Question

For the following fission reaction
${}^{235}U+n\to {}^{140}Ce+{}^{94}Zr+2n$,
Find the disintegration energy.
$\left(M_U=23502u,M_n=1.0u,M_{Ce}=139.9u,M_{Zr}=93.9u\right)$

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

Answer 1

Answer: (A)

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