Question

Consider the nuclear reaction $X^{200}\to A^{110}+B^{80}.$ If the binding energy per nucleon for $X,A$ and $B$ are $7.4\text{MeV},8.2\text{MeV}$ and $8.1\text{MeV}$ respectively, then the energy released in the reaction is

• A. $70\text{MeV}$
• B. $200\text{MeV}$
• C. $190\text{MeV}$
• D. $10\text{MeV}$

Answer 1

The correct option is A $70\text{MeV}$

For X :
Binding energy $=200\times 7.4=1480\text{MeV}$
For A :
Binding energy $=110\times 8.2=902\text{MeV}$
For B :
Binding energy $=80\times 8.1=648\text{MeV}$
By energy conservation
$\therefore$ Energy released= BE of Products – BE of Reactants$=(902+648)–1480=1550–1480=70\text{MeV}$