8.3 Solutions
465
8.54 Inelastic scattering is like an endoergic reaction except that the identity of
particles is unchanged.
Q = E b
1 +
m b
m y
− E a
1 −
m a
m y
∵ θ = 90
◦
Substitute Q = −4.4, m a = 1, m b = 1, m y = 12, E a = 15 MeV to obtain
E b = 8.63 MeV
8.55 In the head-on collision the proton will receive full energy of the incident
neutron, that is 5 MeV and neutron will stop. If the neutron was replaced by
the gamma ray then the proton will be emitted in the forward direction as
before but the gamma ray will be scattered backward at 180
◦ . The kinetic
energy imparted to the proton can be found out by the use of the formula
employed for Compton scattering except now α would mean
α =
hν o
M p c 2 =
E o
Mc
2
.
T =
α E o (1 − cos θ)
1 + α(1 − cos θ)
T max =
2α E o
1 + 2α
=
2E o
2 + Mc
2
E o
= 5 MeV
where we have put θ = 180
◦ . Putting M = 938 MeV, the above equation is
easily solved to yield E o = hν = 51 MeV.
8.56 (a) Q = E b
1 +
m b
m y
− E a
1 −
m a
m y
−
2
m y
√
m a m b E a E b cos θ
Here m a = m b = 1, m y = 10, E a = 5, E b = 3 and θ = 45
0 . Substituting
these values, we find Q = −1.75 MeV. Thus, the excitation energy of
10 B
is 1.75 MeV.
(b) For elastic scattering Q = 0. Substituting the necessary values in the above
equation which is quadratic in
√
E b , we find
√
E b = 2.171 so that E b =
4.715 MeV.
Thus the expected energy of elastically scattered protons will be 4.715 MeV.
8.57 For the reaction X (a, b)Y ,
Q = E b
1 +
m b
m Y
− E a
1 −
m a
m Y
−
2
m Y
m a m b E a E b Cosθ
Here b = p, a = d, Y =
28
Al, θ = 90
o
5.5 = E p
1 +
1
28
− 2.1
1 −
2
28
E p = 7.19 MeV
465
8.54 Inelastic scattering is like an endoergic reaction except that the identity of
particles is unchanged.
Q = E b
1 +
m b
m y
− E a
1 −
m a
m y
∵ θ = 90
◦
Substitute Q = −4.4, m a = 1, m b = 1, m y = 12, E a = 15 MeV to obtain
E b = 8.63 MeV
8.55 In the head-on collision the proton will receive full energy of the incident
neutron, that is 5 MeV and neutron will stop. If the neutron was replaced by
the gamma ray then the proton will be emitted in the forward direction as
before but the gamma ray will be scattered backward at 180
◦ . The kinetic
energy imparted to the proton can be found out by the use of the formula
employed for Compton scattering except now α would mean
α =
hν o
M p c 2 =
E o
Mc
2
.
T =
α E o (1 − cos θ)
1 + α(1 − cos θ)
T max =
2α E o
1 + 2α
=
2E o
2 + Mc
2
E o
= 5 MeV
where we have put θ = 180
◦ . Putting M = 938 MeV, the above equation is
easily solved to yield E o = hν = 51 MeV.
8.56 (a) Q = E b
1 +
m b
m y
− E a
1 −
m a
m y
−
2
m y
√
m a m b E a E b cos θ
Here m a = m b = 1, m y = 10, E a = 5, E b = 3 and θ = 45
0 . Substituting
these values, we find Q = −1.75 MeV. Thus, the excitation energy of
10 B
is 1.75 MeV.
(b) For elastic scattering Q = 0. Substituting the necessary values in the above
equation which is quadratic in
√
E b , we find
√
E b = 2.171 so that E b =
4.715 MeV.
Thus the expected energy of elastically scattered protons will be 4.715 MeV.
8.57 For the reaction X (a, b)Y ,
Q = E b
1 +
m b
m Y
− E a
1 −
m a
m Y
−
2
m Y
m a m b E a E b Cosθ
Here b = p, a = d, Y =
28
Al, θ = 90
o
5.5 = E p
1 +
1
28
− 2.1
1 −
2
28
E p = 7.19 MeV