Atoms & Nuclei

JEE Main level, numerical answer, medium

199 of 239

Question 199 of 239

A nucleus with mass number 242242 and binding energy per nucleon as 7.6MeV7.6MeV breaks into two fragment each with mass number 121121. If each fragment nucleus has binding energy per nucleon as 8.1MeV,8.1MeV, the total gain in binding energy is _______MeV.

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Answer: 121

Binding energy is given by E=∆mc2E=∆m{c}^{2}
where ∆m∆m is the mass defect.
The energy per nucleon of the nucleus having mass number 242242 is 7.6MeV7.6MeV.
The initial binding energy is,
BE=242×7.6MeVBE=242\times 7.6MeV
The energy per nucleon of the nucleus with mass number 121121 is 8.1MeV8.1MeV.
Therefore, binding energy is
BE′=2(121×8.1MeV)BE'=2\left(121\times 8.1MeV\right)
The gain in the binding energy is
BE′−BE=(8.1–7.6)×242MeV=121MeVBE'-BE=(8.1–7.6)\times 242MeV=121MeV
=121MeV=121MeV

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