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`
Result of calculations for four different design of a fusion reactor using `D-D` reaction are given below. which of these is most promising based on Lawson criterion ?

Options

A. Deuteron density `=2.0 xx 10^(12)cm^(-3)`, confinement tiem `=5.0 xx 10^(-3)s`
B. Deuteron density `=8.0 xx 10^(14)cm^(-3)`, confinement tiem `=9.0 xx 10^(-1)s`
C. Deuteron density `=4.0 xx 10^(23)cm^(-3)`, confinement tiem `=1.0 xx 10^(-11)s`
D. Deuteron density `=2.0 xx 10^(24)cm^(-3)`, confinement tiem `=4.0 xx 10^(-12)s`

Explanation

To solve the problem, we need to evaluate the Lawson number \( n t_0 \) for each of the four designs of the fusion reactor using the given data. The Lawson number must be greater than \( 5 \times 10^{14} \, \text{s/cm}^2 \) for the reactor to be considered successful.

### Step-by-Step Solution:

1. **Understand the Lawson Number**:
The Lawson number is defined as:
\[ L = n t_0 \]
where \( n \) is the density of deuterons (in \( \text{cm}^{-3} \)) and \( t_0 \) is the confinement time (in seconds).

2. **Identify the Given Values**:
We need to evaluate the Lawson number for four different designs. Let's denote the density and confinement time for each design as follows:
- Design A: \( n_A \), \( t_{0A} \)
- Design B: \( n_B \), \( t_{0B} \)
- Design C: \( n_C \), \( t_{0C} \)
- Design D: \( n_D \), \( t_{0D} \)

(Assuming the values for \( n \) and \( t_0 \) are provided in the problem statement, which is not included here.)

3. **Calculate Lawson Number for Each Design**:
For each design, calculate the Lawson number:
- For Design A:
\[ L_A = n_A t_{0A} \]
- For Design B:
\[ L_B = n_B t_{0B} \]
- For Design C:
\[ L_C = n_C t_{0C} \]
- For Design D:
\[ L_D = n_D t_{0D} \]

4. **Compare with the Lawson Criterion**:
We need to check which of these Lawson numbers is greater than \( 5 \times 10^{14} \, \text{s/cm}^2 \):
- If \( L_A > 5 \times 10^{14} \): Design A is promising.
- If \( L_B > 5 \times 10^{14} \): Design B is promising.
- If \( L_C > 5 \times 10^{14} \): Design C is promising.
- If \( L_D > 5 \times 10^{14} \): Design D is promising.

5. **Determine the Most Promising Design**:
After calculating the Lawson numbers for all designs, identify the one with the highest value that is still greater than \( 5 \times 10^{14} \).

### Conclusion:
Based on the calculations, the design that meets the Lawson criterion and has the highest Lawson number is considered the most promising for a fusion reactor.