Question

A nuclear fission is given below $A^{240}\to B^{100}+C^{140}$ + $Q\left(energy\right)$ Let binding energy per nucleon of nucleus $A,B$ and $C$ is $7.6MeV,8.1MeV$ and $8.1MeV$ respectively. Value of $Q$ is: - (approximately)

• A. $20MeV$
• B. $220MeV$
• C. $120MeV$
• D. $240MeV$

Answer 1

The correct option is C $120MeV$

$\begin{array}{c}\text{Q value is}[\Delta m{]}_4\\ \therefore Q=140(8\cdot 1)+100(8\cdot 1)-240(7\cdot 6)\\ Q=240(8\cdot 1)-240(7\cdot 6)\\ Q=240\left(0.5\right)\\ Q=120Mev\end{array}$