JEE Challenger
More from Nuclei
PhyMedAdv 2009 • Paper 1

Evaluation of Fusion Reactor Designs Based on Lawson Criterion

Results of calculations for four different designs 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 =2.0×1012 cm−3= 2.0\times 10^{12}\text{ cm}^{-3}, confinement time =5.0×10−3 s= 5.0\times 10^{-3}\text{ s}

B

deuteron density =8.0×1014 cm−3= 8.0\times 10^{14}\text{ cm}^{-3}, confinement time =9.0×10−1 s= 9.0\times 10^{-1}\text{ s}

Correct
C

deuteron density =4.0×1023 cm−3= 4.0\times 10^{23}\text{ cm}^{-3}, confinement time =1.0×10−11 s= 1.0\times 10^{-11}\text{ s}

D

deuteron density =1.0×1024 cm−3= 1.0\times 10^{24}\text{ cm}^{-3}, confinement time =4.0×10−12 s= 4.0\times 10^{-12}\text{ s}

Topics & Concepts

NucleiNuclear Reactions

Step-by-Step Solution

According to the Lawson criterion, for a thermonuclear fusion reactor to achieve net power gain (breakeven or ignition), the product of the ion density nn and the energy confinement time τ\tau, known as the Lawson parameter nτn\tau, must exceed a critical minimum value:
nτ≥(nτ)thresholdn\tau \ge (n\tau)_{\text{threshold}}

For a fusion reaction (such as the D-D reaction), the design having the largest value of the product nτn\tau is the most promising.

Let us evaluate the product nτn\tau for each of the given reactor designs:

  1. Option (A):
    n=2.0×1012 cm−3,τ=5.0×10−3 sn = 2.0 \times 10^{12}\text{ cm}^{-3}, \quad \tau = 5.0 \times 10^{-3}\text{ s}
    nτ=(2.0×1012 cm−3)×(5.0×10−3 s)=1.0×1010 s⋅cm−3n\tau = (2.0 \times 10^{12}\text{ cm}^{-3}) \times (5.0 \times 10^{-3}\text{ s}) = 1.0 \times 10^{10}\text{ s}\cdot\text{cm}^{-3}

  2. Option (B):
    n=8.0×1014 cm−3,τ=9.0×10−1 sn = 8.0 \times 10^{14}\text{ cm}^{-3}, \quad \tau = 9.0 \times 10^{-1}\text{ s}
    nτ=(8.0×1014 cm−3)×(9.0×10−1 s)=7.2×1014 s⋅cm−3n\tau = (8.0 \times 10^{14}\text{ cm}^{-3}) \times (9.0 \times 10^{-1}\text{ s}) = 7.2 \times 10^{14}\text{ s}\cdot\text{cm}^{-3}

  3. Option (C):
    n=4.0×1023 cm−3,τ=1.0×10−11 sn = 4.0 \times 10^{23}\text{ cm}^{-3}, \quad \tau = 1.0 \times 10^{-11}\text{ s}
    nτ=(4.0×1023 cm−3)×(1.0×10−11 s)=4.0×1012 s⋅cm−3n\tau = (4.0 \times 10^{23}\text{ cm}^{-3}) \times (1.0 \times 10^{-11}\text{ s}) = 4.0 \times 10^{12}\text{ s}\cdot\text{cm}^{-3}

  4. Option (D):
    n=1.0×1024 cm−3,τ=4.0×10−12 sn = 1.0 \times 10^{24}\text{ cm}^{-3}, \quad \tau = 4.0 \times 10^{-12}\text{ s}
    nτ=(1.0×1024 cm−3)×(4.0×10−12 s)=4.0×1012 s⋅cm−3n\tau = (1.0 \times 10^{24}\text{ cm}^{-3}) \times (4.0 \times 10^{-12}\text{ s}) = 4.0 \times 10^{12}\text{ s}\cdot\text{cm}^{-3}

Comparing the values:
(nτ)B=7.2×1014 s⋅cm−3>(nτ)C=(nτ)D=4.0×1012 s⋅cm−3>(nτ)A=1.0×1010 s⋅cm−3(n\tau)_{\text{B}} = 7.2 \times 10^{14}\text{ s}\cdot\text{cm}^{-3} > (n\tau)_{\text{C}} = (n\tau)_{\text{D}} = 4.0 \times 10^{12}\text{ s}\cdot\text{cm}^{-3} > (n\tau)_{\text{A}} = 1.0 \times 10^{10}\text{ s}\cdot\text{cm}^{-3}

Since the product nτn\tau is maximum for design (B), it is the most promising based on the Lawson criterion.

Correct Answer: B

PreviousNext