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Question
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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Solution
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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Write notes on Nuclear fusion
Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A
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`"_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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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.)
Why nuclear fusion reaction is also called thermo-nuclear reaction?
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A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of the slab is:
(Given latent heat of fusion of ice = 3.36 × 105 J kg−1)
