Nuclear Physics and RadioactivityHard
Question
A fission reaction is given by 23692U → 14054Xe + 9438 Sr + 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, Ksr ,Kx (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 options(s) is (are) :-
Options
A.x = n, y = n, Ksr = 129 MeV, Kxe = 86 MeV
B.x = p, y = e-, KSr = 129 MeV, Kxe = 86 MeV
C.x = p, y = n, KSr = 129 MeV, Kxe = 86 MeV
D.x = n, y = n, KSr = 86 MeV, Kxe = 129 MeV
Solution
ighter body will have higher KE.U → Xe + Sr + x + y
2 2
Q = 4 + Kxe + Ksr ... (i)
- Q = EB = 236 × 7.5 - 140 × 8.5 - 94 × 8.5
∴ Q = 219 ... (ii)
∴ Kxe + Ksr = 215 MeV
Since, both x & y have same KE
∴ both particles should have same mass & l
∴ Ans. (A)
2 2
Q = 4 + Kxe + Ksr ... (i)
- Q = EB = 236 × 7.5 - 140 × 8.5 - 94 × 8.5
∴ Q = 219 ... (ii)
∴ Kxe + Ksr = 215 MeV
Since, both x & y have same KE
∴ both particles should have same mass & l
∴ Ans. (A)
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