Question

The Earth revolves around the Sun in an orbit of radius \(1.5 \times 10^{11} \, \text{m}\) with orbital speed \(30 \, \text{km/s}\). Find the quantum number that characterizes its revolution using Bohr’s model in this case (mass of Earth \( = 6.0 \times 10^{24} \, \text{kg}\)).

Hint:

Bohr’s model for the quantization of angular momentum helps determine the quantum number for large systems like the Earth by using the formula \( m v r = n \hbar \).

Answer 1

Step 1: According to Bohr’s model, the angular momentum of the Earth in orbit is quantized and is given by:
\[ m v r = n \hbar \] where:
- \( m \) is the mass of the Earth,
- \( v \) is the orbital speed,
- \( r \) is the radius of the orbit,
- \( n \) is the quantum number,
- \( \hbar = \frac{h}{2\pi} \) is the reduced Planck's constant (\( h = 6.626 \times 10^{-34} \, \text{J s} \)).
Step 2: Rearranging the equation for \( n \):
\[ n = \frac{m v r}{\hbar} \] Step 3: Substituting the given values:
- \( m = 6.0 \times 10^{24} \, \text{kg} \),
- \( v = 30 \, \text{km/s} = 3.0 \times 10^4 \, \text{m/s} \),
- \( r = 1.5 \times 10^{11} \, \text{m} \),
- \( \hbar = 1.055 \times 10^{-34} \, \text{J s} \).
\[ n = \frac{(6.0 \times 10^{24}) (3.0 \times 10^4) (1.5 \times 10^{11})}{1.055 \times 10^{-34}} \] \[ n = \frac{2.7 \times 10^{40}}{1.055 \times 10^{-34}} = 2.56 \times 10^{74} \] Thus, the quantum number \( n \) is approximately \( 2.56 \times 10^{74} \).