L 17 QUANTUM

Ver 1.0 © Chua Kah Hean xmphysics 1

XMLECTURE

17 QUANTUM NO DEFINITIONS. JUST PHYSICS.

17.1.0 Photoelectric Effect ................................................................................................................ 2

17.1.1 Experimental Set-Up .......................................................................................................... 3

17.1.2 Saturation Current .............................................................................................................. 3

17.1.3 Stopping Potential .............................................................................................................. 5

17.1.4 I-V Graph ............................................................................................................................ 6

17.2.0 Experimental Contradictions .................................................................................................. 7

17.3.0 Photon Theory ....................................................................................................................... 9

17.3.1 Photoelectric Equation ...................................................................................................... 12

17.3.2 Experimental Evidence ..................................................................................................... 14

17.3.3 KEmax-f Graph ................................................................................................................... 18

17.4.0 Atomic Structure .................................................................................................................. 19

17.4.1 Emission Spectrum........................................................................................................... 24

17.4.2 Absorption Spectrum ........................................................................................................ 28

17.5.0 X-ray Spectra ....................................................................................................................... 31

17.5.1 Characteristic Lines .......................................................................................................... 32

17.5.2 Bremsstrahlung Radiation ................................................................................................ 35

17.6.0 Wave-Particle Duality ........................................................................................................... 38

17.6.1 Momentum of Photon ....................................................................................................... 38

17.6.2 De Broglie Wavelength ..................................................................................................... 39

17.6.3 Heisenberg’s Uncertainty Principle ................................................................................... 40

17.7 Afterword ................................................................................................................................ 42

Appendix A: Electron-Volt ............................................................................................................... 43

Online resources are provided at https://xmphysics.com/quantum

https://xmphysics.com/quantum


17.1.0 Photoelectric Effect

What is light? A wave? Or a stream of particles? In 1801, Thomas Young performed the famous

double slit experiment. The resulting fringe pattern provided irrefutable evidence that light undergo

interference. Furthermore, in 1865, James Maxwell demonstrated (through a set of partial differential

equations) that an electromagnetic wave would travel at 8 13.00 10 m s , the exact same speed as

light. The coincidence is too rich to ignore. Surely, light has got to be an electromagnetic wave. Right?

Photoelectric Effect

Near the end of the 19th century, it was discovered that electrons are emitted from a piece of metal

when ultraviolet light is shone on it. Electrons emitted in this manner were given the name

photoelectrons. And this phenomenon came to be known as the photoelectric effect.

Can we explain this phenomenon? Of course. A piece of metal is full of electrons imprisoned by the

positively charged metal lattice. From the energy perspective, the electrical attraction imposes an

energy barrier that normally confines the electrons inside the metal. Obviously, the ultraviolet light

passes its energy to the electrons in the metal, which allow them to overcome the energy barrier and

escape from the metal as photoelectrons. Easy-peasy. So are we done?

Not so fast. As they say, the devil is in the details.

atomic lattice

ultraviolet

light

electron

photoelectron

http://www.xmphysics.com/17.1


17.1.1 Experimental Set-Up

Let’s take a look at the experimental set-up for studying the photoelectric effect.

At its most basic, we have two metal plates, called the emitter and the collector, housed in an

evacuated tube. Outside the tube, the emitter and collector are connected by a wire (and a micro-

ammeter). When ultraviolet light is shone on the emitter plate, photoelectrons are liberated. Some of

these photoelectrons will arrive at the collector by chance, where they are reabsorbed. As

photoelectrons are continuously liberated at the emitter and reabsorbed at the collector, the micro-

ammeter records a continuous DC current, aka the photoelectric current.

17.1.2 Saturation Current

Not all the photoelectrons arrive at the collector. Some are emitted in the wrong direction, miss the

collector, and land at the tube instead. Others are emitted at very low KE, wander around, and return

to the emitter instead. What can we do to guide these “lost” photoelectrons towards the collector?

uA

collector emitter

evacuated tube

UV

photoelectrons

uA

collector emitter

photoelectrons

photoelectric current

UV

+ + + +

− − − −


As illustrated in the diagram, a bias voltage can be set up between the two plates by a variable DC

power supply. When the collector is connected to the positive terminal of the power supply, it is said

to be in positive bias. As we gradually increase the positive bias, the collector (being positively

charged) becomes increasingly attractive to the (negatively charged) photoelectrons. As more and

more of the “lost” photoelectrons are reabsorbed at the collector, the photoelectric current increases.

However, once every single “lost” photoelectron has been successfully collected, the photoelectric

current saturates at a maximum value. This maximum current is called the saturation current, Is.

When the micro-ammeter shows that the photoelectric current has saturated, we know that the rate

of arrival (of photoelectrons at the collector) is equal to the rate of emission (of photoelectrons at the

emitter). This allows us to use Isat to calculate the rate of emission of photoelectrons N

t since

sat

Q NI e

t t

Worked Example 1

Calculate the rate of emission of photoelectrons for a 1.2 uA saturation current.

Solution

6 19

12 1

1.2 10 (1.60 10 )

7.5 10 s

sat

NI e

t

N

t

N

t


17.1.3 Stopping Potential

If the collector is connected to the negative terminal of the variable power supply, it is said to be in

negative bias. Electrically, the collector (being negatively charged) is now repulsive to the (negatively

charged) photoelectrons. From the energy perspective, a negative bias establishes a potential energy

barrier for the photoelelectrons.

Take for example a negative bias of 2.0 V, which establishes a potential energy barrier of 2.0 eV

between the emitter and the collector1. This is because every photoelectron that travels from the

emitter to the collector must gain 2.0 eV of EPE. This must come at the expense of losing 2.0 eV of

KE. So 2.0 V negative bias will only allow photoelectron that are emitted with at least 2.0 eV of KE to

be re-absorbed at the emitter. All the photoelectrons with 2.0 eV or lower KE do not have sufficient

KE to overcome the energy barrier, and are returned to the emitter.

