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Question

In the nuclear fusion reaction, 21H+31H42He + n given that the repulsive potential energy between the two nuclei is 7.7×1014 J, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant k=1.38×1023J/K]

A
107 K
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B
105 K
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C
103 K
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D
109 K
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Solution

The correct option is D 109 K
To initiate the reaction, the energy of the gas molecules must be equal to the repulsive potential energy.

We know that the kinetic energy of translation per molecules is given by:
K.E.=32kT

In order to initiate the fusion reaction,

32kT=7.7×1014J

T=2×7.7×10143×1.38×10233.7×109K

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