If the binding energy per nucleon in $$_3L{i^7}$$ and $$_2H{e^4}$$ nuclei are respectively $$5.60\,MeV$$ and $$7.06\,MeV,$$ then the energy of proton $$_3L{i^7} + p \to {2_2}H{e^4}$$ is
A.
$$19.6\,MeV$$
B.
$$2.4\,MeV$$
C.
$$8.4\,MeV$$
D.
$$17.3\,MeV$$
Answer :
$$17.3\,MeV$$
Solution :
Total BE of nucleons in $$_3L{i^7} = 7 \times 5.60 = 39.20\,MeV$$
Total BE of nucleons in $$2\left( {_2H{e^4}} \right) = \left( {4 \times 7.06} \right) \times 2 = 56.48\,MeV$$
Therefore, energy of protons in the reaction = difference of BE's
$$ = 56.48 - 39.20 = 17.3\,MeV$$
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