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प्रश्न
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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उत्तर
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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संबंधित प्रश्न
Calculate the energy in fusion reaction:
`""_1^2H+_1^2H->_2^3He+n`, where BE of `""_1^2H`23He=7.73MeV" data-mce-style="position: relative;">=2.2323He=7.73MeV MeV and of `""_2^3He=7.73 MeV`
Write one balanced equation to show Nuclear fission
During a nuclear fission reaction,
Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5 kT equals the Coulomb potential energy at 2 fm.
Calculate the Q-values of the following fusion reactions :-
(a) `""_1^2H + ""_1^2H → ""_1^3H + ""_1^1H`
(b) `""_1^2H + ""_1^2H → ""_2^3H + n`
(c) `""_1^2H + ""_1^3H → _2^4H + n`.
Atomic masses are `m(""_1^2H) = 2.014102 "u", m(""_1^3H) = 3.016049 "u", m(""_2^3He) = 3.016029 "u", m(""_2^4He) = 4.002603 "u".`
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
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\[{}_{1}^{2}\mathrm{H}+{}_{1}^{2}\mathrm{H}\rightarrow{}_{2}^{3}\mathrm{He}+\mathrm{n}+3.27\mathrm{MeV}\]
