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Question
In a nuclear reaction
`"_2^3He + _2^3He -> _2^4He +_1^1H +_1^1H + 12.86 Me V` though the number of nucleons is conserved on both sides of the reaction, yet the energy is released. How? Explain.
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Solution
In a nuclear reaction, the sum of the masses of the target nucleus `"_2^3He` and the bombarding particle `"_2^3He` may be greater or less than the sum of the masses of the product nucleus `"_2^4He` and the `"_1^1He`. So from the law of conservation of mass-energy some energy (12.86 MeV) is evolved in nuclear reaction. This energy is called Q-value of the nuclear reaction. The binding energy of the nucleus on the left side is not equal to the right side. The difference in the binding energies on two sides appears as energy released or absorbed in the nuclear reaction.
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RELATED QUESTIONS
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction :
`""_1^2H+_1^3H->_2^4He+n`
Using the data :
m(`""_1^2H`) = 2.014102 u
m(`""_1^3H`) = 3.016049 u
m(`""_2^4He`) = 4.002603 u
mn = 1.008665 u
1u = 931.5 MeV/c2
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(Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
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