How is it possible to have a microscope able to zoom so ... · electrons. If electrons act as a...

How is it possible to have a microscope able to zoom so much so that

individual atoms can be seen?

www.aip.org/history/einstein/atoms.htm

Discover the answer to this question in this chapter.

Luc Tremblay Collège Mérici, Quebec City

2019 Version 11-Quantum Mechanics 2

The Wavelength Formula

In 1923, Louis de Broglie came to a surprising conclusion: matter can also act as a wave!

He even found the formula giving the wavelength of these matter waves.

De Broglie Wavelength

h

pλ =

where p is the momentum of the particle.

Example 11.1.1

What is the wavelength of an electron travelling at 3 x 106 m/s (1% of the speed of light)?

In this case, the momentum can be calculated with the non-relativistic formula.

Therefore,

34

31 6

6.626 10

9.11 10 3 10

0.243

ms

h

mv

Js

kg

nm

λ

−

−

=

×=

× × ×

=

De Broglie arguments were not based on any experiment. Instead, it consisted of a series

of theoretical arguments on the coherence of physics. To illustrate the idea, here are two of

the evidence presented by de Broglie.

Argument 1: Einstein Relativity

As mentioned in the chapter on relativity, vectors with three components have no

place in relativity. Often, vector quantities are found in four-vectors with

4 components. For example, the relativistic energy and the momentum are the four

components of the energy-momentum four-vector.

, , ,x y z

Ep p p

c



This was never mentioned, but k is also a vector (we would have seen it with a more

advanced study of three-dimensional waves). It is a vector of

magnitude 2π/λ directed towards the direction of propagation of the wave. In

relativity, this vector is part of a four-vector with the following 4 components.

, , ,x y zk k kc

ω

E = hf is a relationship between one of the components of these two four-vectors.

22

E hf

hE f

E

E

c c

ππ

ω

ω

=

=

=

=

�

�

However, if there is a relationship between one of the components of a four-vector,

the same relationship must hold for the other components. Thus, the following

relationship must also be true.

p k= �

This leads to

2

2

hp

h

p

π

π λ

λ

=

=

Argument 2: Bohr’s Postulates

Let’s see what happens with Bohr’s postulate

mvr n= �

If de Broglie formula is used, it becomes



2

2

mvr n

pr n

h hr n

n r

λ π

λ π

=

=

=

=

�

�

This result indicates that the

circumference of the orbit

must be an integer number of

the wavelength of the electron.

This is exactly the condition to

have a standing wave on a

circular path.

www.scinote.org/blog/seeking-scinote-physics-why-dont-electrons-fall-into-the-nucleus

De Broglie formula, therefore, seems to give a justification to Bohr’s postulate by

saying that the allowed orbits are those for which the electron forms a standing wave.

De Broglie presented his theory in a paper in September 1923. Many scientists were

initially skeptical until Einstein came to the same conclusion as de Broglie in January 1925

starting from different premises. Erwin Schrödinger also showed that the trajectory of a

projectile could be explained by a refraction if the projectile is a wave whose wavelength

is given by de Broglie formula. The ideas of de Broglie then seemed to be confirmed.

It only remained to experimentally prove this idea. De Broglie himself mentioned, in his

paper, that the wave nature of matter can be shown by observing interference and

diffraction with electrons. But first, we have to understand the meaning of matter can act

as a wave.



The idea that matter can act as wave is not a simple restatement of what was learned about

mechanical waves, which are waves propagating in matter. For these waves, matter is not

a wave, it only composes the medium that supports the waves.

Also, it does not mean that particles travels following an undulating path.

Common Mistake: Thinking That the Electron

Follows an Oscillating Path

Matter waves do not represent the path of the particles. The electrons do

not move along a wavy path whose wavelength is h/p.

The idea that the matter can act as a wave rather means that particles like electrons can, for

example, undergo diffraction when passing through a small hole. But what happens then?