This explains why as we gradually increase the negative bias, the photoelectric current becomes

smaller and smaller. The negative bias voltage at which the photoelectric current first drops to zero is

called the stopping potential, Vs. This tells us that there is not a single photoelectron with sufficient

KE to over the energy barrier of eVs. This informs that that the maximum KE of the photoelectrons is

max sKE eV

1 The eV (electron-volt) is an energy unit. Just like 1 kWh (1000)(60 60) 3.6 MJ ,

19 191 eV (1)(1.60 10 ) 1.60 10 J .

uA

collector emitter

− − − −

+ + + +


17.1.4 I-V Graph

The data from a photoelectric effect experiment is often presented in an I-V graph.

Note how the photoelectric current eventually saturates as the positive bias increases. At zero bias,

not all the photoelectrons arrive at the collector to be reabsorbed.

Note also how the photoelectric current decreases towards zero gradually as the negative bias

increases. If the photoelectrons were emitted at one uniform KE, the photoelectric current should

plunge to zero abruptly at one particularly negative bias. This informs us that the photoelectrons are

emitted with a range of KE.

Obviously, the main features to look out for are the saturation current Is and the stopping potential Vs.

1. sat

Q NI e

t t allows us to calculate the rate of emission of photoelectrons, which is the number

of photoelectrons emitted per unit time, and

2. maxseV KEallows us to calculate the maximum kinetic energy of the photoelectrons.

These two quantities informed us of the manner in which light is delivering energy to the electrons.

And as we shall see, the experimental findings were rather disturbing.

photoelectric current I

collector potential V

saturation current Is

stopping potential Vs


17.2.0 Experimental Contradictions

As mentioned earlier, physicists thought it was light waves that liberate the electrons from the metal

lattice. But when they looked at the experimental results more carefully, they began to have doubts.

Below are the I-V graphs for different light intensities (but the same light frequency, and using the

same emitter). They show that increasing the light intensity results in higher saturation currents. Is

this a problem? Nah. Since more energy per unit time (per unit area) is being delivered to the emitter,

more photoelectrons are liberated per unit time, resulting in a higher saturation current. This

observation is totally in agreement with the wave model of light.

What raises the eye brows is the I-V graphs for different light intensities sharing the same stopping

potential. This indicates that while a higher light intensity does increase the rate of emission, it does

not increase the maximum KE of emitted photoelectrons. But if a light wave is flooding the emitter

with more energy per unit time, the electrons should on average be more energetic. Not only should

more of them should break free per unit time, those which break free should escape at higher speed

and KE.

So if changing the amplitude of a light wave does not knock out photoelectrons at higher KE, what

about changing the frequency of the light wave?

Below are the I-V graphs for different light frequencies (but still using the same emitter). They show

that higher light frequency requires higher stopping potential. This shows that the maximum KE of

emitted photoelectrons is dependent on the light frequency (but independent of this light intensity).



stopping potential Vs

Is at intensity L3

L3 > L2 > L1

Is at intensity L2

Is at intensity L1



This is very odd because it is well established that the rate of transmission of energy by an EM wave

is proportional to amplitude-square, and independent of the frequency2. So the wave model predicts

that the maximum KE of photoelectrons should increase with the amplitude, not the frequency, of the

light wave. Yet the results so far indicate that the maximum KE of photoelectrons increases with the

frequency, not the amplitude, of the light. This observation flies in the face of the wave model of light!

In fact, it was discovered that if the frequency of the light used is too low, no photoelectric current can

be measured, no matter how strong the light is. This shows that there is a particular threshold

frequency f0 (for a particular type of metal) below which no photoelectric effect can occur, regardless

of the intensity of light. This is extremely odd because as an EM wave, light of any frequency should

be able to deliver energy to the emitter at a high enough rate, provided the amplitude is high enough.

If anything, there should be a threshold amplitude, not a threshold frequency, before photoelectrons

can be knocked out of the emitter. Once again, the experimental results contradict the wave model.

Last but not least, there seems to be zero time lag between the shining of the light and the registration

of a steady photoelectric current. As a wave, light can only deliver energy in a distributed (over space)

and continuous (over time) manner. In the context of the photoelectric effect, it means that the light

energy must be shared all the trillions of electrons hit by the wave front3. It also means that each

electron must accumulate the energy over a duration of time. If we keep reducing the light intensity,

we should be able to prolong the length of time an electron takes to accumulate sufficient energy

before it can be liberated. Yet, results show that photoelectrons are emitted without any delay the

instant the light arrives at the emitter, regardless of how low the light intensity is. The transfer of

energy seems to be instantaneous.

2 Mechanical waves would transfer energy at a rate that is proportional to amplitude-square and frequency-square. EM waves are not mechanical waves. 3 Light diffracts. It is impossible to focus the light beam on one single electron.



Vs at f3

f3 > f2 > f1

Vs at f2

Vs at f1


17.3.0 Photon Theory

Since the wave model of light cannot square with these three experimental observations, perhaps

light is not a wave after all? If not a wave, then what?

In 1805, Albert Einstein made the bold hypothesis that light consists of a stream of particles called

photons. He described a photon as a packet of electromagnetic radiation energy. And the amount of

energy in each packet is proportional to the frequency (and thus wavelength) of the EM radiation

hcE hf

The symbol h represents the Planck’s constant, which has a tiny value of 346.63 10 J sh . So a

photon (even with the very high frequency of X-rays) is a tiny packet of energy.

To familiarise ourselves with Einstein’s idea, let’s contrast the photon model with the wave model.

What is a beam of monochromatic light? The wave model says that it is a light wave with a single

frequency. The photon model says that it is a stream of photons with a single photon energy E hf .

Note also that a light wave delivers energy in a wave-like manner, which is continuous over time and

spread over space. But a stream of photons delivers energy in a particle-like manner, which is lumpy

and spotty.

wave model photon model

photon energy

E=hf

wave energy

spread all over



What about light intensity? As a wave, increasing the light intensity is to increase the amplitude of the

wave. As photons, increasing the light intensity is to increase the number of photons (per unit time

per unit area) in the beam.

