Photons and QM Part II Quiz - PowerPoint Presentation

Slide1

Photons and QM Part II

Quiz

Bremsstrahlung, Photo-emissionCompton scattering bootcamp

1

Are you ready?

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.

2

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).

3

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

-34

Joule-sec or 4.136 x 10

-15

eV

-s

2.41 x 10

14

Hz

No minimum frequency

1.24 x 10

14

Hz

2.0 x 10

14

Hz2.4 x 1015 Hz

4

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 4.136 x 10

-15 eV

-s

2.41 x 10

14

Hz

No minimum frequency

1.24 x 10

14

Hz

2.0 x 10

14

Hz

2.4 x 10

15

Hz

5

hf

=

ϕ

f =

ϕ

/h

f

= 1eV/4.136x10-15eV-s

= 2.41x1014 Hz

h(J)f = ϕ(J) f = ϕ /h

f

=

1.6x10

-19

J

/

6.626x10

-34

J-

s

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

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 !

7

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 V

AC 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

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

Question: X-ray production via bremsstrahlung

In the x-ray apparatus (and photo-emission via electrons), why is there a maximum photon energy ?

Each electron produces at most one photon

(ii)

Ans: the maximum energy corresponds to an electron

giving up all its kinetic energy

to the anode when producing a photon.

10

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

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

X-ray scattering: Compton Scattering

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).

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"

.

13

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

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 mc

2.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

Are you ready for Compton Scattering Bootcamp ?

Next exercise

16

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

? Wha

t about the photon ?

One equation, two unknowns; need another constraint. What should we do ?

Conserve vector momentum (relativistic) !

What is the final energy of the electron ?

17

X-ray scattering: Compton Scattering (cont’d)

Question: Do you remember the law of cosines

? Or definition of dot product ?

The (mc

2

)

2

terms cancel

18

X-ray scattering: Compton Scattering (nearly there)

Divide by c

2

Compare Eqn(*) and Eqn(**)

Divide by p p

’

19

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 is m (here)

? Which mass ?

Ans: the mass of the electron (0.511 MeV)

Question:What would happen with protons ?

20

Ans

:

938 MeV

For next time

Read 38.4 (Heisenberg Uncertainty Principle)Read material in advance

Concepts require wrestling with material21