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 \( = 6.0 \times 10^{24} \, \text{kg}\)).
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Bohr’s model for the quantization of angular momentum helps determine the quantum number for large systems like the Earth by using the formula \( m v r = n \hbar \).
Step 1: Bohr's model posits that the Earth's orbital angular momentum is quantized, expressed as:
\[m v r = n \hbar\]
Where:
- \( m \) denotes the Earth's mass.
- \( v \) represents the orbital velocity.
- \( r \) is the orbital radius.
- \( n \) 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: The equation can be rearranged to solve for \( n \):
\[n = \frac{m v r}{\hbar}\]
Step 3: Inputting the provided values:
- \( m = 6.0 \times 10^{24} \, \text{kg} \)
- \( v = 30 \, \text{km/s} = 3.0 \times 10^4 \, \text{m/s} \)
- \( r = 1.5 \times 10^{11} \, \text{m} \)
- \( \hbar = 1.055 \times 10^{-34} \, \text{J s} \)
The calculation proceeds as:
\[n = \frac{(6.0 \times 10^{24}) (3.0 \times 10^4) (1.5 \times 10^{11})}{1.055 \times 10^{-34}}\]
\[n = \frac{2.7 \times 10^{40}}{1.055 \times 10^{-34}} = 2.56 \times 10^{74}\]
Consequently, the quantum number \( n \) is approximately \( 2.56 \times 10^{74} \).
