QJEEAdv09Ph4b54: JEE Advanced 2009 Physics - Class 12 JEEAdv09Ph4b54 Solution: Results of calculations for four different designs of a fusion…

To determine the most promising design based on the Lawson criterion, we need to calculate the Lawson number ( $nt_0$ ) for each design and check if it exceeds $5 \times 10^{14}$ s/cm $^3$ . We will do this for each of the given options:

Option A:

Deuteron density, $n = 2.0 \times 10^{12}~\mathrm{cm^{-3}}$

Confinement time, $t_0 = 5.0 \times 10^{-3}~\mathrm{s}$

Lawson number, $nt_0 = n \cdot t_0 = (2.0 \times 10^{12}) \cdot (5.0 \times 10^{-3}) = 1.0 \times 10^{10}~\mathrm{s/cm^3}$

Option B:

Deuteron density, $n = 8.0 \times 10^{14}~\mathrm{cm^{-3}}$

Confinement time, $t_0 = 9.0 \times 10^{-1}~\mathrm{s}$

Lawson number, $nt_0 = n \cdot t_0 = (8.0 \times 10^{14}) \cdot (9.0 \times 10^{-1}) = 7.2 \times 10^{14}~\mathrm{s/cm^3}$

Option C:

Deuteron density, $n = 4.0 \times 10^{23}~\mathrm{cm^{-3}}$

Confinement time, $t_0 = 1.0 \times 10^{-11}~\mathrm{s}$

Lawson number, $nt_0 = n \cdot t_0 = (4.0 \times 10^{23}) \cdot (1.0 \times 10^{-11}) = 4.0 \times 10^{12}~\mathrm{s/cm^3}$

Option D:

Deuteron density, $n = 1.0 \times 10^{24}~\mathrm{cm^{-3}}$

Confinement time, $t_0 = 4.0 \times 10^{-12}~\mathrm{s}$

Lawson number, $nt_0 = n \cdot t_0 = (1.0 \times 10^{24}) \cdot (4.0 \times 10^{-12}) = 4.0 \times 10^{12}~\mathrm{s/cm^3}$

Based on the calculations, the Lawson numbers for each option are:

• Option A: $1.0 \times 10^{10}~\mathrm{s/cm^3}$
• Option B: $7.2 \times 10^{14}~\mathrm{s/cm^3}$
• Option C: $4.0 \times 10^{12}~\mathrm{s/cm^3}$
• Option D: $4.0 \times 10^{12}~\mathrm{s/cm^3}$

The most promising design based on the Lawson criterion, which requires the Lawson number to be greater than $5 \times 10^{14}~\mathrm{s/cm^3}$ , is Option B because its Lawson number exceeds the threshold.