Notes: Quantum 1
Reading 1: Waves and Electron Volts
At the turn of the 20th century, scientists were felt pretty confident that light was a wave. They had done a
lot of experiments that confirmed light definitely had a wave nature. In this section, we are going to
explore that actual nature (shape) of light, which turned out to be quite a bit different than they thought
it would.
Before we can do that, we will need to review a few things.
Characteristics of Waves
This should be a review for most of you. We will go into this in much greater detail in the next unit. For now,
you need to understand the key parts of a wave.
Wavelength
The wavelength of a wave is how far apart two peaks of a wave are. Visible light has wavelengths that
are very close together. The symbol for wavelength is 𝜆 (𝑙𝑎𝑚𝑑𝑎). For this reason, we tend to measure light
wavelengths in nanometers (nm, 10-9 m)
The amplitude of the wave is how tall the wave is. We will not be using that a great deal in this section.
Frequency
Frequency (f) is how many times the wave wiggles back and forth every second.
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 =
𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛
𝑠𝑒𝑐𝑜𝑛𝑑
We measure frequency in Hertz (Hz). 1 Hz= 1 vibration/second.
Speed of Light
In this unit, the waves we are dealing with will almost always be light waves. It is therefore important that
we know the speed of light in a vacuum. This is a universal constant so it gets its own letter c.
c= 3.0 x 108 m/s
Some sections draw directly from Knapp’s Notes.
The Wave Equation
For light waves (and lots of other waves, which we will play with later), you can say:
𝑣 = 𝑓𝜆
where v is the speed of the wave, f is the frequency and 𝜆 is the wavelength. Since the speed of light is c
we can say
𝑐 = 𝑓𝜆
If we rearrange and solve for f we get
𝑐
𝜆
= 𝑓 . Notice that the
wavelength and the frequency are inversely proportional to each
other. So if the frequency gets higher, the wavelength gets smaller.
Many times in the course of this unit, you will be given the
wavelength. You need to remember that means that you can
always easily find the frequency as well, because c is constant. The
reverse is true. If you are given the frequency, you can easily
calculate the wavelength.
One last thing about waves: the higher frequency, the more energy the wave has. The longer the
wavelength, the less energy the wave has. This makes a lot of sense when you think about it. If you
wanted to make a slinky wiggle at a really high frequency, you would need to wiggle your arm back and
forth and ton. That would take a lot of energy.
Electron Volts
In the past, we spoke briefly about electron Volts. We have only used them a handful of times. In this
section, however, they will become incredibly useful. What you need to remember is that a) eV are a
measure of energy. b) To convert between electron volts and joules, just multiply by e (1.6 x 10 -19). This
actually makes a lot of sense.
2𝑒𝑉 = 2 ∙ 𝑒 ∙ 𝑉 = 2(1.6 𝑥 10−19 ) = 3.2 𝑥10−19 𝐽
Because we will often be dealing with small amounts of energy in this unit, using the electron volt will
many times be easier.
Reading 2: Black Body Radiation
Problems with Wave Theory
At the turn of the last century, scientists had a problem. Nearly all their research suggested that light was
actually a wave (we will look into this research later in the course). They could back this up with lots of
experiments and math. Unfortunately, they kept running into a few situations where light wasn’t acting like
a wave at all. This made them want to light things on fire.
In this section, we are going to learn about two of the experiments that made scientists question whether
light actually does behave like a wave.
The experiments:
Some sections draw directly from Knapp’s Notes.
-
Black body radiation
Photoelectric Effect
We are also going to talk about how light actually acts and use some of Einstein’s equations to prove it.
Problem Experiment #1Black Body Radiation
WHAT IS BLACK BODY RADIATION?
Look at the desk in from of you. Right now, the lights of the classroom are beating down on that surface.
Some of the light that hits the table bounces off. This is the part you can see. Some of the light is absorbed
by the table. What happens to this energy? Well, the table takes it in and is warmed slightly. The table will
then release heat (thermal radiation). We can’t see this type of radiation, but we can feel it. All objects
basically undergo this same process.
Figure 1: How a normal object acts
The reemitted radiation is not released at the same frequency as the absorbed radiation. It has changed.
That’s why, even though we can see the light that hits the table, we can’t see the radiation that comes
off the table in the form of heat, even thought the radiation that hit was visible light. It has a different
frequency now, a frequency that is out of the range of our eyes.
A black body is a perfect absorber and emitter of light. When we say it is a perfect absorber if we were to
shine a light on it, none of that light would be reflected. It would all be completely absorbed.
Some sections draw directly from Knapp’s Notes.
Figure 2: How a Black Body Acts
This isn’t a real thing. Nothing acts exactly like this. This is one of these physics creations that allow us
to do calculations- like frictionless surfaces (not real), particles without mass (not real) and situations where
gravity doesn’t affect anything (not real). There are no real perfect black bodies in the world. But for some
objects, treating it like a “black body” is a close enough approximation.
Notice that the black body doesn’t just absorb the light and keep it- that would be a black hole, not a
