Question

Suppose a nucleus with mass number A = 240 and `(B.E.)/A` = 7.6 MeV, breaks into two nuclei, each of mass number A = 120 with `(B.E.)/A` = 8.5 MeV. Calculate the energy released in the process.

Answer 1

First, determine the total binding energy `(BE_"initial")` of the original nucleus using its mass number (A = 240) and its binding energy per nucleon (7.6 MeV):

`(BE_"initial")` = `A xx ((BE)/A)`

= 240 × 7.6

= 1824 MeV

Next, calculate the total binding energy `(BE_"final")` of the two daughter nuclei. Each fragment has a mass number of 120 and a binding energy per nucleon of 8.5 MeV:

`(BE_"final")` = 2(120 × 8.5)

= 2 × 1020

= 2040 MeV

The energy released is the difference between the final and initial total binding energies:

Energy released = `BE_"final" - BE_"initial"`

= 2040 − 1824

= 216 MeV

The energy released in this nuclear fission process is 216 MeV.