Q. 274.5( 4 Votes )
Consider the fission 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 14058Ce and 9944Ru. Calculate Q for this fission process. The relevant atomic and particle masses are
m(23892U) = 238.05079 u
m(14058Ce) = 139.90543 u
m(9944Ru) = 98.90594 u
Answer :
Given that,
Mass of a nucleus 
, m1 = 238.05079 u
Mass of a nucleus 
, m2 = 139.90543 u
Mass of a nucleus 
, m3 = 98.90594 u
Mass of a neutron 
, m4 = 1.008665 u
In the fission of 
, 10 beta particles are emitted. The nuclear reaction is,
Energy released during above nuclear fission is given by,
Where,
m’ represents the corresponding atomic masses of the nuclei
= m1 – 92me (∵ atomic number 92)
= m2 – 58me (∵ atomic number 58)
= m3 – 44me (∵ atomic number 44)
= m4
∴ 
= 
= (0.247995)u×c2
But, u = 931.5 MeV/c2
∴ Q = 0.247995×931.5 Mev
= 231.007 MeV
The energy released during given fission reaction is 231.007 MeV.
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