black body. The name “black body” is actually stupid- a black body does actually give off radiation. In
some cases, you can even see this radiation. You just don’t have to worry about the reflected light.
The interesting thing about black bodies is that if you heat them up, they give off radiation. The filament
inside a light bulb is an example of a black body. When you heat it up by running electricity through it, it
gives off bright visible light. Another example could be the burner on an electric stove. As it heats up, it
gives off an orange light.
We call the radiation a black body gives off when you heat it up blackbody radiation. And, while looking
at black body radiation, scientists noticed a big problem.
The Problem With Black Body Radiation
Okay, in this section, don’t worry if every single thing doesn’t make absolute sense. A lot of it will be over
your heads and we are just skimming the surface of some pretty heavy math. Just try to understand the
basic story.
Imagine heating up a black body to 5000 degrees. It would start to give off all types of radiation, some at
higher wavelengths, some at lower wavelengths. Scientists predicted that if light were made of waves, the
black bodies would give off more radiation with shorter wavelengths and less radiation with lower
wavelengths. It would look something like the diagram below.
Some sections draw directly from Knapp’s Notes.
Figure 3: Predicted Black Body Radiation
Their wave theory predicted that the black body would give off radiation with all different frequencies.
The theory said it would give off tons of short wavelength (high frequency) radiation, and less and less
higher wavelength(low frequency) waves. See the diagram above.
If we were to graph that prediction, it would look like the line on the right on the graph below.
When they actually heated up a black body and looked at it, it actually looked like the line on the left.
The black bodies did put off lots of different wavelengths- but it was all clustered around one peak
wavelength. See the diagram below.
Some sections draw directly from Knapp’s Notes.
Figure 4: What Black Body Radiation Actual Looked Like
To the scientists, this broke physics. You do not need to know all the nitty-gritty details, but you should know
this went against all the math predicting that light was a wave. It completely messed with physicists’
heads. It told them everything they thought they knew about light and radiation could be wrong.
Some sections draw directly from Knapp’s Notes.
They tested this out at lots of different temperatures (probably hoping it was wrong). What they saw was
this:
Again, the curve on the right shows
what they predicted.
The huge bumps are what they actually
found. Each bump represents the type
of radiation given off when the black
body was heated to different
temperatures.
You should notice (and know) a few things about these graphs.
1. They seemed to suggest that light doesn’t always act like a wave.
2. At long wavelengths (the right side of the graph) the black body radiation did sort of match the
predictions.
3. At a given temperature, the radiation given off seemed to cluster around a peak wavelength.
4. At different temperatures, the black body gave off different peak wavelengths.
5. The higher the temperature, the shorter the wavelength of the peak wavelength.
6. The area under the curve tells us the intensity of the radiation given off. We won’t go into this
mathematically here, but this basically means that the higher the temperature, the more short
wavelength (high frequency) radiation is given off.
Supplemental: This is not required, but if you would like to play with these ideas a bit,
you can check out the Phet simulation Blackbody Radiation.
To the physicists, this meant that everything they knew could be wrong. Maybe light wasn’t a wave after
all….?
Reading 3: Intro to Photons
The Solution to the Black Body Radiation Problem
Along came a guy named Planck. We won’t go into the details here, but Planck came up with an
equation that actually matched the pattern of radiation the black bodies were giving off. A little while
Some sections draw directly from Knapp’s Notes.
later, Einstein came along and used Planck’s equation to figure out that light doesn’t just act like a wave,
it also acts like a particle. A tiny, magical, massless particle of light that Einstein called a photon.
How can light act like both a wave and a particle? You can visualize it (for now) as tiny particle that is
essentially vibrating. If at this point you feel like your brain is inside a blender and that makes no sense,
don’t worry. We are moving beyond the space that humans can actually visualize. We are moving into a
space that is totally abstract and mathematical. You will get more comfortable with it as we go. Or you’ll
go mad, one of the two.
Anyway, Einstein used Plank’s equation to calculate how much energy each of these tiny photons
(vibrating packets of energy, drops of light) would have if they were vibrating at a given frequency. He
created the equation below.
Planck’s Equation
E  hf
Here, E is energy of one photon, f is the frequency of the photon, and h is known as Planck's constant. It’s
value is:
h  6.63 x 1034 J  s
or
h  4.14  1015 eV  s
You will have both of these values when you take the AP Physics test. You may want to put this in your
calculator, however. You pick which h you want to use depending on whether the problem is using J or
eV (electron volts, another way of measuring energy).
Practice Problem: Energy of a photon
What energy is carried by a photon (piece of light) of electromagnetic radiation that has a frequency of
1.55 x 1017 Hz?
Solution
E  hf
1

