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In a Nuclear Reaction 3 2 H E + 3 2 H E → 4 2 H E + 1 1 H + 1 1 H + 12.86 M E 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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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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