What about light frequency? As a wave, blue light has a higher frequency (and shorter wavelength)

than red light. As photons, each blue photon packs more energy than a red photon (E hf ). In other

words, red light delivers energy in larger instalments than red light.

It may take a while to wrap your head around the new meaning given to the light frequency. It has the

unit of Hz, but there is nothing “per second” about it. It may sound like a weird theory at first, but as

you shall see, this weird theory is totally supported by experimental results. And that’s all that matters

in science.

dim light (wave model) dim light (photon model)

bright light (wave model) bright light (photon model)

red light (wave model) red light (photon model)

blue light (wave model) blue light (photon model)


Worked Example 2

A beam of monochromatic red light of wavelength 750 nm is delivering 5.0 W of power to an emitter

plate. Calculate the rate of arrival of photons Nt at the emitter.

Solution

34 819

9

(6.63 10 )(3.00 10 )2.652 10 J 1.7 eV

750 10red

hcE

19

19 1

5.0 (2.652 10 )

1.89 10 s

t

t

t

hcP N

N

N


17.3.1 Photoelectric Equation

The electrons in the metal are bound to the metallic lattice, so work must be done in order to liberate

them. Some of them are bound less loosely than others, so the amount of required work differs among

them.

The least tightly bound electrons (those nearest to the metal surface) are the ones which requires the

least amount of work to break free from the metal lattice. This minimum work is called the work

function . Each type of metal has its own work. For example, the work function for sodium, aluminium

and gold are about 2.4 eV, 4.1 eV and 5.1 eV respectively.

Ok. We are now ready to relook the photoelectric effect through our new lens given by the photon

model.

So a beam of light is shone at a piece of metal. What do you see? We used to see a wave crashing

into the metal. Now we see discrete energy packets raining down on the metal. It is like trying to

rescue the imprisoned electrons by throwing energy balls at them.

atomic lattice

photons

electron

photoelectron


If an electron is lucky enough to be “hit” by a photon, it can absorb the energy of the photon (hf). After

using part of the energy to break free from the metal lattice (W), it escapes with the balance of the

energy as its kinetic energy (KE). This energy transfer can be represented by the equation

KE hf W

This explains why photoelectrons are emitted with a range of KE. Because the amount of required

work W is different for each electron (depending on how tightly it was bound at the moment it was hit

by the photon). But the minimum W is the work function . So the most energetic photoelectrons

would be having the maximum KE of

maxKE hf

hf

KE W

hf

KEmax

sea of

electrons

inside metal

outside metal

KEmax

KE

W

hf

Why photoelectrons are emitted with a range of KE


17.3.2 Experimental Evidence

The photon theory sounds really sexy. But does it have the support of experimental observations?

17.2.2.0 Emission Rate

Firstly, can the photon theory explain the observation that higher intensity results in higher saturation

current? The picture in our head is that of photons raining down at the metal. Of all the photons arriving

at the metal, only a tiny fraction of the photons are absorbed by electrons4. Nevertheless, higher

intensity means more photons arriving at the metal per unit time. With more photons per unit time,

more photoelectrons must be liberated per unit time. Good job, photon model.

4 The so-called yield rate is seldom more than 0.1 %. This is not surprising since most metal surfaces are highly reflective.

collector emitter

− − − −

+ + + +

collector emitter

− − − −

+ + + +

low intensity light high intensity light


17.2.2.1 KEmax

Next, can the photon theory explain why increasing the intensity of light does not result in more

energetic photoelectrons? Well, the answer is clear if we remember

maxKE hf

Increasing the intensity of light merely increases the number of photons arriving per unit time. It does

not increase the amount of energy packed in each individual photon hf. So while there are more

liberations per unit time, each liberation is still being achieved by the absorption of the same amount

of photon energy. Hence the same KEmax and stopping potential.

To obtain more energetic photoelectrons, one must use a light with higher frequency which delivers

its energy using more energetic photons. Each liberation now involves a larger energy packet. The

least tightly bound electrons get to absorb a larger photon energy hf, pay off the same amount of work

function , and escape with a higher KEmax! Way to go, photon model!

collector emitter

− − − −

+ + + +

collector emitter

− − − −

+ + + +

low frequency light high frequency light

sea of

electrons

inside metal

outside metal

KEmax

hf

Why KEmax increases with frequency f


17.2.2.2 Threshold Frequency

Can the photon theory explain why photoelectric effect cannot occur below the threshold frequency,

regardless of the light intensity? Easy.

The threshold frequency f0 is the frequency at which the photon energy is equal to the work function

of the metal.

0hf

At this frequency, a photon packs just sufficient energy to knock out the least tightly bound electron

with zero KE.

max 0 0KE hf

Below f0, each photon does not pack sufficient energy to liberate even the least tightly bound electron

in the lattice. Hence no photoelectric effect!

collector emitter

− − − −

+ + + +

collector emitter

− − − −

+ + + +

above threshold frequency below threshold frequency

sea of

electrons

inside metal

outside metal

Why no photoelectric effect if hf <

hf0

KEmax

hf


But why does increasing the light intensity and having more photons per unit time not help? Well, to

understand this, one must realize that the chance of an electron being hit by two photons

simultaneously at the same time is practically zero (except at extremely high intensity situations).

Besides, the electron is also unable to hold on to any surplus energy for long: if it does not escape

after absorbing the energy of one photon, it must give up the energy immediately and start from

square one again. As such, an electron is unable to accumulate the energy of multiple photons. Each

photoelectron must make its escape with the energy of one photon only. If every individual photon is

unable to knock out a photoelectron, having more of them per unit time is not going to change the

outcome.

17.1.2.3 Zero Latency

Finally, does the photon theory explain why photoelectric effect is instantaneous and has zero latency?

You bet!

A wave delivers energy continuously to many electrons over a duration of time. A photon delivers its

entire packet of energy to one electron instantaneously. The moment one electron absorbs the energy

of one photon, one photoelectron is liberated (if the photon energy at least equal to the work function).

This explains why a steady photoelectric current (albeit a small one) is registered the moment the

light (no matter how low the intensity) is arrives on the metal.

