The nuclear fusion reaction $${2_1}{H^2}{ \to _2}H{e^4} + {\text{Energy}},$$ is proposed to be used for the production of industrial power. Assuming the efficiency of process for production of power is $$20\% ,$$ find the mass of the deuterium required approximately for a duration of 1 year. Given mass of $$_1{H^2}\,{\text{nucleus}} = 2.0141\,a.m.u$$ and mass of $$_2H{e^4}\,{\text{nuclei}} = 4.0026\,a.m.u$$ and $$1\,a.m.u = 31\,MeV$$
A.
$$165\,kg$$
B.
$$138\,kg$$
C.
$$180\,kg$$
D.
$$60\,kg$$
Answer :
$$138\,kg$$
Solution :
Mass defect $$\Delta m = 2 \times 2.014 - 4.0026 = 0.0256\,a.m.u$$
Energy released when two $$_1{H^2}$$ nuclei fuse $$ = 0.0256 \times 931 = 23.8\,MeV$$
Total energy required to be produced by nuclear reaction in 1 year
$$ = 2500 \times {10^6} \times 3.15 \times {10^7} = 7.88 \times {10^{16}}J$$
No. of nuclei of $$_1{H^2}$$ required
$$ = \frac{{7.88 \times {{10}^{16}}J}}{{23.8 \times 1.6 \times {{10}^{ - 13}}}} \times 2 = 4.14 \times {10^{28}}$$
Mass of Deuterium required
$$ = \frac{{4.14 \times {{10}^{28}}}}{{6.02 \times {{10}^{23}}}} \times 2 \times {10^{ - 3}}kg = 138\,kg$$
Releted MCQ Question on Modern Physics >> Atoms or Nuclear Fission and Fusion
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is