgAccording to Bohr’s model, angular momentum is quantized and given as:

mvr = nh/2π

Where,

h = Planck’s constant = 6.62 x 10⁻

^{34}Jsn = Quantum number

Therefore , n = mvr2π/h

= 2π x (6.0 x 10²

^{4}) x (3 x 10^{4}) x (1.5 x 10^{11}) /(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 10

^{74}.