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Q12.9:
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 \times 10^{11})) m with orbital speed ((3 \times 10^{4})) m/s. (Mass of earth = ((6.0 \times 10^{24})) kg.)

Solution :

Step 1: Given Variables and Fundamental Constants

  • Mass of Earth ($M$): $6.0 \times 10^{24}\ \text{kg}$
  • Radius of orbit ($r$): $1.5 \times 10^{11}\ \text{m}$
  • Orbital speed ($v$): $3.0 \times 10^{4}\ \text{m/s}$
  • Planck's constant ($h$): $6.63 \times 10^{-34}\ \text{J}\cdot\text{s}$

Step 2: Theoretical Framework

According to the Bohr quantization postulate, the angular momentum ($L$) of a revolving body is quantized and must be an integral multiple of the reduced Planck constant ($\hbar = \frac{h}{2\pi}$). The relationship is expressed as:

$L = mvr = n\frac{h}{2\pi}$

[Per Bohr’s Postulate of Angular Momentum Quantization]

Step 3: Derivation of the Quantum Number ($n$)

To find the quantum number $n$, we rearrange the equation to solve for $n$:

$n = \frac{2\pi mvr}{h}$

Step 4: Substitution and Numerical Computation

Substitute the given values into the equation:

$n = \frac{2 \times 3.14159 \times (6.0 \times 10^{24}\ \text{kg}) \times (3.0 \times 10^{4}\ \text{m/s}) \times (1.5 \times 10^{11}\ \text{m})}{6.63 \times 10^{-34}\ \text{J}\cdot\text{s}}$

Calculating the numerator:

$\text{Numerator} = 2 \times 3.14159 \times 6.0 \times 3.0 \times 1.5 \times 10^{(24 + 4 + 11)}$

$\text{Numerator} \approx 169.646 \times 10^{39} = 1.696 \times 10^{41}$

Calculating $n$:

$n = \frac{1.696 \times 10^{41}}{6.63 \times 10^{-34}} \approx 0.2558 \times 10^{75} \approx 2.56 \times 10^{74}$

Step 5: Visualization of Orbital Dynamics

r Sun Earth

Final Solution: The quantum number characterizing the Earth's revolution is $n \approx 2.56 \times 10^{74}$. This extremely large value indicates that for macroscopic systems, the quantum nature of angular momentum is effectively continuous, aligning with the correspondence principle.


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