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