If two deuterium nuclei get close enough to each other, the attraction of the strong nuclear force will fuse them to make an isotope of helium. This process releases a huge amount of energy. The range of nuclear force is `10^(-15)` m. This is the principle behind the nuclear fusion reactor. The deuterium nuclei moves to fast that, it is not possible to contain them by physical walls. Therefore they are confined magnetically (Assume coulomb law to hold even at `10^(-18)` m)
Two deuterium nuclei having same speed undergo a head on collision. Which of the following is closest to the minimum value of v (in km/sec) for which fusion occurs

Options

A. 1000
B. 5000
C. 10000
D. 50000

Answer

To solve the problem of determining the minimum speed (v) required for two deuterium nuclei to undergo fusion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces Involved**:
The fusion of two deuterium nuclei occurs when they come close enough for the strong nuclear force to overcome the electrostatic repulsion between them. The electrostatic repulsion can be calculated using Coulomb's law.

2. **Coulomb's Law**:
The potential energy (U) due to the electrostatic force between two charges (q1 and q2) separated by a distance (R) is given by:
\[ U = \frac{q_1 q_2}{4 \pi \epsilon_0 R} \]
Here, \(q_1\) and \(q_2\) are the charges of the deuterium nuclei, \(R\) is the distance (range of nuclear force), and \(\epsilon_0\) is the permittivity of free space.

3. **Kinetic Energy**:
The kinetic energy (KE) of the two deuterium nuclei moving towards each other is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass of one deuterium nucleus and \(v\) is their speed.

4. **Setting Up the Equation**:
For fusion to occur, the kinetic energy must be equal to the potential energy at the point of closest approach:
\[ \frac{1}{2} mv^2 = \frac{q_1 q_2}{4 \pi \epsilon_0 R} \]
Rearranging gives:
\[ v^2 = \frac{2q_1 q_2}{4 \pi \epsilon_0 m R} \]

5. **Substituting Values**:
- The charge of a deuterium nucleus \(q = 1.6 \times 10^{-19}\) C.
- The permittivity of free space \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\).
- The range \(R = 10^{-15} \, \text{m}\).
- The mass of a deuterium nucleus \(m = 2 \times 1.67 \times 10^{-27} \, \text{kg}\).

Plugging in these values:
\[ v^2 = \frac{2 \times (1.6 \times 10^{-19})^2}{4 \pi (8.85 \times 10^{-12}) (2 \times 1.67 \times 10^{-27}) (10^{-15})} \]

6. **Calculating**:
First, calculate the numerator:
\[ 2 \times (1.6 \times 10^{-19})^2 = 2 \times 2.56 \times 10^{-38} = 5.12 \times 10^{-38} \]
Next, calculate the denominator:
\[ 4 \pi (8.85 \times 10^{-12}) (2 \times 1.67 \times 10^{-27}) (10^{-15}) \approx 3.34 \times 10^{-42} \]
Now, substituting back:
\[ v^2 = \frac{5.12 \times 10^{-38}}{3.34 \times 10^{-42}} \approx 1.53 \times 10^{4} \]

7. **Finding v**:
\[ v \approx \sqrt{1.53 \times 10^{4}} \approx 123.5 \times 10^{2} \approx 1.1 \times 10^{7} \, \text{m/s} \]
Converting to km/s:
\[ v \approx 11,000 \, \text{km/s} \]

8. **Conclusion**:
The minimum speed required for fusion to occur is approximately \(10,000 \, \text{km/s}\).