Question 8 of 8intermediate🔧 ApplyNumerical3 marks

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 1.5×101.5 \times 10¹¹ m with orbital speed 3×103 \times 10⁴ m/s. (Mass of earth = 6.0×106.0 \times 10²⁴ kg.)

Correct Answer

The quantum number n = 2.56×102.56 \times 10⁷⁴

Exercise: EXERCISES | Q: 12.9 | (Page 16)
For More Understanding

Explanation

According to Bohr's model, the angular momentum of an orbiting body is quantised and given by the relation mvr = nh/2π. This quantum condition was proposed by Bohr for the angular momentum of the electron. Applying this same principle to the Earth-Sun system allows us to calculate the quantum number characterising Earth's revolution. The extremely large value of n (approximately 10⁷⁴) indicates that the Earth's orbit corresponds to a very high quantum state, which is consistent with macroscopic objects having extremely large quantum numbers.

Solution Steps

  1. Step 1: Write Bohr's quantisation condition for angular momentum: mvr = nh/2π

  2. Step 2: Rearrange to find quantum number n: n = 2πmvr/h

  3. Step 3: Substitute the given values: m = 6.0×106.0 \times 10²⁴ kg, v = 3×103 \times 10⁴ m/s, r = 1.5×101.5 \times 10¹¹ m

  4. Step 4: Calculate angular momentum: mvr = (6.0×106.0 \times 10²⁴) × (3×103 \times 10⁴) × (1.5×101.5 \times 10¹¹) = 2.7×102.7 \times 10⁴⁰ kg·m²/s

  5. Step 5: Calculate n: n = (2π × 2.7×102.7 \times 10⁴⁰)/(6.626×106.626 \times 10⁻³⁴) = 2.56×102.56 \times 10⁷⁴

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