In the nuclear fusion reaction
`_(1)^(2)H + _(1)^(3)H rarr _(2)^(4)He + n `
given that the repulsive potential energy between the two nuclei is `- 7.7 xx 10^(-14) J` , the temperature at which the gases must be heated the reaction is nearly
[Boltzmann's constant `k = 1.38 xx 10^(-23) J//K]`

Options

A. `10^(9)K`
B. `10^(7)K`
C. `10^(5)K`
D. `10^(3)K`

Explanation

To solve the problem, we need to find the temperature at which the kinetic energy of the gas molecules is sufficient to overcome the repulsive potential energy between the two nuclei in the fusion reaction.

### Step-by-Step Solution:

1. **Understand the Kinetic Energy of Gas Molecules**:
The kinetic energy (KE) of a gas molecule at temperature \( T \) is given by the formula:
\[ KE = \frac{3}{2} k T \]
where \( k \) is Boltzmann's constant.

2. **Set Up the Equation**:
We want the kinetic energy to be equal to the repulsive potential energy, which is given as \( 7.7 \times 10^{-14} \, J \). Therefore, we can set up the equation:
\[ \frac{3}{2} k T = 7.7 \times 10^{-14} \]

3. **Substitute the Value of Boltzmann's Constant**:
We know that \( k = 1.38 \times 10^{-23} \, J/K \). Substituting this value into the equation gives:
\[ \frac{3}{2} (1.38 \times 10^{-23}) T = 7.7 \times 10^{-14} \]

4. **Solve for Temperature \( T \)**:
Rearranging the equation to solve for \( T \):
\[ T = \frac{7.7 \times 10^{-14} \times 2}{3 \times 1.38 \times 10^{-23}} \]

5. **Calculate the Value**:
Now we perform the calculation:
\[ T = \frac{15.4 \times 10^{-14}}{4.14 \times 10^{-23}} \approx 3.73 \times 10^{9} \, K \]

6. **Final Answer**:
The temperature at which the gases must be heated for the reaction to occur is approximately:
\[ T \approx 3.7 \times 10^{9} \, K \]