1. Photons and QM Part II Quiz Bremsstrahlung, Photo-emission
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Quiz
Bremsstrahlung, Photo-emission
Compton scattering bootcamp

2. Increase the intensity of light shining on the metal.
Q16.1
In the photoelectric effect, how can one increase the kinetic energy of an electron coming out the metal surface?
Increase the intensity of light shining on the metal.
Increase the frequency of the light.
Shine the light on the metal surface for longer time.
Polish the metal surface.
B (the energy depends on frequency but not intensity or time the light has been shining (no time delay as in the classical wave picture).

3. Increase the intensity of light shining on the metal.
Q16.1
In the photoelectric effect, how can one increase the kinetic energy of an electron coming out the metal surface?
Increase the intensity of light shining on the metal.
Increase the frequency of the light.
Shine the light on the metal surface for longer time.
Polish the metal surface.
The energy depends on frequency but not intensity or time the light has been shining (no time delay as in the classical wave picture).
B (the energy depends on frequency but not intensity or time the light has been shining (no time delay as in the classical wave picture).

4. Given: Planck’s constant = 6.626x10-34Joule-sec or 4.136 x 10-15 eV-s
Q16.2
In the photoelectric effect for a metal, the electron has to overcome an energy barrier of 1 eV = 1.6x10-19 Joules to escape. What is the minimum frequency of light in order to produce the photoelectric effect for this metal?
Given: Planck’s constant = 6.626x10-34Joule-sec or x eV-s
2.41 x 1014 Hz
No minimum frequency
1.24 x 1014 Hz
2.0 x 1014 Hz
2.4 x 1015 Hz
A hf = phi f = phi/h = 1 eV/4.136 x 10^-15 eV-s = 2.41 x 10^14 Hz or use the version in Joules.

5. Given: Planck’s constant = 6.626x10-34Joule-sec or 4.136 x 10-15 eV-s
Q16.2
In the photoelectric effect, the electron has to overcome an energy barrier of 1 eV = 1.6x10-19 Joules to escape. What is the minimum frequency of light in order to produce the photoelectric effect for this metal?
Given: Planck’s constant = 6.626x10-34Joule-sec or x eV-s
2.41 x 1014 Hz
No minimum frequency
1.24 x 1014 Hz
2.0 x 1014 Hz
2.4 x 1015 Hz
hf = ϕ f = ϕ/h
A hf = phi f = phi/h = 1 eV/4.136 x 10^-15 eV-s = 2.41 x 10^14 Hz or use the version in Joules.
f = 1eV/4.136x10-15eV-s
= 2.41x1014 Hz
h(J)f = ϕ(J) f = ϕ /h
f = 1.6x10-19J/ 6.626x10-34 J-s

6. X-ray production
One mechanism of x-ray production is called “bremsstrahlung” from a German word.
Question: What is the meaning of this German word ?
Ans: “braking radiation”. Here “braking” means deceleration. 6

7. X-ray production via bremsstrahlung
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 the figure on the lower right). [continuous energy spectrum]
The photon model explains this behavior: Higher-energy electrons can convert their energy into higher-energy photons, which have a shorter wavelength.
Notice dependence of energy cutoff on accelerating voltage
Does not depend on material – just Planck’s constant.
Does not depend on material ! 7

8. Question: X-ray production via bremsstrahlung
In the x-ray apparatus below, suppose you increase the number of electrons that are emitted from the cathode per second while keeping the potential difference VAC the same. How will this affect the intensity and minimum wavelength λmin of the emitted x-rays ?
(i) I and λmin will both increase; (ii) I will increase but λmin will remain the same; (iii) I will increase but λmin will decrease; (iv) I will remain the same but λmin will decrease; (v) none of the above. [not-a-clicker]
Each electron produces at most one photon(ii)
We use such an apparatus in PHYS481L 8

9. Example: Applications of x-rays
CAT scan= Computerized Axial Tomography
Many 2-D images taken around the axis of a patient, combined into a single 3-D result 9

10. Question: X-ray production via bremsstrahlung
In the x-ray apparatus (and photo-emission via electrons), why is there a maximum photon energy ?
Ans: the maximum energy corresponds to an electron giving up all its kinetic energy to the anode when producing a photon.
Each electron produces at most one photon(ii) 10

11. Question: X-ray photoemission example (SI units)
Electrons in an x-ray tube accelerate through a potential difference of 10.0 kV before striking a target. If the electron produces one photon on impact with the target, what is the minimum wavelength of the resulting x-rays ? 11

12. Question: X-ray photoemission example (in eV units)
Electrons in an x-ray tube accelerate through a potential difference of 10.0 kV before striking a target. If the electron produces one photon on impact with the target, what is the minimum wavelength of the resulting x-rays ?
N.B e’s cancel in the expression. 12

13. X-ray scattering: Compton Scattering
Question: Why was the discovery of Compton scattering so important ?
The 1927 Nobel Prize in Physics was divided equally between Arthur Holly Compton "for his discovery of the effect named after him" and Charles Thomson Rees Wilson "for his method of making the paths of electrically charged particles visible by condensation of vapor".
Ans: In a classical (non-QM) physics picture: light is a wave and the final energy will be the same as the initial. Compton scattering is possible if light is composed of photons (a.k.a quanta). 13

14. X-ray scattering: Compton Scattering
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 ϕ.
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. 14

15. Pair production (Interesting consequence of special relativity)
When γ rays of sufficiently short wavelength are fired into a metal plate, they can convert into an electron and a positron (positively-charged electron), each of mass m and rest energy mc2.
The photon model explains this: The photon wavelength must be so short (or the energy high enough) so that the photon energy is at least 2mc2. 15

16. Are you ready for Compton Scattering Bootcamp ?
Next exercise
Are you ready for Compton Scattering Bootcamp ? 16

17. X-ray scattering: Compton Scattering (derivation)
Let’s write down total energy conservation using special relativity and QM for the photon.
Question: what is the initial energy of the electron ? What about the photon ?
What is the final energy of the electron ?
One equation, two unknowns; need another constraint. What should we do ?
Conserve vector momentum (relativistic) ! 17

18. X-ray scattering: Compton Scattering (cont’d)
Question: Do you remember the law of cosines ? Or definition of dot product ?
The (mc2)2 terms cancel 18

19. X-ray scattering: Compton Scattering (nearly there)
Divide by c2
Compare Eqn(*) and Eqn(**)
Divide by p p’ 19

20. X-ray scattering: Compton Scattering (result)
Question: How are the momentum and wavelength of a photon related ?
Ans: p=h/λ and p’ =h/λ’
Question:What would happen with protons ?
Ans:
938 MeV
Question : What is m (here) ? Which mass ?
Ans: the mass of the electron (0.511 MeV) 20

21. For next time Read 38.4 (Heisenberg Uncertainty Principle)
Read material in advance
Concepts require wrestling with material