Yay!

photon photoelectron

(electron)

wave model photon model


17.3.3 KEmax-f Graph

So far, the photon model has done well to provide qualitative explanations for the experimental

observations. But it is not physics until we start measuring and verifying the numbers.

So we conduct the photoelectric effect experiment using emitters made of different types of metal.

We measure the maximum kinetic energy (KEmax) at different light frequencies f. We can then plot the

graph of KEmax against f, one for each type of metal. Guess what we get?

Parallel lines! You are staring at the quantitative proof of the photoelectric equation

maxKE hf

The gradient of those parallel lines correspond to the Planck’s constant, 346.63 10 J sh . And the

y-intercepts of each graph corresponds to the work function of each metal!

So what do you think now? Is light a wave or a stream of photons? The photoelectric effect endorses

the photon theory totally. But what about the double-slit experiment? Doesn’t it still lend irrefutable

support to the wave theory? Hmm. Obviously, we are not finished yet.

maximum

KE sodium

aluminium

frequency

−2.4 eV

−4.1 eV

−5.1 eV

gold

no photoelectric effect below

threshold frequency f0


17.4.0 Atomic Structure

In 1896, J. J. Thompson discovered the electron. In 1911, E. Rutherford discovered the nucleus, and

went on to propose the Rutherford model: that each atom contains a miniature solar system, with

electrons orbiting around a massive nucleus.

It’s all beautiful except the one small problem: Maxwell’s Equation predicts that a time-varying electric

field must result in an electromagnetic wave. If the electrons in an atom are going in circles, their

acceleration must result in the emission of EM waves from the atom. The electrons should be losing

energy and spiraling towards the nucleus (just like satellites losing energy must eventually crash down

to Earth). In fact, based on calculations of the rate of emission, the collapse of the atom should be

over and done with in the blink of an eye. The Rutherford’s atom did not sound very stable.

Backtrack to 1885, when J J Balmer discovered the formula to describe the hydrogen spectrum. You

see, most light sources have a continuous spectrum with no break or gap throughout its wavelength

range. For example, when sunlight is passed through a prism, the “rainbow” spectrum is a continuous

spectrum: light of all wavelengths is present.

nucleus

electron

filament lamp

Continuous

spectrum


The light emitted by hot gases, however, have discrete spectra consisting of a set of discrete spectral

line: only a few colours are present.

For example, the visible spectrum from hydrogen5 displays only four wavelengths: 410 nm, 434 nm,

486 nm and 656 nm. What’s so special about these wavelengths?

Amazingly, Balmer (who was a secondary school mathematics teacher) was able to come up with a

formula for those wavelengths!

2 2

1 1 110973731.57( )

2n n for 3,4,5,6,...n

Impressive as the formula may look, it doesn’t explain why hydrogen atoms are emitting light only at

these wavelengths and not others.

5 The hydrogen spectrum was intensely studied because hydrogen is the simplest atom: if you can’t explain

even the hydrogen atom, you have no chance with other atoms.

neon sign

Discrete

spectrum

hydrogen emission spectrum


In 1913, Niels Bohr, after thinking about the instability of Rutherford’s atom, Balmer’s hydrogen

spectrum and Einstein’s photon, proposed the Bohr model, which asserts the electron in the hydrogen

atom is only allowed to orbit at certain radii.

Just like the Earth-satellite system has different amounts of energy depending on the altitude of the

satellite’s orbit, the proton-electron system has different amounts of energy depending on the radius

of the electron’s orbit. If the electron is allowed only certain orbits, then the atom can only have certain

amounts of energy. We say that the energy levels of a hydrogen atom is quantized.

According to Bohr, the hydrogen atom can make a transition from a higher energy level EH to a

lower energy level EL by emitting a photon with energy Ephoton equal to the energy difference

between those two energy levels.

photon H LE E E

If the atom is allowed only certain energy levels, it would imply that it cannot emit photons of just

any energy. Instead it can only emit photons whose energies correspond to the energy difference

between a pair of allowed energy levels. This may explain why the spectrum is discrete and not

continuous!

allowed orbits

E1

(Ground State)

E2

E3

Energy

E1

E2

E3

allowed energy levels

EH

EL excitation

EH

EL de-excitation

photon


Using Balmer’s formula, Bohr “worked backward” 6and deduced that the allowed energy levels for

the hydrogen atom is given by the formula

2

13.6 eVnE

n for 1,2,3,...n

Note that

1. 1n represents the lowest possible energy, aka ground level of the hydrogen atom. If a hydrogen

atom is already at the ground level, it cannot lose any more energy (hence the atom is saved from

the death spiral).

2. All higher levels are called excited levels.

3. The energy levels are all negative in values. This is because the atom is held together by the

attraction between the proton and the electron, making the electric potential energy of the system

negative. Bohr basically took the total energy to be the summation of this EPE and the KE of the

orbiting electron. So the total energy is zero only when the electron is infinitely far from the proton.

4. 13.6 eV is thus the energy required to remove the electron from the hydrogen atom. This is also

called ionization energy of the hydrogen atom.

6 Ok. Bohr actually started from the argument that the angular momentum L of the orbiting electron must be

quantized (as in 2

hL n

), but you don’t have to know such details for the H2 syllabus.

n=1

n=2

n=3

n=4

n=5

n=∞

-13.6 eV (Ground level)

-3.40 eV

-1.51 eV

-0.85 eV

-0.54 eV

0 eV

Balmer series

Lyman series

Paschen series

(Free electron)


Let’s take Bohr’s model out for a test drive. Suppose the electron makes a transition from 3n to

2n . The emitted photon thus has energy

( 1.51) ( 3.40) 1.89 eVphotonE

Since photon

hcE

, the emitted photon’s wavelength can be calculated as follow.

34 819(6.626 10 )(2.998 10 )

1.89(1.602 10 )

656 nm

This explains why the hydrogen spectrum contains a spectral line at 656 nm, and not say at 646 nm

or 666 nm! In fact, all the spectral lines of hydrogen can now be “explained”: The transitions that end

at 2n (which happen to produce photons in the visible spectrum) match up to the Balmer series.

