Question

a neutron moving with velocity v collides with a stationary alpha particle . the velocity of neutron after collision [assume the collision tobe perfectly elastic]...

Answer 1



Q value of reaction is

Q=2Kα−KpQ=2Kα−Kp

Therefore , Q=(2×4×7.06−7×5.6)Q=(2×4×7.06−7×5.6)

=17.28MeV=17.28MeV

Where KpKp & KαKα are kinetic energies of photon and αα particle respectively . By conservation of linear momentum
Ko
2mpKp−−−−−−√=22mαKα−−−−−−√cosθ2mpKp=22mαKαcosθ

Kp=16Kαcos2θ=KαKp=16Kαcos2θ=Kα

Answer 2

mass of alpha particle is 4 times to neutron
according to the condition alpha paticle is at rest so its velocity is zero
now the final velocity of neutron is calculated from this formula
v' = (m1-m2÷ m1+m2) v1
by putting
v' = (1-4÷1+4) v
= -3/5 v
Negative sign shows reverse direction of neutro.