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A fission reaction is given by .(92)^(23...

A fission reaction is given by `._(92)^(236)U rarr ._(54)^(140)Xe + ._(54)^(140)Xe + ._(38)^(94)Sr + x +y`, where `x` and `y` are two particles. Considering `._(92)^(236)U` to be at rest, the kinetic energies of the products are denoted by `K_(xe),K_(Sr),K_(x)(2MeV)` and `K_(y)(2Me V)`, respectively. Let the binding energies per nucleon of `._(92)^(236)U, ._(54)^(140)Xe` and `._(38)^(94)Sr` be `7.5 MeV, 8.5 Me V` and `8.5 MeV`, respectively. Considering different conservation laws, the correct options `(s)` is (are)

A

`x = n, y = n, K_(Sr) = 129 MeV, K_(xe) = 86 MeV`

B

`x = p, y = e^(-), K_(Sr) = 129 MeV, K_(xe) = 86 MeV`

C

`x = p, y = n, K_(Sr) = 129 MeV, K_(xe) = 86 MeV`

D

`x = n, y = n, K_(Sr) = 86 MeV, K_(xe) = 129 MeV`

Text Solution

Verified by Experts

The correct Answer is:
A
`U rarr Xe+Sr+underset(2)(x)+underset(2)(y)`
`Q=4+K_(xe)+K_(sr)` …(i)
`-Q=E_(B)=236xx7.5-140xx8.5-94xx8.5`
`:. Q=219` …(ii)
`:. K_(xe)+K_(sr)=215 MeV`
Since, both x & y have same KE
`:.` both particles should have same mass & lighter body will have higher KE.
`:.` Ans. (A)
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