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Question

A fission reaction is given by 23692U14054Xe+9438Sr+x+y, where x and y are two particles. Considering 23692U to be at rest, the kinetic energies of the products are denoted by KXe,KSrKx(2MeV) and Ky(2MeV), respectively. Let the binding energies per nucleon of 23692U,14054Xe and 9438Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is (are)

A
x=n,y=n,KSr=129MeV,K_{Xe} = 86 MeV$
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B
x=p,y=e,KSr=129MeV, K_{Xe} = 86 MeV$
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C
x=p,y=n,KSr=129MeV, K_{Xe} = 86 MeV$
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D
x=n,y=n,KSr=86MeV, K_{Xe} = 129 MeV$
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Solution

The correct option is A x=n,y=n,KSr=129MeV,K_{Xe} = 86 MeV$
23692U14054Xe+9438Sr+x+y

From conservation of charge, the combined charges of x and y should be zero since LHS has a charge of 92 and RHS has already charge of 92.
Hence, option(C) is ruled out since combined charge of p and n is not zero.
Change in BE=236×7.5+140×8.5+94×8.5=219Mev
x and y share 4 MeV together.
KE of Xe and Sr combined =2194=215 MeV
This difference is equal to combined KE of the products from fission.
Momentum is conserved and is zero since initial momentum is zero.

Case A:
Both x and y have the same KE. They will have same momentum since masses are same.
Sr is lighter, hence should have more KE.
This is satisfied.

Case D:
Sr has less KE, it is not possible.

Case B and C: Since the number of electron and proton before reaction and after reaction should be same but if x is proton and y is electron, law of conservation of protons and electrons changes. Same is the case with option C.
Hence option B and C is incorrect.

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