In addition, the transitions that end at 1n and 3n produce photons in the ultraviolet region and

infrared region respectively. The corresponding spectral lines have been confirmed experimentally,

and are known as the Lyman and Paschen series.

As you can see, the Bohr model was strongly supported by the spectral lines of hydrogen. Even

though there were still many unresolved issues, the Bohr model represented an important milestone

in our understanding of the structure of the atom.


17.4.1 Emission Spectrum

People just love those bright and colorful “neon” signs. They are basically tubes containing gases at

low pressure. When a voltage (at least a few hundred volts) is applied across the ends of such a

discharge tube, some of the gas atoms become ionized, thus enabling a discharge current. The gas

gets heated up and gives off light.

All gaseous atoms display a discrete line spectrum. This is easily confirmed by passing the light

through diffraction gratings. In fact, each element in its gaseous state has a unique line spectrum.

This suggests that atoms of each element has its own discrete energy levels. And the transitions

between each element’s unique set of energy levels give rise to each element’s signature spectral

lines.

watch video at xmphysics.com

He Neon Argon Krypton Xenon

hydrogen

helium

neon

sodium

mercury

http://www.xmphysics.com/17.4.1



Unfortunately, the energy levels of atoms with two or more electrons cannot be described by a simple

formulae like the one for the hydrogen atom. To be honest, Bohr’s model of electrons as particles

orbiting around the nucleus has long gone obsolete7. Nevertheless, its main ideas of quantization and

electronic transition remain true to all atoms.

Worked Example 3

A hypothetical atom has four energy levels: the ground level and 1.90 eV, 2.90 eV and 3.70 eV above

the ground level respectively. Find the wavelengths of the spectral lines that this atom can emit when

excited.

Solution

Firstly, let’s sketch the energy lines for this atom. Note that rightfully, all the energy levels should be

negative in value (since the electrons are still bound to the nucleus). But for convenience, we often

depict the ground level as the reference level of 0 eV and the excited levels as positive energy levels.

7 It’s a long (but truly fascinating) story. But our current understanding is that electrons exist as a wave-like

probability cloud around the nucleus.

E1 (ground level)

E2

E3

E4

1.90 eV

0 eV

2.90 eV

3.70 eV


Suppose the atom is excited to energy level E4. From here, the atom can return to ground level in one

de-excitation, producing one single photon. Or it can “stop” at each intermediate energy level along

the away, returning to ground state in three de-excitations, producing three (less energetic) photons.

Or it can returning to ground state in two “jumps”, producing two photons.

In general, given N energy levels, there are 2

NC possible down transitions (since we are looking for

the number of ways to choose 2 out of N energy lines). For this hypothetical atom, the maximum

possible number of spectral lines is 4

2 6C . They come from the three down transitions ending at E1,

two ending at E2, and one ending at E3.8

8 In reality, momentum and other constraints mean transitions between some energy levels are impossible. So the actual number of spectral lines are much fewer than NC2.

E1 (ground level)

E2

E3

E4

1.90 eV

0 eV

2.90 eV

3.70 eV

E1 (ground level)

E2

E3

E4

1.90 eV

0 eV

2.90 eV

3.70 eV


We can calculate the wavelength of the emitted photon for any de-excitation using

H L

hcE E

Using the de-excitation from E2 to E1 as an example, we have

34 819(6.63 10 )(3.00 10 )

(1.90 0)(1.60 10 )

654 nm

Calculations for all the possible transitions are tabulated below.

Transition Ephoton photon EM region

4→1 3.70 eV 336 nm UV

3→1 2.90 eV 429 nm violet

2→1 1.90 eV 654 nm red

4→2 1.80 eV 691 nm red

3→2 1.00 eV 1240 nm IR

4→3 0.80 eV 1550 nm IR

It turns out that only three spectral lines fall in the visible spectrum. The rest are in the ultraviolet and

infrared regions.

emission spectrum for this hypothetical atom


17.4.2 Absorption Spectrum

The absorption spectrum is basically the reverse of the emission spectrum. Instead of studying the

photon energies emitted by gas atoms, we are studying the photon energies absorbed by gas atoms.


To produce an absorption spectrum, we need to pass a beam of white light through a gas. The beam

of “white” light should contain photons of all possible energies in the continuous spectrum. This light

beam passing through the gas is like photons trying to sneak past gas atoms without getting eaten

up by the gas atoms. Actually, not all photons need to worry. Because the gas atoms are only able to

absorb a photon only if the absorption puts them in another allowed energy level. In other words, the

photons energy must correspond to

H L

hcE E

where EH and EL correspond to two transitioning energy levels.

After being “filtered” by the gas, the light beam is passed through a diffraction grating. On the other

side of the grating we observe an absorption spectrum, which is basically a continuous spectrum (of

the white light) punctured by a few dark lines (absorbed by the gas atoms). Those dark lines are called

the absorption lines. Since each type of gas has its own energy levels, each type of gas produces its

own characteristic absorption lines.

white light

continuous

spectrum

emission

spectrum

hot gas

white light

absorption

spectrum

gas




You may be wondering what happens after the atoms have absorbed those photons. Don’t these

excited atoms eventually (if not immediately) return to ground level and inadvertently emit the same

photons that they absorbed? That would return the “spilled” photons back to the “white” light beam,

wouldn’t it? Well, yes and no. You see, the photons are indeed re-emitted, but they are re-emitted

randomly and uniformly in all directions. So only a tiny fraction will be re-emitted in the original

direction of the “white” light (and pass through the grating). In that sense, scatter may be a more

accurate word than absorb. By the way, to form an absorption line, it is not even necessary for all the

photons of that absorption line to be absorbed by the gas. Even if only a small fraction is scattered,

the dip in intensity is usually enough to cast a dark line against the very bright continuous spectrum.

Worked Example 4

An absorption spectrum is formed by passing white light through a gas consisting of the hypothetical

atoms mentioned in worked example 3. Find the wavelength of the absorption lines.

Solution

In general, all the possible down transitions are also the possible up transitions.

