A nucleus with mass number 242 and binding energy per nucleon as 7 . 6   MeV breaks into two fragment…

A nucleus with mass number 242 and binding energy per nucleon as 7.6 MeV breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as 8.1 MeV, the total gain in binding energy is _______ MeV.

Solution

Binding energy is given by E=mc2

where m is the mass defect.

The energy per nucleon of the nucleus having mass number 242 is 7.6 MeV.

The initial binding energy is,

BE=242×7.6 MeV

The energy per nucleon of the nucleus with mass number 121 is 8.1 MeV.

Therefore, binding energy is

BE'=2121×8.1 MeV

The gain in the binding energy is 

BE'-BE = (8.1  7.6) × 242 MeV=121 MeV

= 121 MeV

Asked in: JEE Main 2023 (08 Apr Shift 1)

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