In a head on collision between an alpha particle and gold nucleus, the minimum distance of approach is `4xx10^(-14)m`. Calculate the energy of of `alpha`-particle. Take Z=79 for gold.

Answer

To calculate the energy of the alpha particle in a head-on collision with a gold nucleus, we can use the formula for the minimum distance of approach in a Coulombic interaction. The formula is given by:

\[ R_0 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z \cdot e \cdot 2e}{E} \]

Where:
- \( R_0 \) is the minimum distance of approach (given as \( 4 \times 10^{-14} \, m \))
- \( Z \) is the atomic number of the gold nucleus (given as \( 79 \))
- \( e \) is the elementary charge (\( 1.6 \times 10^{-19} \, C \))
- \( E \) is the energy of the alpha particle (which we need to find)

### Step 1: Rearranging the formula to find the energy \( E \)

We can rearrange the formula to solve for \( E \):

\[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z \cdot e \cdot 2e}{R_0} \]

### Step 2: Substitute known values into the equation

Now, we will substitute the known values into the equation. We know:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \)
- \( Z = 79 \)
- \( e = 1.6 \times 10^{-19} \, C \)
- \( R_0 = 4 \times 10^{-14} \, m \)

Substituting these values into the equation:

\[ E = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{79 \cdot (1.6 \times 10^{-19}) \cdot 2(1.6 \times 10^{-19})}{4 \times 10^{-14}} \]

### Step 3: Calculate the value of \( E \)

Calculating the constants:

1. Calculate \( 4 \pi \epsilon_0 \):

\[ 4 \pi (8.85 \times 10^{-12}) \approx 1.112 \times 10^{-10} \, C^2/(N \cdot m^2) \]

2. Calculate \( 79 \cdot (1.6 \times 10^{-19})^2 \):

\[ 79 \cdot (1.6 \times 10^{-19})^2 = 79 \cdot 2.56 \times 10^{-38} \approx 2.0224 \times 10^{-36} \]

3. Now substitute back into the energy equation:

\[ E = \frac{1}{1.112 \times 10^{-10}} \cdot \frac{2.0224 \times 10^{-36}}{4 \times 10^{-14}} \]

Calculating the fraction:

\[ \frac{2.0224 \times 10^{-36}}{4 \times 10^{-14}} = 5.056 \times 10^{-23} \]

Now, substituting this back into the equation for \( E \):

\[ E = \frac{5.056 \times 10^{-23}}{1.112 \times 10^{-10}} \approx 4.54 \times 10^{-13} \, J \]

### Step 4: Convert energy from Joules to electron volts

To convert the energy from Joules to electron volts, we use the conversion factor \( 1 \, eV = 1.6 \times 10^{-19} \, J \):

\[ E \approx \frac{4.54 \times 10^{-13}}{1.6 \times 10^{-19}} \approx 2.84 \times 10^6 \, eV \]

### Step 5: Final Result

Thus, the energy of the alpha particle is approximately:

\[ E \approx 2.84 \, MeV \]