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Atoms and Nuclei question

2009 · Shift 1 · Q58
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  5. /2009 · Shift 1 · Q58

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.
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