1000 Solved Problems in Modern Physics

8.3 Solutions 465

8.54 Inelastic scattering is like an endoergic reaction except that the identity of
particles is unchanged.

Q=Eb

(

1 +

mb

my

)

−Ea

(

1 −

ma

my

)

∵θ= 90 ◦

SubstituteQ=− 4. 4 ,ma= 1 ,mb= 1 ,my= 12 , Ea=15 MeV to obtain

Eb= 8 .63 MeV

8.55 In the head-on collision the proton will receive full energy of the incident
neutron, that is 5MeVand 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 nowαwould mean

α=

hνo

Mpc^2

=

Eo

Mc^2

.

T=

αEo(1−cosθ)

1 +α(1−cosθ)

Tmax=

2 αEo

1 + 2 α

=

2 Eo

2 +Mc^2

/

Eo

=5MeV

where we have putθ = 180 ◦. PuttingM= 938 MeV, the above equation is

easily solved to yieldEo=hν=51 MeV.

8.56 (a)Q=Eb

(

1 +

mb

my

)

−Ea

(

1 −

ma

my

)

−

2

my

√

mambEaEbcosθ

Herema=mb= 1 ,my= 10 ,Ea= 5 ,Eb=3 andθ= 450. Substituting

these values, we findQ=− 1 .75 MeV. Thus, the excitation energy of^10 B

is 1.75 MeV.

(b) For elastic scatteringQ=0. Substituting the necessary values in the above

equation which is quadratic in

√

Eb, we find

√

Eb= 2 .171 so thatEb=

4 .715 MeV.

Thus the expected energy of elastically scattered protons will be 4.715 MeV.

8.57 For the reactionX(a,b)Y,

Q=Eb

(

1 +

mb

mY

)

−Ea

(

1 −

ma

mY

)

−

2

mY

√

mambEaEbCosθ

Hereb=p,a=d,Y=^28 Al,θ= 90 o

5. 5 =Ep

(

1 +

1

28

)

− 2. 1

(

1 −

2

28

)

Ep= 7 .19 MeV