SATHEE: Atoms And Nuclei Question 347

Atoms And Nuclei Question 347

Question: In the nuclear fusion reaction $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} H e + n$ given that the repulsive potential energy between the two nuclei is- $\sim 7.7 \times 10^{- 14} J$ , the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann’s Constant $k = 1.38 \times 10^{- 23} J / K$ ]

Options:

A) $10^{7} K$

B) $10^{5} K$

C) $10^{3} K$

D) $10^{9} K$

Show Answer

Answer:

Correct Answer: D

Solution:

average kinetic energy per molecule
$= \frac{3}{2} k T$
This kinetic energy should be able to provide the repulsive potential energy
$∴ \frac{3}{2} k T = 7.7 \times 10^{- 14}$
$\Rightarrow, T = \frac{2 \times 7.7 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}} = 3.7 \times 10^{9} K$

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Atoms And Nuclei Question 347

Question: In the nuclear fusion reaction 12H+13H→24He+n given that the repulsive potential energy between the two nuclei is- ∼7.7×10−14J , the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann’s Constant k=1.38×10−23J/K ]

Options:

Answer:

Solution:

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