The energies of photons that the atom is capable of emitting, are also the energies of photons that

the atom is capable of absorbing.

Transition Ephoton photon EM region

1→4 3.70 eV 336 nm UV

1→3 2.90 eV 429 nm violet

1→2 1.90 eV 654 nm red

2→4 1.80 eV 691 nm red

2→3 1.00 eV 1240 nm IR

3→4 0.80 eV 1550 nm IR

E1 (ground level)

E2

E3

E4

1.90 eV

0 eV

2.90 eV

3.70 eV


This is why the absorption spectrum for a particular gas often “matches” its emission spectrum: the

emission lines in the emission spectrum are also the absorption lines in the absorption spectrum.

This is assuming that all the possible up transitions are taking place. This may not be true if a gas is

at too low a temperature and there are too few atoms existing at the higher energy levels. For example,

suppose that gas in our worked example is so “cool” that at any one time, practically all the atoms are

at the ground state. Since there is hardly any atom existing at E2, E3 and E4, the 2→4, 2→3 and 3→4

absorptions (that require the photons to hit an atom initially at E2, E3 and E4) would not be happening.

The only absorptions occurring would be those that begin from the ground level, i.e. 1→2, 1→3 and

1→4. This would explain why the 691 mm absorption line (2→4) is “missing” from the absorption

spectrum.

emission spectrum

absorption spectrum

E1 (ground level)

E2

E3

E4

1.90 eV

0 eV

2.90 eV

3.70 eV

69

1 n

m

65

4 n

m

42

9 n

m

emission spectrum

65

4 n

m

42

9 n

m

absorption spectrum


17.5.0 X-ray Spectra

X-ray is part of the electromagnetic spectrum. It has wavelength of the order of 10-10 m. This puts the

energy of an x-ray photon in the order of 10 keV.

An x-ray tube is an evacuated tube containing a filament and a target metal (most commonly tungsten).

A voltage is applied to the filament to heat it up until electrons are released9. A very high accelerating

voltage VA (typically a few hundred kV) is applied between the filament and the target metal. This

voltage provides a strong electric field that accelerates the electrons to a very high speed before

smashing them against the target metal. X-ray is produced out of the carnage. Basically, the KE of

the filament electrons is being converted into x-ray photons (and a lot of unwanted heat).

The spectrum of the x-ray radiation is pretty interesting; it is a combination of discrete spectral lines

and a continuous spectrum. The discrete spectral lines are known as the characteristic lines. The

continuous spectrum is called the bremsstrahlung (German for braking) radiation, or simply the

background radiation. The continuous spectrum is actually range bound; the minimum wavelength is

called the cut-off wavelength min.

9 It is similar to photoelectric effect, except that heat is used instead of photons to liberate the bound electrons

accelerating

voltage VA

X-ray tube

X-ray

electrons

target metal filament

K

K

L

L

intensity

wavelength min

bremsstrahlung

characteristic lines


It turns out that x-ray photons are being produced through two different kinds of interaction between

the filament electrons and the target metal atoms.

1. Interactions between the filament electrons and the shell electrons produce characteristic x-ray

photons (depicted by (1) in diagram) which form the characteristic line spectrum.

2. Interactions between the filament electrons and the atomic nuclei produce bremsstrahlung

(German for braking) x-ray photons (depicted by (2) in diagram) which form the background

radiation.

We will now look at these two mechanisms in more detail.

17.5.1 Characteristic Lines

We have learnt how an excited hydrogen atom can emit photons when it de-excites. So can we get

x-ray photons out of hydrogen atoms?

No way. To qualify as x-ray, a photon must carry energy of the order of 10 keV. The ionization energy

of a hydrogen atom is a paltry 13.6 eV. There is no way we can milk x-ray photons out of hydrogen

atoms. X-ray is a game played by big boys such as tungsten atoms.

Take for example tungsten, which has an atomic number of 74 with a shell structure of

K2.L8.M18.N32.O12.P2. What this means is that at ground state, the K-shell (the innermost shell) is

filled by 2 electrons, the L-shell (the next higher shell) is filled by 8 electrons, and so on.

filament

electron (1)

(1a) ejected K-shell electron

(1b) cascading L-shell electron

(1c) characteristic x-ray photon

K filament

electron (2)

L M

(1a)

(2b)

(2a) nucleus

(2b) bremsstrahlung x-ray photon

(1b) (1c)

(2a)


The binding energies for a K, L and M shell electron are 69.5 keV, 10.2 keV and 1.8 keV respectively.

Binding energy is the amount of energy required to dislodge an electron from the atom. So binding

energy is actually an energy deficit rather than energy available.

Consider a K-shell electron, which requires 69.5 keV of energy to be kicked out of the atom. It is held

very tightly by the 74 protons in the nucleus. It rarely gets any visitors, being in the innermost shell,

shielded by the L, M and outer shells. Until the x-ray tube voltage sends in a beam of highly energetic

filament electrons. Let’s assume an accelerating voltage of 100 kV. This means the filament electrons

(which left the anode with negligible KE) would lose 100 keV of EPE and thus gain 100 keV of KE by

the time they come knocking on the K shell electrons. At 100 keV, the filament electron has sufficient

KE to knock out a K-shell electron, leaving behind a vacancy in the K-shell. The race is on for an outer

shell electron to fill this vacancy!

More often than not, it is an L-shell electron that wins the race. In making the transition from L shell

(binding energy 10.2 keV) to K shell (binding energy 69.5 keV), the atom must give up

69.5 10.2 59.3 keV of energy. This energy is released as a 59.3 keV x-ray photon. This is called

a K photon. But this is not the end of the story because a vacancy has just opened up in the L shell.

If this vacancy is filled by a M-shell electron, the transition from M to L shell will produce a photon with

energy 10.2 1.8 8.4 keV . This is called a L photon.

filament

electron

K K

L

M

N

ejected

K-shell

electron

L

M


On occasion, an M-shell (binding energy 1.8 keV) can actually beat an L-shell electron to filling the K-

shell vacancy. In that case, the M to K-shell transition would produce an x-ray photon with energy of

69.5 1.8 67.7 keV . This is called a K photon.

