The binding energy per nucleon in deuterium a | vedclass

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(B) The nuclear fusion reaction is given by: $_1H^2 + _1H^2 	o _2He^4 + \Delta E$.The binding energy per nucleon of a deuteron is $1.1 \, MeV$.Since

Vedclass · Accepted Answer

$23.6$

Vedclass · Answer

(B) The nuclear fusion reaction is given by: $_1H^2 + _1H^2 	o _2He^4 + \Delta E$.The binding energy per nucleon of a deuteron is $1.1 \, MeV$.Since a deuteron has $2$ nucleons,the total binding energy of one deuteron is $2 	imes 1.1 = 2.2 \, MeV$.For two deuterium nuclei,the total initial binding energy is $2 	imes 2.2 = 4.4 \, MeV$.The binding energy per nucleon of a helium nucleus $(He^4)$ is $7.0 \, MeV$.Since a helium nucleus has $4$ nucleons,the total binding energy is $4 	imes 7.0 = 28.0 \, MeV$.The energy released in the fusion process is the difference between the total binding energy of the product and the total binding energy of the reactants:$\Delta E = 28.0 \, MeV - 4.4 \, MeV = 23.6 \, MeV$.

The binding energy per nucleon in deuterium and helium nuclei are $1.1 \, MeV$ and $7.0 \, MeV,$ respectively. When two deuterium nuclei fuse to form a helium nucleus,the energy released in the fusion is ........... $MeV$.

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