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Consider the fission of _238 ^92 U by fast neutrons. In one fission event no neutrons are emitted and the final end products ,after the beta decay of the primary fragments, are _140 ^58 Ce and _99 ^44 Ru. Calculate Q for this fission process.

Q.13.27 Consider the fission of _{92}^{238}\textrm{U} by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are _{58}^{140}\textrm{Ce} and _{44}^{99}\textrm{Ru}. Calculate Q   for this fission process. The relevant atomic and particle masses are

                   m(_{92}^{238}\textrm{U})=238.05079\; u

                   m(_{58}^{140}\textrm{Ce})=139.90543\; u

                    m(_{44}^{99}\textrm{Ru})= 98.90594\; u

 

Answers (1)
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S Sayak

The fission reaction given in the question can be written as

_{92}^{238}\textrm{U}+_{0}^{1}\textrm{n}\rightarrow _{58}^{140}\textrm{Ce}+_{44}^{99}\textrm{Ru}+10e^{-}

The mass defect for the above reaction would be

\Delta m=m_{N}(_{92}^{238}\textrm{U})+m(_{0}^{1}\textrm{n})-m_{N}(_{58}^{140}\textrm{Ce})-m_{N}(_{44}^{99}\textrm{Ce})-10m_{e}

In the above equation, mN represents nuclear masses

\\\Delta m=m(_{92}^{238}\textrm{U})-92m_{e}+m(_{0}^{1}\textrm{n})-m(_{58}^{140}\textrm{Ce})+58m_{e}-m(_{44}^{99}\textrm{Ru})+44m_{e}-10m_{e} \\\Delta m=m(_{92}^{238}\textrm{U})+m(_{0}^{1}\textrm{n})-m(_{58}^{140}\textrm{Ce})-m(_{44}^{99}\textrm{Ru})\\ \Delta m=238.05079+1.008665-139.90543-98.90594\\ \Delta m=0.247995u

but 1u =931.5 MeV/c2

Q=0.247995\times931.5

Q=231.007 MeV

Q value of the fission process is 231.007 MeV

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