Consider the fission of `._92^238U` by fast neutrons. In one fission event , no neutrons are emitted and the final stable end products, after the beta decay of the primary fragments , are `._58^140Ce` and `._44^99Ru`. Calculate Q for this fission process.
Given , `m(._92^238U)=238.05079 u, m(._58^140 Ce)` = 139.90543 u
`m(._44^99Ru)` = 98.90594 u, `m_n`=1.008667 u

Explanation

The fission reaction is `._92^238 U + ._0^1n overset(-beta)to ._58^140 Ce + ._44^99Ru +Q`
`therefore Q=[m(._92^238U) + m(._0^1n) -m(._58^140Ce)-m(._44^99Ru)]u`
= [ 238.05079 + 1.00867 - 139.90543 - 98.90594 x 931.2 MeV
=231.1 MeV