JEE Advanced2015PhysicsElectromagnetic WavesActual
A fission reaction is given by U 92 236 → X e 54 140 + S r 38 94 + x + y , where x and y are two particles. Considering U 92 236 to be at rest, the kinetic energies of the products are denoted by K X e , K S r , K x ( 2 M e V ) and K y 2 M e V , respectively. Let the binding energies per nucleon of U 92 236 , X e 54 140 and S r 38 94 be 7.5 MeV, 8.5 MeV, and 8.5 MeV respectively. Considering different conservat
Options
- Ax = n , y = n , K S r = 129 M e V , K X e = 86 M e V
- Bx = p , y = e - , K S r = 129 M e V , K X e = 86 M e V
- Cx = p , y = n , K S r = 129 M e V , K X e = 86 M e V
- Dx = n , y = n , K S r = 86 M e V , K X e = 129 M e V
Correct answer
A. x = n , y = n , K S r = 129 M e V , K X e = 86 M e V
Step-by-step solution
U → X e + S r + x + 2 y 2 Q = 4 + K X e + K S r ...(i) - Q = E B = 236 × 7.5 - 140 × 8.5 - 94 × 8.5 ∴ Q = 219 ...(ii) ∴ K X e + K S r = 215 M e V Since, both x & y have same KE ∴ both particles should have same mass & lighter body will have higher KE.