 6.63 x 1034 J s 1.55 x 1017   10.3 x 1017 J
s


1.03 x 1016 J
Planck’s Equation Using Wavelength
The energy of a photon can also be calculated as a function of wavelength. Wavelength is related to
frequency by:
v f
v f
This is the equation for the speed of the wave.
so c   f
E
f 
c

E  hf
E
hc

hc

You won’t be provided with this equation, so you need to be able to either derive it (like above) or
memorize it.
Some sections draw directly from Knapp’s Notes.
Practice Problem: Energy of a Photon
A photon has a wavelength of 550 nm. How much energy does this represent in Joules?
Solution
v f
so c   f
f 
c

 6.63 x 10 J s  3.00 x 10
E
550 x 10 m 
34
3
9
E  hf
m
s
E
hc

 0.036 x 1022 J

3.6 x 1024 J
Sorry! The teacher I copied this from has an error above. He the real speed of light is 3 x 108 m/s.
This changes the answer to 3.61 x 10-24 J.
Since the value of Planck’s constant multiplied by the speed of light is itself a constant, we can treat hc as
a constant. (Save us some work!) Two such values, using different units, will be provided to you on the AP
Test:
hc  1.99 x 1025 J  m
hc  1.24 x 103 eV  nm
This makes solving the above problem a lot easier. Just be careful if you program these in. Notice the units
on the second guy are in nm (nanometers). That’s okay because we usually talk about wavelength in nm
so it cancels out.
Multiple Photons
The amount of energy in one photon is given by:
E  hf
If we had multiple photons, how much energy would they have? Well, that’s easy enough to find.
𝐸 = 𝑛ℎ𝑓
Where E is the energy, h is Planck’s constant, and n is the number of photons at the given frequency.
Photons and Power of a Source
Imagine that you have a source of light that is rated at a certain power level. It produces photons of
only one frequency. So, how many photons per second would it produce?
This is pretty simple. Power is simply the rate that energy is produced. The energy is in the form of
photons. All you have to do is calculate the amount of energy produced in one second. Then
determine the amount to energy one photon represents. Then divide the total energy by the energy per
photon. This gives you the number of photons in a second. That last part is really just a dimensional
analysis problem, ain’t it?
Practice Problem
A 505 nm light source produces 0.250 W. How many photons per second does it kick out?
Some sections draw directly from Knapp’s Notes.
SOLUTION
P
E
t
v f
E  Pt
 0.250
J
1.00 s   0.250 J
s
c
E  hf

