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← IIT-JEE 2009 · Paper 1
PhysicsSingle correct+4 / -1Question 54 of 60

IIT-JEE 2009 · Paper 1, Physics question 54

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Paragraph for Question Nos. 53 to 55

Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, 12H^2_1\mathrm H, known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is 12H+12H→23He+n+energy^2_1\mathrm H+{}^2_1\mathrm H\to{}^3_2\mathrm{He}+n+\text{energy}. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of 12H^2_1\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 nn is the density (number/volume) of deuterons, the product nt0nt_0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5×1014 s/cm35\times10^{14}\ \mathrm{s/cm^3}.

It may be helpful to use the following: Boltzmann constant k=8.6×10−5k=8.6\times10^{-5} eV/K; e24πε0=1.44×10−9\dfrac{e^2}{4\pi\varepsilon_0}=1.44\times10^{-9} eVm.

Assume that two deuteron nuclei in the core of fusion reactor at temperature TT are moving towards each other, each with kinetic energy 1.5 kT1.5\,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 4×10−154\times10^{-15} m is in the range

Choose one option
Answer and worked solution

Official answer: A

Energy balance. Total kinetic energy 2×1.5kT=3kT2\times1.5kT=3kT must equal the Coulomb energy at r=4×10−15r=4\times10^{-15} m:
3kT=e24πε0r=1.44×10−94×10−15=3.6×105 eV.3kT=\frac{e^2}{4\pi\varepsilon_0r}=\frac{1.44\times10^{-9}}{4\times10^{-15}}=3.6\times10^5\ \text{eV}.

Temperature. T=3.6×1053×8.6×10−5≈1.4×109T=\dfrac{3.6\times10^5}{3\times8.6\times10^{-5}}\approx1.4\times10^9 K.

(A)