1. Photoelectric Effect

2. Before we begin
It is a good idea to fully recall the following equation(s):
Speed of a wave: v= f𝝀
Speed of an EMR: c= f𝝀
Energy of a wave/photon E = hf
Where ‘v’ and ‘c’ is in m/s, ‘f’ is in hertz or 1/s, ′𝜆′ is in meters, ‘E’ is in joules (j) or electron volts (eV), and ‘h’ is Planck’s constant in joules second or ev s.

3. Photoelectric Effect
We now have an understanding of E = nhf (Planck’s theory), and electrons bouncing from one shell orbital level to another.
During the same time period we saw another piece of evidence that EMR acted like a particle – the photoelectric effect.
This phenomenon started with Physicist Heinrich Hertz
Heinrich Hertz

4. Heinrich Hertz confusion
Heinrich Hertz was doing experiments in 1887 to test if an electric coil (receiving loop) would emit an EMR.
Hertz made a simple receiver of looped wire. At the ends of the loop were small knobs separated by a tiny gap.
The receiver was placed several yards from the oscillator.
Since electromagnetic waves were spreading from the oscillator sparks, they would induce a current in the loop that would send sparks across the gap.
This occurred when Hertz turned on the oscillator, producing the first transmission and reception of electromagnetic waves.
Spark produced by EMR going through the loop
Spark produced by oscillator

5. Heinrich Hertz confusion
During his experiment he noticed a spark was occurring at the receiving loop, but he could hardly see it and was not 100% sure what he was seeing.
Ultimately, Hertz decided to place his experiment inside a dark box and observed it through a glass window.
Funny thing was, the sparks seemed shorter now. He could see the sparks in the dark, but they just weren't as big anymore.
Hertz noticed that when he changed the glass window to one made out of quartz, the sparks got bigger.
The big difference is that the glass blocks UV from outside the box, going into the box, but the quartz does not.

6. Heinrich Hertz confusion
In the end, Hertz concluded that if random UV radiation in the room was able to go into the box (it could only do that if glass did not block it) it would hit the receiving loop coil and help electrons pop off metal spark balls making bigger sparks.
Ultimately, he proved that electrons pop off of the metal when an EMR wave hits the metal sparks on the receiving loop.

7. Albert Einstein on Hertz
Einstein really did not care as much about the spark size as he did about what caused the electron to move.
He knew of Planck’s equation E = nhf and E = hf. I.E. there is energy in the photon, but why are the electrons being moved?
Albert Einstein

8. Einstein’s conclusion
Einstein figured out that it was the frequency of the light hitting the metal that was important.
When the UV light (an EMR) hit the metal of the receiving loop, it had enough energy to knock off electrons.
This was happening because the individual photons of UV had enough energy according to the formula E = hf.
If the metal is exposed to radiation with a frequency less than UV, nothing happened.
Since the frequency of the light is so low, each photon does not have enough energy to knock off the electrons.
This critical minimum frequency that is needed to start knocking off the electrons was named the threshold frequency.
The special symbol used for it in formulas is fo .

9. Einstein’s conclusion
Einstein believed that to give a single electron the energy to move, the metal was hit by a single photon (destroying itself), and transferred its energy to the one electron.
Since the electron is originally attached to the metal, some minimum amount of energy must be needed just to snap it off. Otherwise, electrons would just be dropping off of atoms all the time.
Einstein called this the work function of the metal, since you needed to do work on the electron to break it off.
Every metal has its own work function, since different metals hold on to their electrons with different strengths.
Popping the electrons off starts to happen at a minimum threshold frequency, so that must correspond to the work function.

10. Work Function Equation
The formula for what Einstein had discovered is a modification of Planck's formula.
E=hf (Planck’s equation)
W=hfo (Work Function Equation)
WHERE
W = work function (J) or (eV)
h = Planks constant – x J · s or 4.14 x eV· s where 1 eV = ·10-19 J
fo = threshold frequency (Hz)
Just like Planck's formula, you can use the value for Planck's constant in electron volts and get your final answer in electron volts.
In fact, it is very common to give the value for the work function in electron volts.
A maximum threshold wavelength would indicate the max wavelength needed to release an electron, anything bigger will not release an electron.