f 

1.99 x 1025 J  m
E
505 x 106 m
hc

 0.00394 x 1019 J
 3.94 x 1022 J




1
  0.0635 x 1022 photons 
0.250 J 
J
 3.94 x 1022



photon


6.35 x 1020 photons
There is also a mistake here. For some reason, he thinks nano is x10-6 It is actually x 10-9 This changes the
answer to 7.47 x 1038 photons
VIDEO LECTURE: Pause here to watch AP Physics B Lessons with Ms.Twu: Modern Physics 4:
Photon Energy Problem
Momentum and Light
Normally, to find the momentum of an object in Newtonian or Classical Physics, you would multiply the
mass by the velocity. What do you do to find the momentum of a photon, which doesn’t have any mass?
Well, again we rely on Einstein, who came up with the following equation.
The momentum for a photon is given by this equation:
E  hf  pc
Where E is the energy, momentum is p and c is the speed of light 3x 108 m/s.
Practice Problem: Momentum of a Photon
What is the momentum of a photon that has a wavelength of 455 nm?
Solution
v f
p
h

f 
c

E  hf
 6.63 x 1034

hc

E  pc

kg  m 2  s 
1


6
s2
 455 x 10 m 
Some sections draw directly from Knapp’s Notes.
pc 
hc

p  0.0146 x 1028
kg  m
s

1.46 x 1026
kg  m
s
There is also a mistake here. For some reason, he thinks nano is x10-6 It is actually x 10-9 This changes the
answer to 1.46 x 10-27 kg m/s.
VIDEO LECTURE: AP Physics B Lessons with Ms.Twu: Modern Physics 5: Photon Momentum
Reading 4: The Photoelectric Effect
Problem Experiment #2: The Photoelectric Effect
Black body radiation was one experiment that contradicted the scientists’ theory of light as a wave. There
were others. Probably the most famous was something called the photoelectric effect.
What is the photoelectric effect?
It is going to take a little while to fully explain the photoelectric effect, but the basic idea is this. Imagine
we have a sheet of metal and we shine a light on it. The light hits the metal, gives electrons trapped in the
metal some energy and they go flying off.
This is the photoelectric effect in a nutshell. It is going to more complicated than this, but that is a basic
description. We are going to see that the electrons didn’t pop off when the scientists expected them to,
and this freaked them out. But before we get there, we have to have a more in depth understanding of
the photoelectric effect.
Setting Up A Photoelectric Experiment
Why did scientist care about this at all? This can tell scientist exactly how much energy a photon has.
They set up an apparatus like the one below.
Some sections draw directly from Knapp’s Notes.
We have the plate (on the left) with the light shining on it.
The battery is basically there to help accelerate the
particles to the right once they are released from the
metal. Once electrons break free of the metal and make it
to the opposite plate, a small current will start to flow
through the circuit. We can measure the flow of that circuit
to see exactly how many electrons are being released.
Then, we can change the brightness (intensity) of the light
bulb and the frequency of the light bulb and see if this
changes how many electrons break off from the metal.
Figure 5: Basic Photoelectric Apparatus
The Work Function
The amount of energy binding the electrons to the metal is called the work function. The symbol for this is
the Greek letter .
  Work Function
Work Function for some
Different Metals
In other words, this is the amount of energy the electron is going to
need just to break the bond that holds it to the metal.
Different metals have different work functions. Here is a useful table.
You would never need to memorize any of these, but it will be helpful
in future problems.
If you just want to break the bond and free the electron, you need the
light hitting it (the photon) to give you enough energy.
Remember the energy of a photon?
𝐸𝑝ℎ𝑜𝑡𝑜𝑛 = ℎ𝑓
Metal
Work Function
(eV)
Sodium (Na)
2.28
Aluminum (Al)
4.08
Copper (Cu)
Cobalt (Co)
4.70
3.90
Zinc (Zn)
4.31
Silver (Ag)
4.73
Platinum (Pt)
6.35
Lead (Pb)
4.14
You would need a certain frequency of light shining on an atom to
Iron (Fe)
4.50
break the bond. We call this frequency the cutoff frequency, or the
threshold frequency. They call it that because if we drop below this frequency, you won’t have enough
energy to break the bonds.
𝜙 = ℎ𝑓𝑐𝑢𝑡𝑜𝑓𝑓
Practice Problem Cutoff Frequency
What is the cutoff wavelength (threshold wavelength) for a copper metal surface?
Solution
C 
hc

