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प्रश्न
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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उत्तर
Nuclear fission: It is the phenomenon of splitting of a heavy nucleus (usually A > 230) into two or more lighter nuclei.
`""_92U^235+_0n^1->_56Ba^141+_36Kr^92+3_0n^1+Q`
Here, the energy released per fission of `""_92U^235` is 200.4 MeV.
Nuclear fusion: It is the phenomenon of fusion of two or more lighter nuclei to form a single heavy nucleus.
The mass of the product nucleus is slightly less than the sum of the masses of the lighter nuclei fusing together. This difference in masses results in the release of tremendous amount of energy.
Example:
`""_1H^1+_1H^1->_1H^2+e^++v+0.42 MeV`
`""_1H^2+_1H^2->_2He^3+n+3.27MeV`
`""_1H^2+_1H^2->_1H^3+_1H^1+4.03MeV`
`""_1^2H+_1^3H->_2^4He+n`
∴ Δm = (2.014102 + 3.016049) − (4.002603 + 1.008665 )
=0.018883 u
Energy released, Q = 0.018883 × 931.5 `MeV"/"_(c^2`
= 17.589 MeV
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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`
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(A) only total energy is conserved.
(B) only total momentum is conserved.
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