11. Example #1
Determine the threshold frequency of a material with a work function of 10 eV.
b) Using the value for Planck’s constant that is in eV· s.
W =hfo
fo= W/h
fo= 10/4.14 x 10-15
fo= x 1015 = 2.4 x 1015 Hz
a) Using the value for Planck’s constant in Joules Second
1ev = x J 10 ev x
x = x 10-18J
W =hfo
fo= W/h
fo= x 10-18J /6.626 x 10-34
fo= x 1015 = 2.4 x 1015 Hz

12. Example #2
Determine the work function of a metal in Joules if the threshold wavelength is 1.10 x 10-7 m.
W =hfo  recall that fo = c/λ
W = hc/λo
W = (6.626 x J · s) (3.00 x 108 m/s)/1.10 x 10-7 m
W= x J = 1.81 x J
Convert this into eV
1ev = x J x x J
x = eV = 1.13 x 101 eV

13. Work Function Different metals have different work functions Metal
Work Function in eV
Potassium
2.24
Sodium
2.36
Zink
3.57
Uranium
3.90
Copper
4.16
Cobalt
5.00
Platinum
6.00

14. Example #3
The work function for copper is 4.16 eV. If a wave length of 250 nm hits copper, will that be enough energy to knock off an electron?
E = 1240 eV nm λ
E = 1240 ev nm nm
E = 4.96 eV, therefore this is a strong enough EMR to knock loose an electron.

15. Photoelectric Effect
To prove that Einstein was correct about the need of a frequency threshold to knock an electron off a metal, Robert Millikan in set up the Photoelectric Effect.
Miliken set up an experiment where he had two pieces of metal inside a vacuum tube. The ends of the tube were attached to a wire which had an ammeter attached to them.
He would then shot EMR light onto a piece of metal.
The result, the ammeter moved and thus showed electron flow.
Wave theory could not predict this, because it assumed that the particles in the atom would vibrate and give off an EMR, it would not knock off an electron.
The Photoelectric Effect PHET Simulation.

16. Photoelectric Effect

17. Photoelectric Formula
This leads to our big equation – the Photoelectric Formula.
hf = KEmax + W ALSO SEEN AS Kemax = hf – hf0
(Note: W = hf0)
WHERE :
hf = The original input energy from the incoming photon.
W = The energy needed to snap off the electron from the metal. This is the work function.
KE max = Any left over energy becomes kinetic energy used by the electron to go zipping across the tube – i.e the velocity of the electron.
Note : Mass of electron is 9.11x kg

18. Example #4
The threshold frequency (work function) of silver is 1.14 x 1015 Hz. An EMR with a wavelength of x 10-7 m strikes a piece of pure silver. Determine the speed of the electrons that are emitted. Mass of an electron is 9.11x kg.
First we'll figure out how much kinetic energy the electrons are getting. We will have to substitute formulas into the one we've built so far.
hf = KEmax + W
hc/λ = KEmax + hfo
KE max = hc/λ − hfo
KE max =( (6.626 x10-34 )(3.00 x 108) / 2.50 x10-7) − ((6.63 x10-34 )(1.14 x1015 ))
KE max = x J = 3.98 x J

19. Example #4 Continued KE max = 1/2mv2 v = 2 (𝐾𝐸𝑚𝑎𝑥)/𝑚
V = 2(3.98 x 10−20 J)) / 9.11x 10−31 kg
v = 295, = 2.96 x 105 m/s

20. Example #5
Gold has a work function of 4.82 eV. A block of gold is illuminated with ultraviolet light (λ = 160 nm). What is the speed of the electron.
hf = KEmax + W  KE = hf –w
W = 4.82 eV
hf (UV Light energy)= ?
KE = ?
First we need to solve for the energy of the UV Light and there are two ways to do this. Option one, we can use eV nm in our ”hf” equation or we can find the frequency of the UV Light and use 4.14 x eV s in our “hf” equation.
Once we know the energy of the UV Light, we subtract it from the Work Function and solve for the speed of the electron.

21. Example 5 Option # 1 Option #2 hf = 1240 eV nm 160 nm = 7.75 eV
solve for the hf of UV Light (energy)= ?
Option # 1
Option #2
hf = 1240 eV nm nm = 7.75 eV
Solve for the frequency of a 160 nm wavelength.
f = 3.00 x 108 m/s = x /s (hz) x 10-9 nm
hf = (4.14 x ev s)(1.875 x /s)
= 7.76 eV

22. Example 5 KE = hf –w KE = 7.75 eV – 4.82 eV KE = 2.95 eV
To solve for velocity the KE HAS TO BE IN JOULES
Recall 1 eV = ·10-19 J
Therefore eV = ·10-19 J eV x
x = 4.73 x J

23. Example 5 KE = ½mv2 v = 2 (𝐾𝐸𝑚𝑎𝑥)/𝑚
v = 2 (4.73 x 10−19 J)/ 9.11x 10−31 kg
v = 1,019,028 m/s = 1.02 x 106 m/s

24. Assignment
Page 736 Q 5 – 7 and Page 738 Q 12 – 14
Handout Questions