I hope you get the drift. Whenever a shell electron is knocked out of the atom, a “chain reaction” is

triggered in which electrons from higher shells “cascade” down to fill up the vacancies. During these

down transitions, characteristic x-ray photons with energies corresponding to the energy gaps

between the transitioning shells are released.

photonE E

This explains why this cascading action produces a line spectrum. Furthermore, since each metal has

its own unique energy structure, each metal and will produce its own signature line spectrum. This is

why they are called characteristic lines.

K-shell -69.5 keV

-10.2 keV

-1.8 keV

0 keV (Free electron)

Energy

L-shell

M-shell

K

K

L

K

K

intensity

frequency

L


17.5.2 Bremsstrahlung Radiation

When we were discussing the problem of the Rutherford’s atom, we mentioned that a time-varying

electric field results in EM radiation. In fact, when we pushes a high frequency AC current in copper

wires, it emits radio waves. This is how antennas work! So can we produce x-rays by oscillating an

electron vigorously?

Nope. The magnitude of acceleration needed for a charge to emit x-rays is incredibly large. We just

don’t have the AC power supply to provide that kind of acceleration to electrons. But the nuclei of a

tungsten atom can.

The nuclei of a tungsten atom is 74e of (positive) charge squeezed into a tiny volume. It is close to

the proverbial point charge whose field strength approaches infinity at close distance. So in the x-ray

tube, some of the filament electrons, being so energetic and penetrative, actually find themselves in

the proximity of the tungsten nuclei (after evading the shields formed by the electron shells). At such

humongous acceleration, bremsstrahlung photons are emitted.

The bremsstrahlung photon energy depends on the degree of interaction between the electron and

the nucleus. The closer the electron swings past the nucleus, the more KE it loses, and the more

energetic the x-ray photon produced. Since the filament electron can lose any fraction of its energy

during an interaction with the nucleus, it can lose anything between 0 keV and 100 keV of energy

(assuming an accelerating voltage of 100 kV). This implies that a bremsstrahlung photon can have

any energy between 0 keV and 100 keV.

filament

electrons

(1)

(2) (3)

(1) distant interaction produces low energy x-ray photon.

(2) close interaction produces moderate energy x-ray photon.

(3) maximum energy of x-ray photon is equal to energy of field electron.


This is the reason why the bremsstrahlung radiation has a continuous spectrum. Furthermore, the

cut-off frequency (and cut-off wavelength) can be calculated easily by equating the x-ray photon

energy to the KE of the filament electron.

max

min

A

hchf eV

Just one more thing. Some students are curious why the continuous spectrum is often depicted with

a minimum frequency, suggesting that low frequency photons are not emitted. Actually, the

bremsstrahlung process produces more low frequency photons than high frequency photons. Just

that in practice, the longer wavelength radiation are naturally absorbed and filtered away by the walls

of the x-ray tube.

intensity

frequency

cut-off

frequency fmax

low frequency

filtered away

by tube


Worked Example 5

The K and L shell ionization energies of copper are 8979 eV and 951 eV respectively. For a x-ray

tube that uses accelerating voltage VA of 12 kV, determine

a) the wavelength of the K line

b) minimum cut-off wavelength.

Solution

a)

34 819

10

(6.63 10 )(3.00 10 )(8979 951)(1.60 10 )

1.55 10 m

hcE

b)

min

34 819

10

(6.63 10 )(3.00 10 )(12000)(1.60 10 )

1.04 10 m

A

hceV

K

intensity

wavelength/10-10 m 1.04

(1.55×10-10 m)


17.6.0 Wave-Particle Duality

By the 1920s, the photoelectric effect and the atomic spectra (plus the blackbody radiation which is

not in the H2 syllabus) had already provided solid evidence that light consists of particles. Yet the

double-slit interference pattern persisted as strong vindication that light consists of waves.

To a classical physicist, these two description of light are mutually exclusive. So (at least) one of them

must be wrong. To some quantum physicist in the 1920s, these two descriptions are complementary

to each other: whether an object behaves as a particle or as a wave depends on your choice of

apparatus and experiments for looking at it. This concept is known as the wave-particle duality.

17.6.1 Momentum of Photon

In two papers published in 1909 and 1916, Einstein pointed out that for photons to be full-fledged

particles, they must also carry momentum

hp

But how can photons, which have no mass, have momentum? Well well well. In 1922, Compton

directed a stream of x-ray photons at carbon atoms. If the photons do have momentum, then their

momentum should decrease after colliding with the electrons (initially bound) in the carbon atoms.

This was indeed confirmed by the observation that the wavelengths of the x-ray increased after being

scattered (by the carbon atoms). Furthermore, calculations based on PCOM and PCOE confirmed

that the numbers checked out. So h

p

was validated.

Worked Example 6

A laser beam is incident normally on a mirror. The beam has wavelength 680 nm and power 14 mW.

Calculate the force exerted by the laser on the mirror if it is 100% reflective.

Solution

Momentum of each laser photon, 34

28 1

9

6.63 109.75 10 kg m s

680 10photon

hp

Energy of each laser photon, 34 8

19

9

(6.63 10 )(3.00 10 )2.925 10 J

680 10photonE

Number of photons incident on mirror per unit time, 16 1

19

0.0144.786 10 s

2.925 10t

photon

PN

E

Force exerted 16 28 11(4.786 10 )(2 9.75 10 ) 9.33 10 Nt

dpN p

dt



17.6.2 De Broglie Wavelength

In 1924 Prince Louis de Broglie advanced (in his one-page doctoral thesis) a crazy hypothesis. He

was convinced that the wave-particle paradox should not be confined to photons only. Instead, the

chaos should be extended to all particles, including electrons and protons.

If light, which we assumed to be waves, can behave like particles with momentum h

p

in some

situations, then it is only fair that electrons, which we assume to be particles, should behave in some

situations like waves with wavelength

h

p

In 1927, G P Thomson designed an experiment to verify this claim. He knew of experiments where

diffraction patterns are formed by shining x-ray through very thin gold foils10. He figured all he had to

do, was to replace the x-ray with a beam of mono-energetic electrons. He chose electrons with KE of

25 keV, which according to de Broglie’s formula, should behave like a wave of wavelength

127.7 10 m . This is to match the wavelength of the x-rays normally used in such experiments.


