# 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 - Physics

Numerical

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)

#### 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 quantized 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 quanta number that characterizes the Earth’s revolution is 2.6 × 1074.

Concept: Bohr’s Model for Hydrogen Atom
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 12 Atoms
Exercise | Q 12.10 | Page 436
NCERT Class 12 Physics Textbook
Chapter 12 Atoms
Exercise | Q 10 | Page 436

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