Looking for classes? Ksquare Career Institute, Bengaluru →
A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is:
A
9.4 MeV
B
804 MeV
C
216 MeV
D
0.9 MeV
Detailed Solution
$X^{240} \rightarrow Y^{120} + Z^{120}$
Gain in binding energy = BE of products – BE of reactant
$= (120 \times 8.5 \times 2) - (240 \times 7.6)$ MeV
$= 2040 - 1824 = 216$ MeV