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Atoms and Nuclei question

2009 · Shift 1 · Q58
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Atoms and Nuclei question

2009 · Shift 1 · Q58

JEE AdvancedPhysicsAtoms and NucleiMCQ+3 / −1
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, 12_1^212​ H, known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is 12_1^212​ H + 12_1^212​ H →23\to_2^3→23​ He + nnn+ energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of 12_1^212​ H 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 temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t0t_0t0​ before the particles fly away from the core. If nnn is the density (number/volume) of deuterons, the product nt0nt_0nt0​ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 ×\times× 10 14^{14}14 s/cm 3^33. It may be helpful to use the following : Boltzmann constant k=8.6×10−5k = 8.6 \times {10^{ - 5}}k=8.6×10−5 eV/K; e24πε0=1.44×109{{{e^2}} \over {4\pi {\varepsilon _0}}} = 1.44 \times {10^9}4πε0​e2​=1.44×109 eVm.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?
  1. A
    Deuteron density = 2.0×1012 cm−32.0\times10^{12}~\mathrm{cm^{-3}}2.0×1012 cm−3; Confinement time = 5.0×10−3 s5.0\times10^{-3}~\mathrm{s}5.0×10−3 s.
  2. B
    Deuteron density = 8.0×1014 cm−38.0\times10^{14}~\mathrm{cm^{-3}}8.0×1014 cm−3; Confinement time = 9.0×10−1 s9.0\times10^{-1}~\mathrm{s}9.0×10−1 s.
  3. C
    Deuteron density = 4.0×1023 cm−34.0\times10^{23}~\mathrm{cm^{-3}}4.0×1023 cm−3; Confinement time = 1.0×10−11 s1.0\times10^{-11}~\mathrm{s}1.0×10−11 s.
  4. D
    Deuteron density = 1.0×1024 cm−31.0\times10^{24}~\mathrm{cm^{-3}}1.0×1024 cm−3; Confinement time = 4.0×10−12 s4.0\times10^{-12}~\mathrm{s}4.0×10−12 s.
View written solutionFree

Correct answer: B

The user wants me to identify the most promising fusion reactor design based on the Lawson criterion.

Step-by-Step Solution:

  1. Understand the Lawson Criterion: The problem provides the Lawson criterion for a successful fusion reactor. It states that the product of the deuteron density (nnn) and the confinement time (t0t_0t0​), known as the Lawson number, must be greater than a certain threshold. The criterion is given as: nt0>5×1014 s/cm3nt_0 > 5 \times 10^{14} \text{ s/cm}^3nt0​>5×1014 s/cm3 A higher Lawson number indicates a more promising design. The most promising design will be the one with the highest Lawson number, especially if it exceeds the given threshold.

  2. Calculate the Lawson Number for each option: We will calculate the product nt0nt_0nt0​ for each of the four designs provided.

    • Option A: Deuteron density, n=2.0×1012 cm−3n = 2.0 \times 10^{12} \text{ cm}^{-3}n=2.0×1012 cm−3 Confinement time, t0=5.0×10−3 st_0 = 5.0 \times 10^{-3} \text{ s}t0​=5.0×10−3 s Lawson number, nt0=(2.0×1012 cm−3)×(5.0×10−3 s)nt_0 = (2.0 \times 10^{12} \text{ cm}^{-3}) \times (5.0 \times 10^{-3} \text{ s})nt0​=(2.0×1012 cm−3)×(5.0×10−3 s) nt0=10.0×1012−3 s/cm3=10×109 s/cm3=1.0×1010 s/cm3nt_0 = 10.0 \times 10^{12-3} \text{ s/cm}^3 = 10 \times 10^9 \text{ s/cm}^3 = 1.0 \times 10^{10} \text{ s/cm}^3nt0​=10.0×1012−3 s/cm3=10×109 s/cm3=1.0×1010 s/cm3

