Question

Consider a fission reaction
${}_{92}{U}^{236}\to X^{117}+Y^{117}{+}_0{n}^1{+}_{0}n1$
ie., two nuclei of same mass number 117 are formed plus two neutrons. The binding energy per nucleon of X and Y is 8.5MeV where as of $U^{236}$ is 7.6MeV. The total energy liberated will be about-

• A. 2 MeV
• B. 20 MeV
• C. 200 MeV
• D. 2000MeV

Answer 1

The correct option is C 200 MeV

$\begin{array}{c}\text{Given-}\ast \text{Reaction -}\\ {}_{92}{U}^{236}⟶X^{117}+Y^{117}{+}_0{n}^1{+}_0{n}^1\end{array}$

$\begin{array}{c}\text{Binding energy BE}\\ BE=BE\text{per nucleons}X\text{Num of nucleons.}\\ \begin{array}{cc}Q& =\mid B{E}_{\text{Reactant}}-BE\text{product}\mid \\ & =236\times 8.5-(117\times 7.6+117\times 7.6)\text{MeV}\\ & =195.5MeV\\ \text{which is about}200MeV\text{among all other options.}& \\ \text{Hence, option (c) is correct.}& \end{array}\end{array}$