With diffraction, the wave spreads out after its passage through a hole. Does that mean that

an electron spreads out after its passage through a small hole? How can a particle spread?

The Copenhagen Interpretation

To understand what the wave represents, let’s look at

what happens if Young’s experiment is done with

electrons. If electrons act as a wave, there should be an

interference pattern with maxima and minima on the

screen, exactly as with light. (To allow the scientists to

record the position where an electron hits the screen, a

special screen is used. This screen becomes luminous at

the spot where the electron hits it.)

www.blacklightpower.com/theory-2/theory/double-slit/



This is the result that was obtained when this experiment was done (in 1989).

www.hitachi.com/rd/portal/research/em/doubleslit.html

The interference pattern bears a striking resemblance to the one obtained in Young’s

experiment performed with light. In addition, the spacing between the bright fringes fits

with the wavelength of the wave given by the de Broglie formula.

The result of this experiment shows that many particles are hitting the screen at locations

where there is a maximum of interference, so at the places where the amplitude of the wave

is maximum. Few particles hit the screen at locations where there is a minimum of

interference, so at the places where the amplitude of the wave is very small.

This suggests the following interpretation, made by Max Born in 1926, known as the

Copenhagen interpretation.

Copenhagen Interpretation

The square of the amplitude at a location is proportional to the probability of

finding the particle at this location.

Thus, more particles hit the screen at the places where there is a maximum of interference

because the amplitude is greater at these places and there is more chance of finding the

particle at these locations.

The symbol used for the amplitude of the wave is ψ. So, for the double-slit experiment

with electrons, the following figure shows the link between ψ ² and the number of electrons

arriving on the screen.



www.chegg.com/homework-help/questions-and-answers/physics-archive-2011-november-20

It is obvious that more electrons hit the screen at the positions where ψ ² is great.

This interpretation is valid for every kind of particles, including photons. Note that Einstein

had suggested something similar in 1905 when he suggested for the first time that light was

composed of photons. He proposed that the intensity of light on a surface, which is

proportional to the square of the wave amplitude, should be proportional the number of

photons arriving on this surface.

Experimental Proofs

It was impossible to perform the double-slit experiment when de Broglie published its

results but they still managed to confirm the ideas of De Broglie with interference as soon

as 1926.

At that time, Clinton Davisson and Lester Germer, and, independently, George Paget

Thomson and Alexander Reid obtain interference

pattern with electrons. In the Davisson and Germer

experiment, electrons were passing through a nickel

crystal. The regular spacing of the atoms inside the

crystal is such that it acts as a grating. If the electrons

act as a wave, an interference pattern is supposed to be

seen. The wavelength of the wave can even be found

from the spacing between the maximum of interference

with dsinθ = mλ. The image to the right shows the

interference pattern obtained when electrons pass

through a manganese and aluminum crystal.

scienceblogs.com/gregladen/2011/10/05/there-can-be-no-such-creature/



The bright spots are the positions where many electrons hit the screen, and these correspond

to interference maxima. Thus, there is an interference pattern, and this confirms that

electrons act like waves.

The image to the right shows

the diffraction patterns

obtained when X-rays and

electrons with the same

wavelength pass through a

piece of aluminum foil. The

patterns are circular because

the foil is made up of several

small randomly oriented

crystals. The similarity of

the two images is striking.

This shows that electrons

behave like a wave (X-rays

in this case). www.pems.adfa.edu.au/~s9471553/level1/Teaching/Physics1BWaves/Physics1BWaves.html

Davisson’s and Thomson’s experiments thus confirmed that electrons can act as a wave

and that the wavelength is actually given by de Broglie formula. For this discovery, they

all received a Nobel Prize (Broglie in 1929 and Davisson and Thomson in 1937) (quasi-

interesting fact: Thomson received the Nobel Prize for showing that electrons are waves

while his father, J.J. Thomson, had received it in 1906 for showing that electrons are

particles in 1897!)

An experiment made in 1945 also showed that interference patterns can also be obtained

with neutrons.

