In a Rutherford scattering experiment, assume that an incident alpha particle (radius 1.80 fm) is headed directly toward a target gold nucleus (radius 6.23 fm). What energy must the alpha particle have to just barely "touch" the gold nucleus?

Answer

To solve the problem of determining the energy that an alpha particle must have to just barely touch a gold nucleus, we will use the concept of electrostatic potential energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Situation
An alpha particle is a helium nucleus consisting of 2 protons and 2 neutrons, and it is positively charged. The gold nucleus has an atomic number of 79, meaning it has 79 protons and is also positively charged. When the alpha particle approaches the gold nucleus, they will repel each other due to their positive charges.

### Step 2: Determine the Charges
- The charge of the alpha particle (Q1) is:
\[ Q_1 = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]
- The charge of the gold nucleus (Q2) is:
\[ Q_2 = 79e = 79 \times 1.6 \times 10^{-19} \, \text{C} = 1.264 \times 10^{-17} \, \text{C} \]

### Step 3: Calculate the Distance Between the Centers
The distance \( R \) between the centers of the alpha particle and the gold nucleus is the sum of their radii:
\[ R = R_{\text{alpha}} + R_{\text{gold}} = 1.80 \, \text{fm} + 6.23 \, \text{fm} = 8.03 \, \text{fm} = 8.03 \times 10^{-15} \, \text{m} \]

### Step 4: Use the Formula for Electrostatic Potential Energy
The electrostatic potential energy (U) at the distance \( R \) is given by:
\[ U = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{R} \]
Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 5: Substitute the Values
Substituting the values into the formula:
\[ U = \frac{1}{4 \pi (8.85 \times 10^{-12})} \frac{(3.2 \times 10^{-19})(1.264 \times 10^{-17})}{8.03 \times 10^{-15}} \]

Calculating \( \frac{1}{4 \pi \epsilon_0} \):
\[ \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \]

Now substituting:
\[ U \approx 9 \times 10^9 \times \frac{(3.2 \times 10^{-19})(1.264 \times 10^{-17})}{8.03 \times 10^{-15}} \]

### Step 6: Calculate the Result
Calculating the numerator:
\[ (3.2 \times 10^{-19})(1.264 \times 10^{-17}) \approx 4.038 \times 10^{-36} \]
Now calculating \( U \):
\[ U \approx 9 \times 10^9 \times \frac{4.038 \times 10^{-36}}{8.03 \times 10^{-15}} \approx 4.52 \times 10^{-11} \, \text{J} \]

### Step 7: Convert to Electron Volts
To convert joules to electron volts:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ U \approx \frac{4.52 \times 10^{-11}}{1.6 \times 10^{-19}} \approx 282.5 \, \text{MeV} \]

### Final Answer
The energy that the alpha particle must have to just barely touch the gold nucleus is approximately **282.5 MeV**.

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