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Standard XII

Physics

Nuclear Fission

Consider the ...

Question

Consider the fission reaction
${_{92} U}^{236} ⟶ X^{117} + Y^{117} + {_{0} n}^{1} + {_{0} n}^{1}$
ie., two nuclei of same mass number $117$ are formed plus two neutrons. The binding energy per nucleon of $X$ and $Y$ is $8.5 M e V$ where as of $U^{236}$ is $7.6 M e V$ . Find the total energy liberated.

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Solution

$_{92}^{236} U \to K^{117} + Y^{117} +_{0}^{1} n +^{1} n + Q$
Q: energy released
Binding energy $_{92}^{236} U = 7.6 \times 236 = 1793.6 M e V$
Binding energy per nuclear of $K^{117} = 8.5 \times 117 = 994.5$
Binding energy per nucler of $Y^{117} = 8.5 M e V$
Total binding energy $Y^{117} = 8.5 \times 117 = 994.5 M e V$
Substituting energy values in equation we get,
$1793.6 = 994.5 + 994.5 + Q$
$Q = (-) 195.4 M e V$
As the Q-value is negative the fission is not energetically possible.

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Consider the fission reaction92U236⟶X117+Y117+0n1+0n1ie., two nuclei of same mass number 117 are formed plus two neutrons. The binding energy per nucleon of X and Y is 8.5MeV where as of U236 is 7.6MeV. Find the total energy liberated.