Question

A nucleus with mass number A=240 and BE/A=7.6 MeV breaks into two fragments each of A=120 with BE/A=8.5 MeV. Calculate the released energy?

Or Calculate the energy in fusion reaction:
${}_2^{1}H{+}_2^{1}H{\to }_2^{3}He+n$ where BE of ${}_2^{1}H$=2.23 MeV and of ${}_3^{2}He$=7.73 MeV.

Answer 1

Let the parent nucleus be $X$ which breaks into two fragments $Y$

The nuclear reaction : ${}^{240}X\to 2$ ${}^{120}Y$

$\therefore$ Binding energy of $X$, $E_{B_X}=240\times 7.6=1824$ MeV

$\therefore$ Binding energy of $Y$, $E_{B_Y}=120\times 8.5=1020$ MeV

Energy released, $E=2\left(E_{B_Y}\right)-E_{B_X}=2\times 1020-1824=216$ MeV

OR

The fusion reaction : ${}_1^{2}H$ $+$ ${}_1^{2}H\to$ ${}_2^{3}He+n$

Energy released in this reaction, $E=E_{B_{He}}-2\left(E_{B_H}\right)$

$\therefore$ $E=7.73-2\times 2.23=3.27$ MeV