a nuetron travelling with velocity v collides elasticaly with the nucleus of an atom of mass number A at rest the fraction of total energy retained by nurtron
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1
let mass no neutron = Mn
mass no proton = Mp
we know,
mass no = mass no of proton + mass no of neutron = Mp + Mn
we also know,
mass of proton ≈ mass of neutron =m(let)
so, generally we can say that ,
mass of nucleons =2Am
====================================
collision is elastic so,
linear momentum conserve , and energy will also conserve.
from law of conservation of linear momentum ,
mV + 2Am×0 = mV1 + 2AmV2
V = V1 +2AV2 --------(1)
from law of conservation of energy,
Kinetic energy initial = kinetic energy final
1/2mv² + 1/2(2Am)x 0= 1/2mV1² + 1/2(2Am)V2²
V² =V1² +2AV² ---------(2)
(V1+2AV2)² = V1² +2AV2²
4A²V2² +4AV1.V2 =2AV2²
V2² (2A² -A) +2AV1.V2 =0
V2(2A -1) +2V1 =0
V2(2A -1)= -2V1
V2 = -2/(2A-1)V1
put this in equation (1)
V =V1 -4A/(2A-1)V1
=V1(2A-1-4A)/(2A-1)
=V1(1+2A)/(1-2A)
V1 =(1-2A)/(1+2A)V
V2 =2/(1+2A)V
hence,
neutron have retain energy = 1/2m(1-2A)²/(1+2A)²V²
initial energy of neutron = 1/2mV²
so,
fraction of energy retained by neutrons =
final energy of neutrons /initial energy of neutrons
=(1-2A)²/(1+2A)²
mass no proton = Mp
we know,
mass no = mass no of proton + mass no of neutron = Mp + Mn
we also know,
mass of proton ≈ mass of neutron =m(let)
so, generally we can say that ,
mass of nucleons =2Am
====================================
collision is elastic so,
linear momentum conserve , and energy will also conserve.
from law of conservation of linear momentum ,
mV + 2Am×0 = mV1 + 2AmV2
V = V1 +2AV2 --------(1)
from law of conservation of energy,
Kinetic energy initial = kinetic energy final
1/2mv² + 1/2(2Am)x 0= 1/2mV1² + 1/2(2Am)V2²
V² =V1² +2AV² ---------(2)
(V1+2AV2)² = V1² +2AV2²
4A²V2² +4AV1.V2 =2AV2²
V2² (2A² -A) +2AV1.V2 =0
V2(2A -1) +2V1 =0
V2(2A -1)= -2V1
V2 = -2/(2A-1)V1
put this in equation (1)
V =V1 -4A/(2A-1)V1
=V1(2A-1-4A)/(2A-1)
=V1(1+2A)/(1-2A)
V1 =(1-2A)/(1+2A)V
V2 =2/(1+2A)V
hence,
neutron have retain energy = 1/2m(1-2A)²/(1+2A)²V²
initial energy of neutron = 1/2mV²
so,
fraction of energy retained by neutrons =
final energy of neutrons /initial energy of neutrons
=(1-2A)²/(1+2A)²
duragpalsingh:
nice
Answered by
2
M1 = m = mass of neutron.
M2 = mass of nucleus of atom = 2A * m approximately.
(assuming number of neutrons = number of protons in the nucleus)
u1 = v , u2 = 0 , v1 = ? v2 = ?
Collision is elastic. So energy is conserved. As M2 is much heavier than M1, the neutron will rebound back. So v1 will be negative.
Conservation of Linear momentum:
M1 u1 + M2 u2 = M1 v1 + M2 v2
m v = m v1 + 2Am v2
v1 + 2A v2 = v ----- (1)
Conservation of energy:
1/2 M1 u1² + 1/2 M2 u2² = 1/2 M2 v1² + 1/2 M2 v2²
m v² = m v1² + 2A m v2²
v1² + 2A v2² = v² ---- (2)
=> 2A v2² = (v - v1) (v+v1) = 2A v2 (v+ v1)
v2 = 2v/(2A+1)
v1 = -(2A-1) v / (2A+1)
Fraction of KE of M1 after collision = 1/2 m v1² / [1/2 m v²]
= v1²/v²
= (2A-1)² / (2A+1)²
We can verify easily, if A =0, then fraction is 1. No collision.
A = 1/2, equal masses... So v1 = 0.
A=1, atom is twice as heavy. Fraction is 1/9 ie., v1 = v/3.
M2 = mass of nucleus of atom = 2A * m approximately.
(assuming number of neutrons = number of protons in the nucleus)
u1 = v , u2 = 0 , v1 = ? v2 = ?
Collision is elastic. So energy is conserved. As M2 is much heavier than M1, the neutron will rebound back. So v1 will be negative.
Conservation of Linear momentum:
M1 u1 + M2 u2 = M1 v1 + M2 v2
m v = m v1 + 2Am v2
v1 + 2A v2 = v ----- (1)
Conservation of energy:
1/2 M1 u1² + 1/2 M2 u2² = 1/2 M2 v1² + 1/2 M2 v2²
m v² = m v1² + 2A m v2²
v1² + 2A v2² = v² ---- (2)
=> 2A v2² = (v - v1) (v+v1) = 2A v2 (v+ v1)
v2 = 2v/(2A+1)
v1 = -(2A-1) v / (2A+1)
Fraction of KE of M1 after collision = 1/2 m v1² / [1/2 m v²]
= v1²/v²
= (2A-1)² / (2A+1)²
We can verify easily, if A =0, then fraction is 1. No collision.
A = 1/2, equal masses... So v1 = 0.
A=1, atom is twice as heavy. Fraction is 1/9 ie., v1 = v/3.
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