1.24 x 103 eV  nm

4.70 eV
 0.264 x 103 nm 
264 nm
Energy In Photoelectric Problems
If you gave the electron the exact work function of energy (the exact amount of energy need to break
the bond), the bond would break, but the electron wouldn’t have any left over energy to get to the
opposite plate on the apparatus and start the current flowing. It will need more for that. How much kinetic
Some sections draw directly from Knapp’s Notes.
energy will it need to just barely get to the other side? If we want the electron to just barely make it to the
other plate without any kinetic energy to spare, we want all the kinetic to turn into potential by the time it
reaches the other plate. This is like throwing a ball up and having it just kiss the ceiling. In other words, we
want all the kinetic energy it has leaving the plate to turn it potential.
𝑈𝑓𝑖𝑛𝑎𝑙 = 𝐾𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑈 = 𝑞𝑉
We learned this back in electrostatics
𝑈 = 𝑒𝑉
Since we are dealing with an electron
𝑲𝒎𝒂𝒙 = 𝒆𝑽
This is the amount of kinetic energy it would need to just make it to the opposite plate
So how much energy will you need to give the electron to break the bond and make it to the opposite
plate?
𝐸𝑛𝑒𝑟𝑔𝑦 = 𝜙 + 𝐾𝑚𝑎𝑥
𝐸𝑛𝑒𝑟𝑔𝑦 = 𝜙 + 𝑒𝑉
So if 𝐸 > 𝜙 + 𝑒𝑉 you will see a current flow through the circuit, because the electron received enough
energy to break the bond and make it to the opposite plate, maybe even with some energy to spare.
If 𝐸 < 𝜙 + 𝑒𝑉 you won’t see a current flow at all, because the electrons did not receive enough energy to
break the bond and make it to the opposite plate.
What supplies this energy? The photons. Remember the energy of a photon?
𝐸 = ℎ𝑓
The energy of one photon
So we can say:
𝒉𝒇 = 𝝓 + 𝒆𝑽
You can think of this equation this way:
ℎ𝑓 = 𝜙 + 𝑒𝑉
𝐸𝑛𝑒𝑟𝑔𝑦 𝑓𝑟𝑜𝑚 𝑝ℎ𝑜𝑡𝑜𝑛 = 𝑒𝑛𝑒𝑔𝑦 𝑡𝑜 𝑏𝑟𝑒𝑎𝑘 𝑏𝑜𝑛𝑑 + 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑜 𝑚𝑎𝑘𝑒 𝑖𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑝𝑙𝑎𝑡𝑒
Notice that if we were to increase the voltage on the current (increase V) it would take more energy for
the electron to make it to the opposite plate and cause the current to flow. If you make the voltage too
high, the electrons will not be able to make it between the two plates. You would need more energy.
We can use this idea to measure exactly how much energy a photon has. We can adjust the voltage in
the circuit, making it larger and larger, until the electrons stop making it across the gap and the current
stops. Using this number, we could calculate the exact amount of energy the electron received from the
photon. We call the voltage where the electrons stop flowing the stopping voltage (Vstopping).
Photoelectric Equation
𝒉𝒇 = 𝝓 + 𝒆𝑽𝒔𝒕𝒐𝒑𝒑𝒊𝒏𝒈
Some sections draw directly from Knapp’s Notes.
Video Animation: Pause here to watch “The Photoelctric Effect Explained” You only need to
watch up to 4:30, though the whole video is useful. We will cover the rest of the video in the
notes below.
Practice Problem
Note: They are not asking for the frequency required to have enough energy to break the bonds and
make it to the opposite plate. They are just asking about breaking the bonds. How much energy would
that take? What frequency of photon has this energy?
Solution
Practice Problem: The Photoelectric Effect

