Calculate the Coulomb barrier height for two Li nuclei that are fired at each other with the same initial kinetic energy K.

Answer

To calculate the Coulomb barrier height for two Lithium nuclei that are fired at each other with the same initial kinetic energy \( K \), we can follow these steps:

### Step 1: Understand the Coulomb Barrier
The Coulomb barrier is the potential energy barrier due to the electrostatic repulsion between two positively charged nuclei. When two nuclei approach each other, they experience a repulsive force due to their positive charges.

### Step 2: Set Up the Energy Conservation Equation
Initially, both Lithium nuclei have kinetic energy \( K \). As they approach each other, this kinetic energy is converted into potential energy at the point of closest approach (the Coulomb barrier). Thus, we can write:

\[ 2K = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{R} \]

where \( q_1 \) and \( q_2 \) are the charges of the Lithium nuclei, \( R \) is the distance of closest approach, and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Determine the Charge of Lithium Nuclei
For Lithium, which has an atomic number \( Z = 3 \), the charge of each nucleus is given by:

\[ q = Z \cdot e = 3 \cdot e \]

where \( e \) is the elementary charge (\( e \approx 1.602 \times 10^{-19} \) C).

### Step 4: Substitute the Charges into the Equation
Substituting \( q_1 \) and \( q_2 \) into the energy conservation equation gives:

\[ 2K = \frac{1}{4\pi \epsilon_0} \frac{(3e)(3e)}{R} \]

This simplifies to:

\[ 2K = \frac{9e^2}{4\pi \epsilon_0 R} \]

### Step 5: Solve for the Coulomb Barrier Height
Rearranging the equation to solve for the Coulomb barrier height \( V \):

\[ V = \frac{9e^2}{8\pi \epsilon_0 R} \]

### Step 6: Determine the Radius \( R \)
The radius \( R \) of a nucleus can be estimated using the formula:

\[ R = R_0 A^{1/3} \]

where \( R_0 \approx 1.2 \times 10^{-15} \) m and \( A \) is the mass number of the Lithium nucleus (for Lithium-7, \( A = 7 \)).

Calculating \( R \):

\[ R = 1.2 \times 10^{-15} \times 7^{1/3} \approx 1.2 \times 10^{-15} \times 1.913 = 2.2956 \times 10^{-15} \text{ m} \]

### Step 7: Substitute \( R \) Back into the Equation
Now substitute \( R \) back into the equation for \( V \):

\[ V = \frac{9(1.602 \times 10^{-19})^2}{8\pi (8.85 \times 10^{-12}) (2.2956 \times 10^{-15})} \]

### Step 8: Calculate the Value
Calculating the above expression will yield the Coulomb barrier height in joules, which can be converted to electron volts (1 eV = \( 1.602 \times 10^{-19} \) J).

### Final Result
After performing the calculations, you will find that the Coulomb barrier height for two Lithium nuclei is approximately \( 1.41 \) MeV.

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