Examples

Example 11.2.1

The kinetic energy of an electron is 350 keV. What is its wavelength?

The wavelength is found with

h

pλ =

Therefore, the momentum of the electron is needed.

First possibility: calculate the speed

The mass energy of the electron is



( )2

2 31 8

14

9.11 10 3 10

8.199 10

511.8

ms

mc kg

J

keV

−

−

= × ⋅ ×

= ×

=

The speed of the electron is found with

( )

( )

2

2

21

350 1 511.8

1.6839

11.6839

1

0.8046

k

u

c

E mc

keV keV

u c

γ

γ

γ

= −

= −

=

=−

=

Therefore, the wavelength is

34

31 8

6.626 10

1.6839 9.11 10 0.8046 3 10

0.00179

ms

h

mu

Js

kg

nm

λγ

−

−

=

×=

⋅ × ⋅ ⋅ ×

=

Second possibility: calculate the momentum

The momentum can be found with

( ) ( )

2 2 2 2 4

2 22 2

13

22

511.8 350 511.8

693.4

1.111 10

3.703 10kgm

s

E p c m c

keV keV p c keV

pc keV

pc J

p

−

−

− =

+ − =

=

= ×

= ×

Then, the wavelength is

34

22

6.626 10

3.703 10

0.00179

kgm

s

h

p

Js

nm

λ

−

−

=

×=

×

=



To obtain a diffraction pattern with these electrons moving at this speed, they must

pass through a really small hole.

In this last example, you could have been tempted to use E = hf (where E is the relativistic

energy) to find the frequency, and then use v = λf to find the wavelength. This method

would have been wrong because, with what was learned here, E = hf cannot be used for

something else than a photon.

Common Mistake: using E = hf for something else

than a photon

The formula E = hf is valid for every particle, but there are a few

subtleties if it is applied to particles that do not travel at the speed of light. For

reasons not given here, (but given here: http://physique.merici.ca/waves/proof-

Ehf.pdf) λ = h/p is valid for any particle (including photons), but not E = hf if v = λf

is used to find the frequency. As you have no formula to find f for matter in these

notes, E = hf cannot be used for something else than a photon.

Example 11.2.2

What is the wavelength of a baseball (m = 145 g) travelling at 15 m/s?

The wavelength is

34

34

6.626 10

0.145 15

3 10

ms

h

mv

Js

kg

m

λ

−

−

=

×=

×

= ×

In this case, there is no chance to obtain a diffraction pattern with baseballs. The balls

would have to pass through a hole whose diameter is about 10-34 m, which is

impossible (the atomic nucleus has a diameter of about 10-14 m).

Since matter can act as a wave, everything that was learned about waves can be applied to

matter.



Example 11.2.3

Electrons travelling at 5000 m/s pass through a circular hole whose diameter is 0.1 mm. A

diffraction pattern is then observed on a screen located 2 m from the hole. What is the

diameter of the central diffraction maximum?

The central maximum ends at the first minimum. The angle of this first minimum is

given by

1.22sin

a

λθ =

To find this angle, the wavelength of the electrons is needed. This wavelength is

34

31

6.626 10

9.11 10 5000

145.5

ms

h

mv

Js

kg

nm

λ

−

−

=

×=

× ×

=

Therefore, the angle of the first minimum is

9

3

1.22sin

1.22 145.5 10sin

0.1 10

a

m

m

λθ

θ−

−

=

⋅ ×=

×

sin 0.001775

0.1017

θ

θ

=

= °

On the screen, the distance between the centre of the central maximum and the end of

the central maximum is

( )

tan

tan 0.10172

3.55

y

L

y

m

y mm

θ =

° =

=

Therefore, the diameter of the central maximum is 7.10 mm.



Wave-Particle Duality

In the previous chapter, we were wondering if light is a wave or a particle. Obviously, the

problem is broader because matter also has the same properties.

So the question is: Are matter and light waves or particles?

