The distance of closest approach of an alpha particle to the nucleus of gold is `2.73 xx 10^(-14)` m. Calculate the energy of the alpha particle.

Answer

To calculate the energy of the alpha particle based on the distance of closest approach to the nucleus of gold, we can follow these steps:

### Step 1: Understand the Concept
The distance of closest approach (R₀) is the point at which the kinetic energy of the alpha particle is completely converted into potential energy due to electrostatic repulsion between the positively charged alpha particle and the positively charged gold nucleus.

### Step 2: Identify Given Values
- Distance of closest approach, \( R_0 = 2.73 \times 10^{-14} \, \text{m} \)
- Charge of the alpha particle, \( Q_1 = 2e \) (where \( e = 1.6 \times 10^{-19} \, \text{C} \))
- Charge of the gold nucleus, \( Q_2 = Ze = 79e \) (for gold, \( Z = 79 \))
- Coulomb's constant, \( K = \frac{1}{4\pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

### Step 3: Write the Expression for Potential Energy
The potential energy (U) at the distance of closest approach is given by:
\[ U = \frac{K \cdot Q_1 \cdot Q_2}{R_0} \]
Substituting the values:
\[ U = \frac{(9 \times 10^9) \cdot (2e) \cdot (79e)}{R_0} \]

### Step 4: Substitute the Values
Substituting \( e = 1.6 \times 10^{-19} \, \text{C} \) and \( R_0 = 2.73 \times 10^{-14} \, \text{m} \):
\[ U = \frac{(9 \times 10^9) \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (79 \cdot 1.6 \times 10^{-19})}{2.73 \times 10^{-14}} \]

### Step 5: Calculate the Potential Energy
Calculating the numerator:
\[ U = \frac{(9 \times 10^9) \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot (79 \cdot 1.6 \times 10^{-19})}{2.73 \times 10^{-14}} \]
\[ = \frac{(9 \times 10^9) \cdot (2 \cdot 79 \cdot (1.6)^2 \times 10^{-38})}{2.73 \times 10^{-14}} \]
Calculating \( 2 \cdot 79 \cdot (1.6)^2 \):
\[ = 2 \cdot 79 \cdot 2.56 = 404.48 \]
Now substituting back:
\[ U = \frac{(9 \times 10^9) \cdot (404.48 \times 10^{-38})}{2.73 \times 10^{-14}} \]
Calculating the above:
\[ U = \frac{(9 \cdot 404.48) \times 10^{-29}}{2.73 \times 10^{-14}} \approx \frac{3640.32 \times 10^{-29}}{2.73 \times 10^{-14}} \approx 1.33 \times 10^{-15} \, \text{J} \]

### Step 6: Convert to Electron Volts
To convert joules to electron volts:
\[ E = \frac{U}{e} = \frac{1.33 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 8.33 \times 10^{3} \, \text{eV} = 8.33 \, \text{MeV} \]

### Final Answer
The energy of the alpha particle is approximately \( 8.33 \, \text{MeV} \).

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