JEE MAIN - Physics (2003 - No. 3)
In the nuclear fusion reaction
$$${}_1^2H + {}_1^3H \to {}_2^4He + n$$$
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 $$k = 1.38 \times {10^{ - 23}}\,J/K$$ ]
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 $$k = 1.38 \times {10^{ - 23}}\,J/K$$ ]
$${10^7}\,\,K$$
$${10^5}\,\,K$$
$${10^3}\,\,K$$
$${10^9}\,\,K$$
Explanation
The average kinetic energy per molecule $$ = {3 \over 2}kT$$
This kinetic energy should be able to provide the repulsive potential energy
$$\therefore$$ $${3 \over 2}kT = 7.7 \times {10^{ - 14}}$$
$$ \Rightarrow T = {{2 \times 7.7 \times {{10}^{ - 14}}} \over {3 \times 1.38 \times {{10}^{ - 23}}}} = 3.7 \times {10^9}$$
This kinetic energy should be able to provide the repulsive potential energy
$$\therefore$$ $${3 \over 2}kT = 7.7 \times {10^{ - 14}}$$
$$ \Rightarrow T = {{2 \times 7.7 \times {{10}^{ - 14}}} \over {3 \times 1.38 \times {{10}^{ - 23}}}} = 3.7 \times {10^9}$$
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