As it turned out, concentric rings were formed on the photographic plate! It is the exact same

interference pattern (including the positions of the rings) as what would have been produced if x-ray

of 127.7 10 m were used instead of the electron beam. The diffraction of electrons served as the

first evidence of the wave nature of electrons11.

10 The regular arrangement of the atoms of the gold crystal act as a natural three-dimensional diffraction grating. 11 Davisson and Germer performed another experiment in which the diffraction pattern is formed by reflecting a low power electron beam on nickel crystal. They won the Nobel prize together with Thompson.

thin

gold foil

electron

beam

photographic

plate

diffraction

pattern




17.6.3 Heisenberg’s Uncertainty Principle

In 1927, Heisenberg arrived at a peculiar looking equation (while plowing through the mathematics of

quantum mechanics).

x p ⪆ h

here indicate standard deviations. And x and p represent the uncertainties in the position and

momentum of a particle respectively. So the inequality seemed to suggest that the product of these

two uncertainties (for a particle) cannot be smaller than a minimum value that is of the order of the

planck’s constant h.

First, let’s remind ourselves that h is a very tiny number. For everyday measurements, x and p (due

to instrumental precision and procedural uncertainties) are humongous compared to h. So this

inequality does not pose any meaningful constraint when you are measuring objects like a ball or a

chicken. On the other hand, if you are measuring elementary particles like photons or electrons, you

are likely to be dealing with x and p whose products are of the same order as h. Now the minimum

limit imposed by this inequality is an imposing one. Because decreasing either of one of x or p will

must cause the other one to increase. The inequality also rules out the possibility of either one of

them being completely zero.

So that’s the mathematics. But what’s the meaning of it all, physically? Heisenberg originally explained

this as a consequence of the measuring process: Measuring position accurately would disturb

momentum and vice versa. He offered the "gamma-ray microscope" thought experiment as an

illustration: He imagined determining the position and momentum of an electron by illuminating it with

gamma rays. But our new found quantum knowledge informs us that we are actually bouncing gamma

photons off the electron, inadvertently changing the momentum of the electron. To reduce the

uncertainty in the position measured, we could reduce the wavelength of the gamma radiation. This

comes from our understanding of optics, and is related to the Rayleigh’s criterion. But de Broglie’s

equation tells us that reducing the wavelength increases the momentum of those gamma photons.

The electron will be hit harder and undergo a larger change due to the large momentum of the photon,

thus increasing the uncertainty in the momentum measured.

Today, it is understood that Heisenberg’s original interpretation of the Heisenberg’s Uncertainty

Principle (as it came to be known), is at best incomplete, and at worst totally mistaken. Today, we

believe that the Heisenberg’s uncertainty does not arise because of a disturbance caused by a



measurement. The HUP is inherent in the wave-particle duality nature of the particle itself. As the

complete explanation will be too complicated, I will only offer an analogy here.

Let’s contrast a wave and a particle. A (continuous sinusoidal) wave has a definite wavelength and

thus momentum (h

p

), but a completely undefined position. On the other hand, a particle has a

definite position, but a completely undefined wavelength and thus momentum.

Because of wave-particle duality, nothing is completely a wave nor a particle. Instead, everything is

more like a wave packet. If the packet were narrower, the uncertainty in its position would be smaller,

but the uncertainty in its wavelength12 and thus momentum will be higher. Conversely, if the packet

were broader, the uncertainty in its wavelength and thus momentum will be smaller, but the

uncertainty in its position would be larger.

Every elementary particle exists like a wave packet. When it is not interacting with other objects, it

has neither a completely defined position nor momentum. Instead, it exists with a range of possible

positions and possible momentums. The x and p of its probabilistic existence is limited by the HUP

x p ⪆ h

Well, I will stop here and let you read up on the topic on your own if you’re interested.

12 Because a narrower pulse is formed by summation of higher number and higher frequency sinusoidal waves. Look up Fourier Transform if you’re interested.

∆p=0

∆x=∞

∆x=0

∆p=∞

larger ∆p

smaller ∆x

smaller ∆p

larger ∆x


17.7 Afterword

The concepts in Section 17.5 are presented as what’s required by the H2 syllabus. They do not

represent the most up-to-date understanding of the subject by current physicists13. They are more like

snapshots of some of the major experimental observations and theoretical interpretations during the

early days of modern physics.

Expectedly, many students are unconvinced about many concepts, especially the wave-particle

duality and heisenberg’s uncertainty principle (HUP). Unfortunately, to fully explain the current state

of understanding is not a trivial undertaking. If you’re interested, I strongly encourage you to read

“How to Teach Quantum Physics to Your Dog” by Chad Orzel. It is written in a humorous but

nevertheless conceptually accurate manner, which should help you attain some degree of closure.

13 For example, the Bohr’s model of electrons in circular orbits has been superseded by the Schrodinger’s

equations’ probabilistic “clouds”, the wave-particle duality has been clarified by the wave-function, and the

HUP has been updated to be xp ≥ h/4


Appendix A: Electron-Volt

In the field of quantum physics, where electrons serve as the protagonists, the electron-volt (eV) is

the preferred unit over the joule (J).

Just like 1 kWh is

1 kWh (1000)(60 60) 3.6 MJ

1 eV is

19 191 eV (1)(1.60 10 ) 1.60 10 J

Remember we learnt in electric field that W qV ?

How much KE does a photoelectron lose if it travels from the emitter to collector with a negative bias

of 3.0 V?

19(1.60 10 )(3.0) 3.0 eVW qV

This implies that only photoelectrons with initial 3 eVKE can arrive at the collector. Those with

initial 3 eVKE will be turned back before reaching the collector. This is why a stopping potential of

3.0 V indicates that KEmax is 3.0 eV. It is that straight-forward!


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L 17 QUANTUM