A sodium photoelectric surface with work function 2.3 eV is illuminated by electromagnetic radiation and
emits electrons. The electrons travel toward a negatively charged cathode and complete the circuit shown
above. The potential difference supplied by the power supply is increased, and when it reaches 4.5 V, no
Some sections draw directly from Knapp’s Notes.
electrons reach the cathode.

(a) For the electrons emitted from the sodium surface, calculate the following.
i.
The maximum kinetic energy.
K Max  qV


K Max  1.6 x 1019 C  4.5 V  
7.2 x 1019 J


1 eV
K Max  7.2 x 1019 J 
 
19
1.60
x
10
J


ii.
or
4.5 eV
The speed at this maximum kinetic energy.
1
K  mv 2
2
2K
m
v
2

19 kg  m
2  7.2 x 10

2

s



31
9.11 x 10 kg

1.26 x 106
m
s
(b) Calculate the wavelength of the radiation that is incident on the sodium surface.
c f
E  hf
K Max  h

c


hc
K Max  
hc
c
E  h  





c
c
  h
K Max    h


 K Max   
f 
c
1.24 x 103 eV  nm

4.5 eV  2.3 eV
 0.182 x 103 nm 
182 nm
(c) Calculate the minimum frequency of light that will cause photoemission from this
sodium surface.
Some sections draw directly from Knapp’s Notes.
KMax  hf  
0  hf  
f 
KMax  0
But
hf  
2.3 eV
4.14 x 1015 eV  s
f 
for the minimum frequency, so

