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Atoms & Nuclei

JEE Main level, single correct, medium

203 of 239

Question 203 of 239

Nucleus $A$ is having mass number $220$ and its binding energy per nucleon is $5.6MeV$ . It splits in two fragments $B$ and $C$ of mass numbers $105$ and $115$ . The binding energy of nucleons in $B$ and $C$ is $6.4MeV$ per nucleon. The energy $Q$ released per fission will be:

Show answer and solution

Answer: (D) $176MeV$

The total binding energy of nucleus

A

is,

{E}_{A}=220\times 5.6=1232MeV

The total binding energy of

B

and

C

combined is,

{E}_{B}+{E}_{C}=105\times 6.4+115\times 6.4=1408MeV

Therefore, the energy released per fission is,

Q={E}_{B}+{E}_{C}-{E}_{A}=1408-1232=176MeV

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