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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Physics
Chapter
Nuclear Physics

JEE Advanced 2009 · Paper 1 · Physics · Nuclear Physics

Paragraph: Scientists are working hard to develop nuclear fusion...

Paragraph: Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, 12H{ }_1^2 \mathrm{H} known as deuteron and denoted by DD can be thought of as a candidate for fusion reactor. The DD - DD reaction is 12H+12H→23He+n+{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+n+ energy. In the core of fusion reactor. A gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of 14H{ }_1^4 \mathrm{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_0 before the particles fly away from the core. If n is the density (number/volume) of deutrons, the product t 0_0 is called Lawson number. In one of the criteria, aa reactor is termed successful if Lawson number is greater than 5×1014scm−35 \times 10^{14} \mathrm{scm}^{-3}. It may be helpful to use the following : Boltzmann constant k=8.6×10−5eV/Kk=8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K}; e24πε0=1.44×10−9eVm\frac{e^2}{4 \pi \varepsilon_0}=1.44 \times 10^{-9} \mathrm{eVm}Question: Assume that two deuteron nuclei in the core of fusion reactor at temperature TT are moving towards each other, each with kinetic energy 1.5kT1.5 \mathrm{kT}, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature TT required for them to reach a separation of 4×10−15 m4 \times 10^{-15} \mathrm{~m} is in the range

Options

  1. A1.0×109 K<T<2.0×109 K1.0 \times 10^9 \mathrm{~K} < T < 2.0 \times 10^9 \mathrm{~K}
  2. B20×109 K<T<3.0×109 K20 \times 10^9 \mathrm{~K} < T < 3.0 \times 10^9 \mathrm{~K}
  3. C3.0×109 K<T<4.0×109 K3.0 \times 10^9 \mathrm{~K} < T < 4.0 \times 10^9 \mathrm{~K}
  4. D4.0×109 K<T<5.0×109 K4.0 \times 10^9 \mathrm{~K} < T < 5.0 \times 10^9 \mathrm{~K}

Correct answer

A. 1.0×109 K<T<2.0×109 K1.0 \times 10^9 \mathrm{~K} < T < 2.0 \times 10^9 \mathrm{~K}

Step-by-step solution

From conservation of mechanical energy, we have Ui+Ki=Uf+Uf0+2(1.5kT)=14πε0⋅(e)(e)d+0Substitutingthevalues,wegetT=1.4\begin{gathered}U_i+K_i=U_f+U_f \\ 0+2(1.5 k T)=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(e)(e)}{d}+0 \end{gathered} Substituting the values, we get T=1.4\times10910^9\mathrm{~K}
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