In a fission reaction,
$$_{92}^{236}U{ \to ^{117}}X{ + ^{117}}Y + n + n$$
the binding energy per nucleon of $$X$$ and $$Y$$ is $$8.5\,MeV$$ whereas of $$^{236}U$$ is $$7.6\,MeV.$$ The total energy liberated will be about
A.
$$2000\,MeV$$
B.
$$200\,MeV$$
C.
$$2\,MeV$$
D.
$$1\,MeV$$
Answer :
$$200\,MeV$$
Solution :
Binding energy of fissioned nucleus $$ = 236 \times 7.6\,MeV$$
Binding energy of products $$ = 117 \times 8.5 + 117 \times 8.5$$
$$ = 2 \times 117 \times 8.5$$
Hence, net binding energy = binding energy of products — binding energy of fissioned nucleus
$$\eqalign{
& = 234 \times 8.5 - 236 \times 7.6 \cr
& = 1989 - 1793.6 \cr
& = 195.4\,MeV \cr
& \approx 200\,MeV \cr} $$
Thus, in per fission of uranium nearly $$200\,MeV$$ energy is released.
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