The earth revolves round the sun due to gravitatinal attraction. Suppose that the sun and the earth are point particle with their existing masses and that Bhor's quantization rule for angular monentum is valid in the case of gravitation (a) Calculate the minimum radius the earth can have for its orbit. (b) What is the value of the principle quantum number n for the present radius ? Mass of the earth = `6.0 xx 10^(24)` kg, mass of the sun = `2.0 xx 10^(30)` kg, earth-sun distance = `1.5 xx 10^(11)m`.

Explanation

To solve the problem, we will follow the steps outlined in the video transcript. We will calculate the minimum radius of the Earth's orbit around the Sun using Bohr's quantization rule for angular momentum and then find the principal quantum number for the current radius of the Earth's orbit.

### Step-by-Step Solution

#### Part (a): Calculate the Minimum Radius

1. **Bohr's Quantization Condition**: According to Bohr's model, the angular momentum \( L \) of a particle in orbit is quantized and given by:
\[ L = m v r = n \frac{h}{2\pi} \]
where:
- \( m \) = mass of the Earth (\( 6.0 \times 10^{24} \) kg)
- \( v \) = orbital speed of the Earth
- \( r \) = radius of the orbit
- \( n \) = principal quantum number
- \( h \) = Planck's constant (\( 6.63 \times 10^{-34} \) Js)

2. **Square the Equation**: Squaring both sides gives:
\[ m^2 v^2 r^2 = n^2 \frac{h^2}{4\pi^2} \]
(Let’s call this Equation 1)

3. **Centripetal Force and Gravitational Force**: The gravitational force provides the necessary centripetal force for the Earth's orbit:
\[ \frac{G M_{sun} m_{earth}}{r^2} = \frac{m_{earth} v^2}{r} \]
where:
- \( G \) = gravitational constant (\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \))
- \( M_{sun} \) = mass of the Sun (\( 2.0 \times 10^{30} \) kg)

4. **Solve for \( v^2 \)**: From the above equation, we can isolate \( v^2 \):
\[ v^2 = \frac{G M_{sun}}{r} \]
(Let’s call this Equation 2)

5. **Substitute \( v^2 \) into Equation 1**: Substitute Equation 2 into Equation 1:
\[ m^2 \left(\frac{G M_{sun}}{r}\right) r^2 = n^2 \frac{h^2}{4\pi^2} \]
Simplifying gives:
\[ m^2 G M_{sun} r = n^2 \frac{h^2}{4\pi^2} \]

6. **Solve for \( r \)**: Rearranging to find \( r \):
\[ r = \frac{n^2 h^2}{4\pi^2 m^2 G M_{sun}} \]

7. **Calculate Minimum Radius**: For the minimum radius, we take \( n = 1 \):
\[ r = \frac{1^2 \cdot (6.63 \times 10^{-34})^2}{4 \pi^2 (6.0 \times 10^{24})^2 (6.67 \times 10^{-11}) (2.0 \times 10^{30})} \]

8. **Plug in the Values**: Calculate the above expression to find \( r \):
\[ r \approx 2.29 \times 10^{-138} \, \text{m} \]

#### Part (b): Calculate the Principal Quantum Number \( n \)

1. **Using Equation 1**: Rearranging Equation 1 to find \( n \):
\[ n^2 = \frac{m^2 v^2 r^2}{\frac{h^2}{4\pi^2}} \]

2. **Substituting \( v^2 \)**: Substitute \( v^2 \) from Equation 2:
\[ n^2 = \frac{m^2 \left(\frac{G M_{sun}}{r}\right) r^2}{\frac{h^2}{4\pi^2}} \]
Simplifying gives:
\[ n^2 = \frac{4\pi^2 m^2 G M_{sun} r}{h^2} \]

3. **Plug in the Values**: Use the current radius of the Earth’s orbit \( r = 1.5 \times 10^{11} \, \text{m} \) to find \( n \):
\[ n^2 = \frac{4 \pi^2 (6.0 \times 10^{24})^2 (6.67 \times 10^{-11}) (2.0 \times 10^{30}) (1.5 \times 10^{11})}{(6.63 \times 10^{-34})^2} \]

4. **Calculate \( n \)**: After calculating, we find:
\[ n \approx 2.5 \times 10^{74} \]

### Final Answers
- (a) Minimum radius \( r \approx 2.29 \times 10^{-138} \, \text{m} \)
- (b) Principal quantum number \( n \approx 2.5 \times 10^{74} \)