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Atoms and Nuclei question

2009 · Shift 1 · Q57
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  5. /2009 · Shift 1 · Q57

Atoms and Nuclei question

2009 · Shift 1 · Q57

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.Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 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 T required for them to reach a separation of 4 ×\times× 10 −15^{-15}−15 m is in the range
  1. A
    1.0×109K<T<2.0<109K1.0 \times {10^9}K \lt T \lt 2.0 \lt {10^9}K1.0×109K<T<2.0<109K
  2. B
    2.0×109K<T<3.0<109K2.0 \times {10^9}K \lt T \lt 3.0 \lt {10^9}K2.0×109K<T<3.0<109K
  3. C
    3.0×109K<T<4.0<109K3.0 \times {10^9}K \lt T \lt 4.0 \lt {10^9}K3.0×109K<T<4.0<109K
  4. D
    4.0×109K<T<5.0<109K4.0 \times {10^9}K \lt T \lt 5.0 \lt {10^9}K4.0×109K<T<5.0<109K
View written solutionFree

Correct answer: A

  1. Given data
  • Reaction: D+D→23He+n+energyD + D \to {}^3_2He + n + \text{energy}D+D→23​He+n+energy
  • Each deuteron has kinetic energy at large separation: K=1.5kTK = 1.5kTK=1.5kT
  • So for two deuterons moving toward each other, total initial kinetic energy is Ktotal=2(1.5kT)=3kTK_{\text{total}} = 2(1.5kT) = 3kTKtotal​=2(1.5kT)=3kT
  • Required closest separation: r=4×10−15 mr = 4 \times 10^{-15}\,\text{m}r=4×10−15m
  • Coulomb constant in useful units: e24πε0=1.44×10−9 eV m\frac{e^2}{4\pi\varepsilon_0} = 1.44 \times 10^{-9}\,\text{eV m}4πε0​e2​=1.44×10−9eV m
  • Boltzmann constant: k=8.6×10−5 eV/Kk = 8.6 \times 10^{-5}\,\text{eV/K}k=8.6×10−5eV/K
  1. Condition for minimum temperature

For the deuterons to just reach separation rrr, their initial total kinetic energy must equal the Coulomb potential energy at that distance.

Since each deuteron has charge +e+e+e, the electrostatic potential energy is U(r)=e24πε0rU(r) = \frac{e^2}{4\pi\varepsilon_0 r}U(r)=4πε0​re2​

So, 3kT=e24πε0r3kT = \frac{e^2}{4\pi\varepsilon_0 r}3kT=4πε0​re2​

  1. Calculate Coulomb potential energy

Substitute r=4×10−15r = 4 \times 10^{-15}r=4×10−15 m: U=1.44×10−94×10−15 eVU = \frac{1.44 \times 10^{-9}}{4 \times 10^{-15}}\,\text{eV}U=4×10−151.44×10−9​eV U=1.444×106 eVU = \frac{1.44}{4} \times 10^6\,\text{eV}U=41.44​×106eV U=0.36×106 eV=3.6×105 eVU = 0.36 \times 10^6\,\text{eV} = 3.6 \times 10^5\,\text{eV}U=0.36×106eV=3.6×105eV

  1. Find temperature

Using 3kT=3.6×105 eV3kT = 3.6 \times 10^5\,\text{eV}3kT=3.6×105eV T=3.6×1053×8.6×10−5T = \frac{3.6 \times 10^5}{3 \times 8.6 \times 10^{-5}}T=3×8.6×10−53.6×105​

First compute denominator: 3×8.6×10−5=2.58×10−43 \times 8.6 \times 10^{-5} = 2.58 \times 10^{-4}3×8.6×10−5=2.58×10−4

Thus, T=3.6×1052.58×10−4T = \frac{3.6 \times 10^5}{2.58 \times 10^{-4}}T=2.58×10−43.6×105​ T≈1.395×109 KT \approx 1.395 \times 10^9\,\text{K}T≈1.395×109K

  1. Match with option

1.0×109 K<T<2.0×109 K1.0 \times 10^9\,\text{K} < T < 2.0 \times 10^9\,\text{K}1.0×109K<T<2.0×109K

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

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