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
A.
$$23.6\,MeV$$
B.
$$26.9\,MeV$$
C.
$$13.9\,MeV$$
D.
$$19.2\,MeV$$
Answer :
$$23.6\,MeV$$
Solution :
The nuclear reaction of process is $$2_1^2H \to _2^4He$$
Energy released = $$4 \times \left( 7 \right) - 4\left( {1.1} \right) = 23.6\,MeV$$
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