Consider the fission `._(92)U^(238)` by fast neutrons. In one fission event, no neutrons are emitted and the final stable and products, after the beta decay of the primary fragments are `._(58)Ce^(140)` and `._(44)Ru^(99)`. Calculate Q for this fission process, The relevant atomic and particle masses are:
`m(._(92)U^(238))=238.05079u, m(._(58)Ce^(140))=139.90543 u, m(._(34)Ru^(99))=98.90594 u`

Explanation

In the fission of `""_(92)^(238) U,10 beta^(-)` particles decay from the parent nucleus. The nuclear reaction can be written as:
`""_(92)^(238)U+""_(0)^(1)n rarr ""_(58)^(140) Ce +""_(454)^(99)Ru+10 ""_(-1)^(0)e`
It is given that:
Mass of a nucleus `""_(92)^(238)U,m_(1)=238.05079 u`
Mass of a nucleus `""_(58)^(140)Ce,m_(2)=139.90543 u`
Mass of a nucleus `""_(44)^(99)Ru,m_(3)=98.90594 u`
Mass of a nutron `""_(0)^(1)n,m_(4)=1.008665u`
Q-value of the above equation,
`Q=[m'(""_(92)^(238)U)+m(""_(0)^(1)n)-m'(""_(58)^(140)Ce)-m'(""_(44)^(99)Ru)-10m_(e)]c^(2)`
Where,
m'=Represents the corresponding atomic masses of the nuclei
`m'(""_(92)^(238)U)=m_(1)-92m_(e)`
`m'(""_(58)^(140)Ce)=m_(2)-58m_(e)`
`m'(""_(44)^(99)Ru)=m_(3)-44m_(e)`
`m(""_(0)^(1)n)=m_(4)`
`Q=[m_(1)-92m_(e)+m_(4)-m_(2)+58m_(e)-m_(3)+44m_(e)-10m_(e)]c^(2)`
`=[m_(1)+m_(4)-m_(2)-m_(3)]c^(2)`
`=[238.0507+1.008665-139.90543-98.90594]c^(2)`
`=[0.247995c^(2)]u`
But 1 u`=931.5 MeV//c^(2)`
`thereforeQ=0.247995xx931.5=231.007 MeV`
Hence, the Q-value of the fission process i s231.007 MeV.