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Consider the fission of .(92)^(238)U by ...

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

Text Solution

Verified by Experts

For this fission reaction,
`._(92)U^(238)+ ._(0)n^(1)to ._(58)Ce^(140)+ ._(44)Ru^(99)+Q`
mass defect `Delta M =` mass of `U^(238)+` mass of n - (mass of `Ce^(140)+` mass of `Ru^(99)`)
`= 238.05079+1.00867-(139.90543+98.90594)`
= 0.24809 u
`therefore Q = 0.24809xx931 MeV = 230.97 MeV`
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