The binding energy per nucleon is maximum in case of
A.
$$_2H{e^4}$$
B.
$$_{26}F{e^{56}}$$
C.
$$_{56}B{a^{141}}$$
D.
$$_{92}{U^{235}}$$
Answer :
$$_{26}F{e^{56}}$$
Solution :
The binding energy curve has a broad maximum in the range $$A = 30$$ to $$A = 120$$ corresponding to average binding energy per nucleon $$= 8\,MeV.$$ The peak value of the maximum is $$8.8\,MeV/N$$ for $$_{26}F{e^{56}}.$$
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