Scientists are working hard to develop nuclear fusion reactor Nuclei of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+` energy. In the core of fusion reactor, a gas of heavy hydrogen of `_(1)^(2) H` is fully ionized into deuteron nuclei and electrons. This collection of `_1^2H` nuclei and electrons is known as plasma . The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually , the temperature in the reactor core are too high and no material will can be used to confine the to plasma for a time `t_(0)` before the particles fly away from the core. If `n` is the density (number volume ) of deuterons , the product` nt_(0) `is called Lawson number. In one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)`
it may be helpfull to use the following boltzmann constant
`lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm`
Assume that two deuteron nuclei in the core of fusion reactor at temperature energy `T` are moving toward each other, each with kinectic energy `1.5 kT` , when the seperation between them is large enough to neglect coulomb potential energy . Also neglate any interaction from other particle in the core . The minimum temperature `T` required for them to reach a separation of `4 xx 10^(-15) m ` is in the range

Options

A. `1.0 xx10^(9)K lt T lt 2.0xx10^(9)K`
B. `2.0xx10^(9)K lt T lt 3.0xx10^(9)K`
C. `3.0xx10^(9) K lt T lt 4.0xx10^(9)K`
D. `4.0xx10^(9)K lt T lt 5.0xx10^(9)K`

Explanation

To solve the problem, we need to find the minimum temperature \( T \) required for two deuteron nuclei to reach a separation of \( 4 \times 10^{-15} \) m. We will use the conservation of energy principle, where the kinetic energy of the deuterons is converted into potential energy as they approach each other.

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy**:
Each deuteron has a kinetic energy given by \( 1.5 kT \), where \( k \) is the Boltzmann constant. For two deuterons, the total kinetic energy \( KE \) will be:
\[ KE = 2 \times 1.5 kT = 3kT \]

2. **Potential Energy Calculation**:
The potential energy \( U \) when the two deuterons are at a separation \( r \) is given by the formula:
\[ U = \frac{Q_1 Q_2}{r} \]
For deuterons, the charge \( Q \) is equal to the elementary charge \( e \). Thus, the potential energy becomes:
\[ U = \frac{e^2}{r} \]

3. **Substituting Values**:
We know from the problem that the separation \( r = 4 \times 10^{-15} \) m. The value of \( e^2/(4\pi \epsilon_0) \) is given as \( 1.44 \times 10^{-9} \) eV·m. Thus, we can express the potential energy as:
\[ U = \frac{1.44 \times 10^{-9}}{4 \times 10^{-15}} = 3.6 \times 10^{5} \text{ eV} \]

4. **Equating Kinetic and Potential Energy**:
By conservation of energy, the total kinetic energy must equal the potential energy at the point of closest approach:
\[ 3kT = U \]
Substituting for \( U \):
\[ 3kT = 3.6 \times 10^{5} \text{ eV} \]

5. **Solving for Temperature \( T \)**:
Rearranging gives us:
\[ T = \frac{3.6 \times 10^{5}}{3k} \]
Substituting \( k = 8.6 \times 10^{-5} \text{ eV/K} \):
\[ T = \frac{3.6 \times 10^{5}}{3 \times 8.6 \times 10^{-5}} = \frac{3.6 \times 10^{5}}{2.58 \times 10^{-4}} \approx 1.39 \times 10^{9} \text{ K} \]

6. **Determining the Range**:
The calculated temperature \( T \approx 1.39 \times 10^{9} \text{ K} \) lies in the range of \( 10^{9} \text{ K} \).

### Conclusion:
The minimum temperature \( T \) required for the two deuteron nuclei to reach a separation of \( 4 \times 10^{-15} \) m is approximately \( 1.39 \times 10^{9} \text{ K} \).