Consider the nuclear fission reaction ${ }_0^1 n+{ }_{92}^{235} U \longrightarrow{ }_{56}^{144} Ba+{ }_{36}^{89} Kr+3{ }_0^1 n$. Assuming all the kinetic energy is carried away by the fast neutrons only and total binding energies of ${ }_{92}^{235} U, { }_{56}^{144} Ba$ and ${ }_{36}^{89} Kr$ to be $1800 \ MeV, 1200 \ MeV$ and $780 \ MeV$ respectively,the average kinetic energy carried by each fast neutron is (in $MeV$):
- A$200$
- B$180$
- C$67$
- D$60$
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$A$ nucleus with mass number $240$ breaks into two fragments each of mass number $120$. The binding energy per nucleon of the unfragmented nucleus is $7.6 \, MeV$,while that of the fragments is $8.5 \, MeV$. The total gain in the binding energy in the process is (in $MeV$):
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View SolutionMention the use of controlled nuclear fission. Mention the use of uncontrolled nuclear fission.
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View Solution$A$ nuclear reactor generates power by fission of ${}_{92}U^{235}$ into two equal fragments of ${}_{46}Pd^{116}$ with the emission of two $\gamma$-rays of $5.2 \text{ MeV}$ each and three neutrons. The average binding energies per nucleon of ${}_{92}U^{235}$ and ${}_{46}Pd^{116}$ are $7.2 \text{ MeV}$ and $8.2 \text{ MeV}$ respectively. The usable energy released per fission event is ......... $\text{MeV}$.
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View SolutionFind the value of $x$ and $y$ in the following nuclear reaction:
${}_{92}^{235} U + {}_{0}^{1} n \rightarrow {}_{x}^{133} Sb + {}_{41}^{y} Nb + 4 {}_{0}^{1} n$
${}_{92}^{235} U + {}_{0}^{1} n \rightarrow {}_{x}^{133} Sb + {}_{41}^{y} Nb + 4 {}_{0}^{1} n$
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 that of $^{236}U$ is $7.6\, MeV$. The total energy liberated will be about . . . . . . $MeV$.
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