The average separation between the proton and the electron in a hydrogen atom in ground state is `5.3xx10^-11m. a. Calculate the Coulomb force between tehm at this separation. b. When the atom goes into its first excited state the average separation between the roton and the electron increases to four times its value in the ground state.What is th Coulomb force in this state?

Explanation

To solve the problem, we will break it down into two parts as described in the question.

### Part A: Calculate the Coulomb Force between the Proton and Electron in the Ground State

1. **Identify the Given Values:**
- Average separation (distance) between the proton and electron in the ground state, \( r_1 = 5.3 \times 10^{-11} \, \text{m} \)
- Charge of the proton, \( q_1 = +1.6 \times 10^{-19} \, \text{C} \)
- Charge of the electron, \( q_2 = -1.6 \times 10^{-19} \, \text{C} \)
- Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

2. **Use Coulomb's Law:**
The formula for the Coulomb force \( F \) between two charges is given by:
\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]

3. **Substitute the Values:**
\[ F = 9 \times 10^9 \frac{(1.6 \times 10^{-19})^2}{(5.3 \times 10^{-11})^2} \]

4. **Calculate the Force:**
- First, calculate \( (1.6 \times 10^{-19})^2 \):
\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \]
- Next, calculate \( (5.3 \times 10^{-11})^2 \):
\[ (5.3 \times 10^{-11})^2 = 2.809 \times 10^{-21} \]
- Now substitute these values into the force equation:
\[ F = 9 \times 10^9 \frac{2.56 \times 10^{-38}}{2.809 \times 10^{-21}} \approx 8.2 \times 10^{-8} \, \text{N} \]

### Part B: Calculate the Coulomb Force in the First Excited State

1. **Determine the New Separation:**
- The average separation in the first excited state is four times that in the ground state:
\[ r_2 = 4 \times r_1 = 4 \times (5.3 \times 10^{-11}) = 2.12 \times 10^{-10} \, \text{m} \]

2. **Use Coulomb's Law Again:**
\[ F_2 = k \frac{|q_1 \cdot q_2|}{r_2^2} \]

3. **Substitute the New Values:**
\[ F_2 = 9 \times 10^9 \frac{(1.6 \times 10^{-19})^2}{(2.12 \times 10^{-10})^2} \]

4. **Calculate the New Force:**
- Calculate \( (2.12 \times 10^{-10})^2 \):
\[ (2.12 \times 10^{-10})^2 = 4.4944 \times 10^{-20} \]
- Substitute into the force equation:
\[ F_2 = 9 \times 10^9 \frac{2.56 \times 10^{-38}}{4.4944 \times 10^{-20}} \approx 5.125 \times 10^{-9} \, \text{N} \]

### Summary of Results:
- **Part A:** The Coulomb force in the ground state is approximately \( 8.2 \times 10^{-8} \, \text{N} \).
- **Part B:** The Coulomb force in the first excited state is approximately \( 5.125 \times 10^{-9} \, \text{N} \).