    • Option B: Deuteron density, n=8.0×1014 cm−3n = 8.0 \times 10^{14} \text{ cm}^{-3}n=8.0×1014 cm−3 Confinement time, t0=9.0×10−1 st_0 = 9.0 \times 10^{-1} \text{ s}t0​=9.0×10−1 s Lawson number, nt0=(8.0×1014 cm−3)×(9.0×10−1 s)nt_0 = (8.0 \times 10^{14} \text{ cm}^{-3}) \times (9.0 \times 10^{-1} \text{ s})nt0​=(8.0×1014 cm−3)×(9.0×10−1 s) nt0=72.0×1014−1 s/cm3=72×1013 s/cm3=7.2×1014 s/cm3nt_0 = 72.0 \times 10^{14-1} \text{ s/cm}^3 = 72 \times 10^{13} \text{ s/cm}^3 = 7.2 \times 10^{14} \text{ s/cm}^3nt0​=72.0×1014−1 s/cm3=72×1013 s/cm3=7.2×1014 s/cm3

    • Option C: Deuteron density, n=4.0×1023 cm−3n = 4.0 \times 10^{23} \text{ cm}^{-3}n=4.0×1023 cm−3 Confinement time, t0=1.0×10−11 st_0 = 1.0 \times 10^{-11} \text{ s}t0​=1.0×10−11 s Lawson number, nt0=(4.0×1023 cm−3)×(1.0×10−11 s)nt_0 = (4.0 \times 10^{23} \text{ cm}^{-3}) \times (1.0 \times 10^{-11} \text{ s})nt0​=(4.0×1023 cm−3)×(1.0×10−11 s) nt0=4.0×1023−11 s/cm3=4.0×1012 s/cm3nt_0 = 4.0 \times 10^{23-11} \text{ s/cm}^3 = 4.0 \times 10^{12} \text{ s/cm}^3nt0​=4.0×1023−11 s/cm3=4.0×1012 s/cm3

    • Option D: Deuteron density, n=1.0×1024 cm−3n = 1.0 \times 10^{24} \text{ cm}^{-3}n=1.0×1024 cm−3 Confinement time, t0=4.0×10−12 st_0 = 4.0 \times 10^{-12} \text{ s}t0​=4.0×10−12 s Lawson number, nt0=(1.0×1024 cm−3)×(4.0×10−12 s)nt_0 = (1.0 \times 10^{24} \text{ cm}^{-3}) \times (4.0 \times 10^{-12} \text{ s})nt0​=(1.0×1024 cm−3)×(4.0×10−12 s) nt0=4.0×1024−12 s/cm3=4.0×1012 s/cm3nt_0 = 4.0 \times 10^{24-12} \text{ s/cm}^3 = 4.0 \times 10^{12} \text{ s/cm}^3nt0​=4.0×1024−12 s/cm3=4.0×1012 s/cm3

  3. Compare the results with the Lawson Criterion: The success criterion is nt0>5×1014 s/cm3nt_0 > 5 \times 10^{14} \text{ s/cm}^3nt0​>5×1014 s/cm3.

    • A: 1.0×1010 s/cm31.0 \times 10^{10} \text{ s/cm}^31.0×1010 s/cm3 (Does not meet the criterion)
    • B: 7.2×1014 s/cm37.2 \times 10^{14} \text{ s/cm}^37.2×1014 s/cm3 (Meets the criterion, as 7.2×1014>5×10147.2 \times 10^{14} > 5 \times 10^{14}7.2×1014>5×1014)
    • C: 4.0×1012 s/cm34.0 \times 10^{12} \text{ s/cm}^34.0×1012 s/cm3 (Does not meet the criterion)
    • D: 4.0×1012 s/cm34.0 \times 10^{12} \text{ s/cm}^34.0×1012 s/cm3 (Does not meet the criterion)

  4. Conclusion: Only the design in Option B has a Lawson number that exceeds the success threshold. It also has the highest Lawson number among all the options. Therefore, it is the most promising design based on the given criterion. The other physical constants provided in the problem are not needed for this specific question.

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