Consider an `alpha-` particle just in contact with a `._(92)U^(238)` nucleus. Calculate the coulombic repulsion energy (i.e., the height of the coulombic barrier between `U^(238)` and alpha particle) assuming that the distance between them is equal to the sum of their radii

Options

A. `23.851 xx 10^(4) eV`
B. `26.147738 xx 10^(4) eV`
C. `25.3522 xx 10^(4) eV`
D. `20.2254 xx 10^(4) eV`

Explanation

To calculate the Coulombic repulsion energy between an alpha particle and a Uranium-238 nucleus, we can follow these steps:

### Step 1: Determine the radii of the Uranium-238 nucleus and the alpha particle.

The radius \( R \) of a nucleus can be estimated using the formula:
\[ R = R_0 A^{1/3} \]
where \( R_0 \) is approximately \( 1.3 \times 10^{-15} \) m, and \( A \) is the mass number of the nucleus.

For Uranium-238 (\( _{92}^{238}U \)):
\[ R_U = 1.3 \times 10^{-15} \times (238)^{1/3} \]

Calculating \( (238)^{1/3} \):
\[ (238)^{1/3} \approx 6.2 \quad \text{(approximately)} \]
Thus,
\[ R_U \approx 1.3 \times 10^{-15} \times 6.2 \approx 8.06 \times 10^{-15} \text{ m} \]

For the alpha particle (\( _{2}^{4}He \)):
\[ R_{\alpha} = 1.3 \times 10^{-15} \times (4)^{1/3} \]

Calculating \( (4)^{1/3} \):
\[ (4)^{1/3} \approx 1.587 \quad \text{(approximately)} \]
Thus,
\[ R_{\alpha} \approx 1.3 \times 10^{-15} \times 1.587 \approx 2.06 \times 10^{-15} \text{ m} \]

### Step 2: Calculate the total distance between the alpha particle and the Uranium nucleus.

The total distance \( R \) between the alpha particle and the Uranium nucleus when they are just in contact is the sum of their radii:
\[ R = R_U + R_{\alpha} \approx 8.06 \times 10^{-15} + 2.06 \times 10^{-15} \approx 1.01 \times 10^{-14} \text{ m} \]

### Step 3: Calculate the Coulombic repulsion energy.

The Coulombic repulsion energy \( U \) between two charged particles is given by:
\[ U = \frac{k \cdot Q_1 \cdot Q_2}{R} \]
where:
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( Q_1 \) is the charge of the Uranium nucleus, which is \( +92e \) (where \( e \approx 1.6 \times 10^{-19} \, \text{C} \)),
- \( Q_2 \) is the charge of the alpha particle, which is \( +2e \).

Calculating the charges:
\[ Q_1 = 92 \times 1.6 \times 10^{-19} \approx 1.472 \times 10^{-17} \text{ C} \]
\[ Q_2 = 2 \times 1.6 \times 10^{-19} \approx 3.2 \times 10^{-19} \text{ C} \]

Now substituting into the formula:
\[ U = \frac{(8.99 \times 10^9) \cdot (1.472 \times 10^{-17}) \cdot (3.2 \times 10^{-19})}{1.01 \times 10^{-14}} \]

Calculating \( U \):
\[ U \approx \frac{(8.99 \times 10^9) \cdot (4.704 \times 10^{-36})}{1.01 \times 10^{-14}} \approx \frac{4.23 \times 10^{-26}}{1.01 \times 10^{-14}} \approx 4.18 \times 10^{-12} \text{ J} \]

### Final Result
The Coulombic repulsion energy (the height of the Coulombic barrier) between the Uranium-238 nucleus and the alpha particle is approximately:
\[ U \approx 4.18 \times 10^{-12} \text{ J} \]