Question

For the following fission reaction
${ }^{235} U+n \rightarrow{ }^{140} C e+{ }^{94} Z r+2 n$,
Find the disintegration energy.
$\left(M_{U}=23502 \,u, M_{n}=1.0 \,u, M_{C e}=139.9 \,u, M_{Z r}=93.9\, u\right)$

• A. 205 MeV
• B. 198 MeV
• C. 123 MeV
• D. 89 MeV

Answer 1

Answer: (A)

The reaction is, ${ }^{235} U +n \rightarrow{ }^{140} Ce +{ }^{94} Zr +2 n$
The mass lost in this reaction (mass defect) is
$\Delta m=M_{U}+M_{n}-\left(M_{C e}+M_{Z r}+2 M_{n}\right)$
$\Delta m=[235.02+1.0-(139.9+93.9+2)] u$
$\Delta m=(236.02-235.8) u$
$\Delta m=0.22 \,u$
$[\because 1 \,amu =931\, MeV ]$
$\therefore $ The disintegration energy
$\Delta E=0.22 \times 931=204.82 \,MeV$
$\cong 205.00 \,MeV$