Question 25.9: Fission Energy Find the energy released for the fission proc...
Fission Energy
Find the energy released for the fission process given in the text, with uranium splitting into molybdenum and tin. The relevant atomic masses are: M\left({ }^{235} U \right)=235.0439 u ; M\left({ }^{102} Mo \right)=101.9103 u ; M\left({ }^{131} Sn \right)= 130.9169 u.
ORGANIZE AND PLAN The energy equivalent of the mass defect is
\Delta m c^{2}=\left[M\left({ }^{1} n\right)+M\left({ }^{235} U \right)-\left(M\left({ }^{102} Mo \right)\right.\right. \left.\left.+M\left({ }^{131} Sn \right)+3 M\left({ }^{1} n\right)\right)\right] c^{2}.
In addition to the masses given, we’ll need the neutron mass and the conversion between u.c² and MeV.
\text { Known: } M\left({ }^{1} n\right) \equiv 1.0087 u ; u \cdot c^{2} \equiv 931.5 MeV.
Using the known masses,
\Delta m c^{2}=[1.0087 u +235.0439 u -(101.9103 u +130.9169 u +3(1.0087 u ))] c^{2}=0.1993 u \cdot c^{2}.
Converting to MeV,
\Delta m c^{2}=0.1993 u \cdot c^{2} \times \frac{931.5 MeV }{ u \cdot c^{2}}=186 MeV.
REFLECT This is close to Meitner’s 200-MeV estimate. The actual energy depends on the masses of the specific fission products. Most of this appears as kinetic energy of the fission products, with the rest as neutron kinetic energy and gamma radiation.