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Atoms and Nuclei question

2003 · Shift 0 · Q123
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Atoms and Nuclei question

2003 · Shift 0 · Q123

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
In the nuclear fusion reaction 12H+13H→24He+n{}_1^2H + {}_1^3H \to {}_2^4He + n12​H+13​H→24​He+n given that the repulsive potential energy between the two nuclei is ∼7.7×10−14J\sim 7.7 \times {10^{ - 14}}J∼7.7×10−14J, the temperature at which the gases must be heated to initiate the reaction is nearly [ Boltzmann's Constant k=1.38×10−23 J/Kk = 1.38 \times {10^{ - 23}}\,J/Kk=1.38×10−23J/K ]
  1. A
    107  K{10^7}\,\,K107K
  2. B
    105  K{10^5}\,\,K105K
  3. C
    103  K{10^3}\,\,K103K
  4. D
    109  K{10^9}\,\,K109K
View written solutionFree

Correct answer: D

  1. Use thermal energy to overcome the Coulomb barrier

For fusion to begin, the thermal energy of nuclei should be of the order of the repulsive potential energy:

kT≈7.7×10−14 JkT \approx 7.7 \times 10^{-14}\,\text{J}kT≈7.7×10−14J

Given:

k=1.38×10−23 J/Kk = 1.38 \times 10^{-23}\,\text{J/K}k=1.38×10−23J/K

So,

T=7.7×10−141.38×10−23T = \frac{7.7 \times 10^{-14}}{1.38 \times 10^{-23}}T=1.38×10−237.7×10−14​

  1. Calculate the temperature

T=7.71.38×10−14+23T = \frac{7.7}{1.38} \times 10^{-14+23}T=1.387.7​×10−14+23

T≈5.58×109 KT \approx 5.58 \times 10^9\,\text{K}T≈5.58×109K

So the required temperature is of the order of

109 K10^9\,\text{K}109K

  1. Check the options
  • A: 107 K10^7\,\text{K}107K
  • B: 105 K10^5\,\text{K}105K
  • C: 103 K10^3\,\text{K}103K
  • D: 109 K10^9\,\text{K}109K

Hence, the nearest correct option is:

D\boxed{D}D​

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