Calculate the distance of closest approach for a proton with energy 3 MeV approaching a gold nucleus.

Answer

To calculate the distance of closest approach for a proton with energy 3 MeV approaching a gold nucleus, we can follow these steps:

### Step 1: Identify the Charges
The charge of the gold nucleus (Z = 79) can be calculated as:
\[ Q_{\text{nucleus}} = Z \cdot e = 79 \cdot e \]
where \( e \) is the charge of a proton (or electron), approximately \( 1.6 \times 10^{-19} \, \text{C} \).

### Step 2: Initial Kinetic Energy
The initial kinetic energy (KE) of the proton is given as:
\[ KE = 3 \, \text{MeV} = 3 \times 10^6 \, \text{eV} \]
To convert this to joules, we use the conversion \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE = 3 \times 10^6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 4.8 \times 10^{-13} \, \text{J} \]

### Step 3: Potential Energy at Closest Approach
At the distance of closest approach \( r \), the kinetic energy will be converted into potential energy (PE). The potential energy between two point charges is given by:
\[ PE = \frac{k \cdot Q_1 \cdot Q_2}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( Q_1 = e \) (charge of proton), and \( Q_2 = 79e \) (charge of gold nucleus). Thus,
\[ PE = \frac{k \cdot e \cdot 79e}{r} = \frac{79k e^2}{r} \]

### Step 4: Conservation of Energy
By conservation of energy, the initial kinetic energy is equal to the potential energy at the closest approach:
\[ KE = PE \implies 4.8 \times 10^{-13} \, \text{J} = \frac{79 \cdot k \cdot e^2}{r} \]

### Step 5: Solve for r
Rearranging the equation to solve for \( r \):
\[ r = \frac{79 \cdot k \cdot e^2}{4.8 \times 10^{-13}} \]

### Step 6: Substitute Values
Substituting \( k \) and \( e \):
\[ r = \frac{79 \cdot (9 \times 10^9) \cdot (1.6 \times 10^{-19})^2}{4.8 \times 10^{-13}} \]

Calculating \( e^2 \):
\[ e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]

Now substituting this back into the equation:
\[ r = \frac{79 \cdot (9 \times 10^9) \cdot (2.56 \times 10^{-38})}{4.8 \times 10^{-13}} \]

### Step 7: Final Calculation
Calculating the numerator:
\[ 79 \cdot 9 \cdot 2.56 \approx 1825.44 \times 10^{-29} \, \text{N m}^2 \]
Now divide by \( 4.8 \times 10^{-13} \):
\[ r \approx \frac{1825.44 \times 10^{-29}}{4.8 \times 10^{-13}} \approx 3.80 \times 10^{-15} \, \text{m} \]

### Step 8: Convert to Fermi
Since \( 1 \, \text{Fermi} = 10^{-15} \, \text{m} \):
\[ r \approx 38.0 \, \text{Fermi} \]

### Final Answer
The distance of closest approach for a proton with energy 3 MeV approaching a gold nucleus is approximately:
\[ \boxed{38.0 \, \text{Fermi}} \]