Ncert Exercises & Solutions

12.10 In accordance with the Bohr’s model, find the quantum number that characterizes 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.)

Hint: In Bohr's model of atom, the angular momentum is quantized.
Step 1: Find the relation between the angular momentum and the quantum number.

The radius of the orbit of the Earth around the Sun, r=1.5x

10^{11}

m
The orbital speed of the Earth, v = 3 x 10⁴ m/s
Mass of the Earth, m=6.0 x

10^{24}

kg
According to Bohr's model, angular momentum is quantized and given as:

mvr = \frac{nh}{2 π}

where,

h = Planck's constant =

6.62 \times 10^{- 34} J-s

n = Quantum number
Step 2: Find the quantum number corresponding to the earth revolution around the sun.

∴ n = \frac{(mvr) 2 π}{h}

= \frac{2 π \times 6 \times 10^{24} \times 3 \times 10^{4} \times 1.5 \times 10^{11}}{6.62 \times 10^{- 34}}

= 25.61 \times 10^{73} = 2.6 \times 10^{74}

Hence, the quantum number that characterizes the Earth's revolution is

2.6 \times 10^{74}

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