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JEE Advanced · Physics · 23. EM Waves

A fission reaction is given by U92236Xe54140+Sr3894+x+y , where x and y are two particles. Considering U92236 to be at rest, the kinetic energies of the products are denoted by KXe, KSr, Kx(2 MeV) and Ky2 MeV , respectively. Let the binding energies per nucleon of U92236, Xe54140  and Sr3894 be 7.5 MeV, 8.5 MeV, and 8.5 MeV respectively. Considering different conservation laws, the correct option(s) is (are)

  1. A \(x=n, y=n, K_{S r}=129 MeV , K_{X e}\)\(=86 MeV\)
  2. B \(x=p, y=e^{-}, K_{S r}=129 MeV , K_{X e}\)\(=86 MeV\)
  3. C \(x=p, y=n, K_{S r}=129 MeV , K_{X e}\)\(=86 MeV\)
  4. D \(x=n, y=n, K_{S r}=86 MeV , K_{X e}\)\(=129 MeV\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(x=n, y=n, K_{S r}=129 MeV , K_{X e}\)\(=86 MeV\)

Step-by-step Solution

Detailed explanation

UXe+Sr+x+2 y2
Q=4+KXe+KSr ...(i)
\(-Q=E_B=236 \times 7.5-140 \times 8.5\)\(-94 \times 8.5\)
Q=219 ...(ii)
KXe+KSr=215 MeV
Since, both x & y have same KE
both particles should have same mass & lighter body will have higher KE.
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