bei48482_FM

Since p m for the electron and positron,

h 2
mc^2 cos

Because c 1 and cos 1,

h 

2

 mc^2

But conservation of energy requires that h 2

 mc^2. Hence it is impossible for pair produc-

tion to conserve both energy and momentum unless some other object is involved in the process

to carry away part of the initial photon momentum.

Example 2.6

An electron and a positron are moving side by side in the xdirection at 0.500cwhen they an-

nihilate each other. Two photons are produced that move along the xaxis. (a) Do both photons

move in the xdirection? (b) What is the energy of each photon?

Solution

(a) In the center-of-mass (CM) system (which is the system moving with the original particles),

the photons move off in opposite directions to conserve momentum. They must also do so in

the lab system because the speed of the CM system is less than the speed cof the photons.

(b) Let p 1 be the momentum of the photon moving in the xdirection and p 2 be the momen-

tum of the photon moving in the xdirection. Then conservation of momentum (in the lab

system) gives

p 1 p 2  2

 m 

0.590 MeV/c

Conservation of energy gives

p 1 c p 2 c 2

 mc^2 1.180 MeV

and so p 1  p 2 1.180 MeV/c

Now we add the two results and solve for p 1 and p 2 :

(p 1 p 2 )(p 1 p 2 ) 2 p 1 (0.5901.180) MeV/c

p 1 0.885 MeV/c

p 2 (p 1 p 2 )p 1  0.295 MeV/c

The photon energies are accordingly

E 1 p 1 c0.885 MeV E 2 p 2 c0.295 MeV

Photon Absorption

The three chief ways in which photons of light, x-rays, and gamma rays interact with

matter are summarized in Fig. 2.27. In all cases photon energy is transferred to elec-

trons which in turn lose energy to atoms in the absorbing material.

2(0.511 MeV)



 1 (0.500)^2

2 mc^2



 1  2 c^2

2(0.511 MeV/c^2 )(c^2 )(0.500c)c^2



 1 (0.500)^2

2(mc^2 )( c^2 )



 1 ^ c^2



c

82 Chapter Two

bei48482_ch02.qxd 1/16/02 1:53 PM Page 82