Compton in Radioactivity class notes.ppt

Nuclear physics
Radioactivity class notes

1. Copyright © 2012 Pearson Education Inc.
PowerPoint
®
Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Photons: Light Waves
Behaving as Particles

2. Copyright © 2012 Pearson Education Inc.
The photoelectric effect
•Blue light striking a cesium target causes the cesium to emit electrons.
Red light does not, no matter how intense. This violates the wave-
description of light.
•Einstein’s explanation: Light comes in photons. To emit an electron, the
cesium atom must absorb a single photon whose energy exceeds the
ionization energy of the outermost electron in cesium. A blue photon has
enough energy;
a red photon does not.
0
eV hf 

3. Copyright © 2012 Pearson Education Inc.
Einstein’s explanation of the photoelectric effect
•A photon contains a discrete amount of
energy. For light of frequency f and
wavelength , this energy is E = hf or
E = (hc)/, where h is Planck’s constant
6.626 × 10
−34
J-s = 4.136 × 10
−15
eV-s.
•1 eV = 1.602 × 10
−19
J-s
•This explains how the energy of an emitted
electron in the photoelectric effect depends
on the frequency of light used.
•Different materials can have a different
threshold frequency, but once that threshold
is exceeded, the dependence is the same.
•The momentum of a photon of wavelength 
is p = E/c = h/. Can write it in terms of
wavenumber k = 2/, or wave-
vector , where
E h  
0
eV hf 
p k
p k
34
= 1.055 10 J-s
2
h


 

4. Copyright © 2012 Pearson Education Inc.
The photoelectric effect—examples
•Example 38.1—Laser-pointer photons.
Let’s explore this idea of photons and get an idea of the numbers of photons
in a typical laser pointer. Say a red laser pointer (wavelength  = 650
nm) has 0.5 mW of output power. Each photon has an energy E =
hc/, = 6.63×10
-34
*3×10
8
/6.50×10
-7
= 3.06×10
-19
J. Since 0.5 mW
= 5×10
-3
J/s, this is 5×10
-3
J/s / 3.06×10
-19
J/photon = 1.63×10
16

photons/s.
•Example 38.3—Determining  and h experimentally.
In a photoelectric experiment you measure stopping potentials V
0 for light
of three different wavelengths, 1.0 V for 600 nm, 2.0 V for 400 nm,
and 3.0 V for 300 nm. Determine the work function  and Planck’s
constant h.
To do this, we need the frequencies
Any two of these is enough for a straight line.
0
eV hf 
8
14 14 15
7
3 10
5 10 ;7.5 10 ;10 Hz
6,4,3 10
c
f



    

2 1 2 1
1 1 1 1
2 1 2 1
( ) ( )
;
e V V e V V
h hf eV f eV
f f f f

 
    
 

5. Copyright © 2012 Pearson Education Inc.
X-ray production
•The reverse process of the photoelectric effect—instead of a photon
being absorbed on interaction with matter and emitting an electron, we
have an electron interacting with matter and emitting a photon.
•An experimental arrangement for making x rays is shown in Figure
38.7 at lower left. The greater the kinetic energy of the electrons that
strike the anode, the shorter the minimum wavelength of the x rays
emitted by the anode (see Figure 38.8 at lower right).
•The photon model explains this behavior: Higher-energy electrons can
convert their energy into higher-energy photons, which have a shorter
wavelength (see Example 38.4).
max min
min max
hc hc
E
eV


  
max min
For 50 kV, 24.8 pmV  

6. Copyright © 2012 Pearson Education Inc.
X-ray scattering: The Compton experiment
•In the Compton experiment, x rays are
scattered from electrons. The scattered x
rays have a longer wavelength than the
incident x rays, and the scattered
wavelength depends on the scattering
angle .
•Explanation: When an incident photon
collides with an electron, it transfers some
of its energy to the electron. The scattered
photon has less energy and a longer
wavelength than the incident photon.
These are elastic collisions, so both
momentum and energy are conserved.
(1 cos )
h
mc
    
2
(cons. of energy)
(cons. of momentum)
e
e
pc mc pc E  
 p p P

7. Copyright © 2012 Pearson Education Inc.
X-ray scattering: The Compton experiment
•Example 38.5—Compton Scattering: You use 0.124-nm x-ray photons in a
Compton-scattering experiment. (a) At what angle is the wavelength of the
scattered photons 1.0% longer than the incident x-rays? (b) How much
energy went into the scattering particle?
(a) Since ′ is 1% longer, ′ –  = 0.01, so solving for ,
(b) The photon energy lost in the scattering is
That is the same energy gained by the particle.
Note max change in wavelength is for “head-on” collision ( = 180),
(1 cos )
h
mc
    
0.01 (1 cos ) arccos 1 0.01 60 .7
omc mc
h h
   
 
     
 
 
0.01
( )
1.01
hc hc hc hc
E  
   
     
 
  
34 8
15
10
6.63 10 J-s 3 10 m/s
0.01 1.59 10 J 991eV
1.24 10 m
E



 
    

2h
mc
  

8. Copyright © 2012 Pearson Education Inc.
Pair production
•When gamma rays of sufficiently short wavelength are fired into a metal
plate, they can convert into an electron and a positron (positively-charged
electron, or anti-matter electron), each of mass m and rest energy mc
2
.
•The photon model explains this: The photon wavelength must be so short that
the photon energy is at least 2mc
2
. Note that the particles have to have
exactly the same energy and angle in order to conserve momentum.
•The reverse process can occur—the positron and electron can come together
and annihilate to produce a photon (or more than one photon)
•Example 38.6: An electron and positron,
initially far apart, move toward each other at the
same speed. They collide head-on, annihilating
each other and producing two photons. Find the
energies, wavelengths and frequencies of the two
photons if the initial kinetic energies of the two are
(a) negligible, or (b) both 5.00 MeV.
(a) In this case, the energies of the particles are each
just their rest masses mc
2
, so each photon has that energy.