h
 0.555 x 1015 Hz

5.55 x 1014 Hz
Watch and Solve: AP Physics B Lessons with Ms.Twu: Modern Physics 9: Photoelectric Effect
Problem 1. Try the problem yourself first. Then watch her solve it. Take it a step at a time. Pause
the video frequently if you need to.
The “Problem” with the Photoelectric Effect
Remember that at this point scientists were pretty sure that light was a wave. The photoelectric effect
ended up showing that, in some ways, light acts like a particle. Now that we know how the experiment
was set up (see above) let’s explore why this happened.
First of all, scientists noticed something that makes a lot of sense. The brighter the light they shone on the
metal plate, the more electrons were knocked free, and the more current you would get in the circuit.
Brighter (More intense) light  more electrons make it across the gap  more current.
This makes a lot of sense, right? The more energy you throw at the metal plate, the more electrons you
would knock free. This did not mess with their wave theory.
What did mess with their wave theory was this: if you shone a lower frequency (longer wavelength) light
on a metal plate, this wouldn’t happen. Remember that the lower the frequency, the longer the
wavelength, the less energy a wave has. They knew shining a low frequency light on the plate would
mean less energy would hit the plate per second, but they thought the electron would eventually collect
enough energy to break free anyway. But this didn’t happen at all. It didn’t matter how bright of a low
frequency wave they shone- none of the electrons would break free.
I’ll give you an analogy to help you make sense of this. Imagine you have a bunch of people trapped in
jail, and bail is $5. The people in jail are the electrons and five dollars is the energy required to break free.
Imagine that you poured a whole bunch of pennies on these people (low frequency light). They would
eventually collect enough pennies to equal five bucks, they’d pay their bail and be free right? The wave
theory of light suggested this.
In reality, it was like you poured buckets and buckets of pennies on these people, and no one had
enough money (energy) to break free. They just stayed there, trapped. This frustrated the scientists to no
end.
They found that they actually give the people (electrons) five dollar bills. Then, suddenly, the people
(electrons) would start breaking free. In other words, they had to send down higher frequency light, where
each photon carried more energy.
Some sections draw directly from Knapp’s Notes.
The wave theory predicted that it didn’t matter what type of money (frequency of the energy) you gave
the people, as long as you gave them enough to equal $5. This didn’t work.
In reality, you had to send down a certain type of money ($5 bills, a higher frequency of light) in order for
electrons (people) to break free.
Wave theory
-Frequency of light doesn’t matter
-Intensity (brightness) of the light does
In reality
-Frequency does matter- the higher the frequency,
the faster the electrons are ejected
-Intensity (brightness) only matters if energy has a
high enough frequency. Then increasing the
intensity will increase the # of electrons emmitted
This strongly suggested to them that light was quantized- that it existed in discrete packets that could not
be broken down any further. Photons with low frequency were like pennies. Photons with higher frequency
were like five dollar bills. You couldn’t get lower frequency photons to do the job of higher frequency
photons. Light was acting like a particle with a set value (1 cent, $5) at different frequencies.
This made the scientists do this:
Light was acting like a wave in some experiments, and acting like a particle in others.
Graphs and the Photoelectric Effect
Let’s talk about some different graphs associated with the photoelectric effect.
Stopping Voltage Vs. Frequency
If you were to shine different frequencies of light on a metal and measure the stopping voltage for these,
you would create a graph like the one below.
Key features of the graph:
1. The x intercept is the cutoff frequency See the
explanation for why below
2. The slope is 𝒉/𝒆 See the explanation for why
below
Here’s how we know this:
Start with what we know
ℎ𝑓 = 𝜙 + 𝑒𝑉
ℎ𝑓
𝜙
𝑉= 𝑒 −𝑒
Rearrange so V is in terms of f
So if V=0 at the x intercept
ℎ𝑓 = 𝜙
Some sections draw directly from Knapp’s Notes.
This is the equation for the cutoff frequency (see the section on work function if you forgot this)
If we graph V in terms of f, and compare it to y=mx+b the slope is h/e.
Practice Problem: 1987B6
a. For a frequency of light that has a stopping potential of 3 volts, what is the maximum kinetic energy of the
ejected photoelectrons?
b. From the graph and the value of the electron charge, determine an experimental value for Planck's constant.
c. From the graph, determine the work function for the metal.
d. On the axes above, draw the graph for a different metal surface with a threshold frequency of 6.0 x 10 14 hertz.
Some sections draw directly from Knapp’s Notes.
Solution
Some sections draw directly from Knapp’s Notes.
Kinetic Energy Vs. Stopping Voltage
More often, for photoelectric problems, you will see kinetic energy vs. frequency graphed. The graphs will
look like this:
This one is even easier.
Key Features:
1.
2.
The x intercept is the cutoff (threshold) frequency.
The slope is h, Planck’s constant.
How we know this:
ℎ𝑓 = 𝜙 + 𝑒𝑉
Kmax = eV
ℎ𝑓 = 𝜙 + 𝐾𝑚𝑎𝑥
Rearrange for Kmax
𝐾𝑚𝑎𝑥 = ℎ𝑓 − 𝜙
If there is no kinetic energy, Kmax=0 and
ℎ𝑓 = 𝜙
This is the equation for the cutoff frequency (see the
section on work function if you forgot this)
If we graph K in terms of f, and compare it to y=mx+b the slope is h.
Some sections draw directly from Knapp’s Notes.