Question

Consider the nuclear reaction $X^{200}\to A^{110}+B^{90}$ if the binding energy per nucleon for X, A and B is 7.4 MeV, 8.2 MeV and 8.2 Mev respectively, what is the energy released?

• A. 200 MeV
• B. 160 MeV
• C. 110 MeV
• D. 90 MeV

Answer 1

The correct option is B 160 MeV

Total energy of $X=2\omega \times 7.4MeV$

Total energy of $A=110\times 8.2MeV$

Total energy of $B=90\times 78.2MMeV$

So, $2\omega \times 7.4=(110\times 8.2)+(90\times 8.2)$

Hence $-$energy $=-2\omega \times 7.4+(110\times 8.2)+(90\times 8.2)=160MeV$