Unfortunately, there is no simple answer to this question. We can only observe that they

sometimes act like waves, sometimes like particles. Generally, they act like particles when

the wavelength is small (so when the energy is high) and like waves when the wavelength

is great (so when the energy is low). For macroscopic objects such as a baseball, the wave

aspect of matter can never be seen.

When matter acts like a particle, it only acts like a particle, and not at all like a wave. For

example, a collision with another particle is described by the equations of a collision

between particles, and this collision is impossible to describe with the wave theory, exactly

as what is happening with light in the Compton effect. The result isn’t just a simple

approximation of the wave theory for long wavelengths because the result is simply

impossible to explain if it is assumed that matter is a wave. However, when matter acts like

a wave, it only acts like a wave, and not at all like a particle. It is impossible to explain the

result of an electron diffraction experiment if it is assumed that matter is made of particles.

This is the wave-particle duality: both theories are needed to explain all the observations.

Both aspects of this duality can never be seen at the same time. Matter act like particles or

like waves, never with both aspects at the same time. That’s what Bohr’s complementarity

principle is saying: particle and wave aspects complement each other. These two aspects

are necessary to explain all the observations and they complement each other because they

are never used at the same time to explain the same observation. Only the wave aspect is

present during electron diffraction and only the corpuscular aspect is present in a particles

collision. A situation cannot be analyzed using both wave and particle. It’s one or the other.

A World of Probabilities

With the Copenhagen interpretation, physics is not deterministic anymore. In a

deterministic physics, it is possible, in theory, to calculate exactly what will happen at a

later time if the position and the speed of every atom in the universe are known. Everything

is predetermined.

This quality is lost with the Copenhagen interpretation. If an electron is sent through two

slits, the formulas of quantum mechanics only give the probability that the electron hits the

screen at a specific place. It is impossible to know with any certainty where the electron

will hit. Only the probability of hitting a specific location is known. It is, therefore,

impossible to predict exactly what will happen. Only the probabilities can be known.



Einstein was strongly opposed to this idea, and that’s why he said that “God does not play

dice with the universe”.

In 1925, Erwin Schrödinger further develop the ideas of de Broglie to arrive at an equation

that can be used to calculate the amplitude of the wave as a function of the position. The

result depends on the energy of the particle and on the potential energy, which also varies

with the position. This equation is the Schrödinger equation.

( )2

2 2

20

d mE U

dx

ψψ+ − =

�

There is a slightly more complex version in three dimensions. Rest assured, you will not

have to solve this differential equation to find the amplitude of the wave. However, the

solutions of this equation for simple or important situations will be shown.

Infinite Potential Well

In this situation, a particle is trapped in a region of space between x = 0 and x = L. This

particle cannot come out of this region, regardless of the energy given to it. As the

behaviour of the wave is described by Schrödinger equation, the situation must be

described with potential energies. In this case, the energies are

The solutions of the Schrödinger equation are relatively simple when the potential energy

is constant. The solutions are:

- A sine wave between x = 0 and x = L.

- 0 for x < 0 and x > L.

Moreover, as the wave function must be continuous, the sine wave must be zero at x = 0

and x = L



Therefore, a situation quite similar to the one obtained with standing waves is obtained.

The possible waves are (the first three actually)

Harris Benson, Physique 3 : Ondes, optique et physique moderne, ERPI, 2009

The following values of wavelengths are thus possible.

Possible Wavelengths for a Particle Confined in a One-Dimensional Box

2n

L

nλ =

As the wavelength is

h

pλ =

The possible momenta for the particle are given by

2n

n

h nhp

Lλ= =

To find the energy, the kinetic energy is needed. To easily find it, this link between

momentum and kinetic energy when the speed is much smaller the speed of light (found in

mechanics) is used. 21

2k

pE

m=

Therefore, the energy is

2

10

2

n

n k

pE E U

m= + = +

Using the value of the momentum, the final result is



Possible Energies for a Particle Confined in a One-Dimensional Box

2 2

28n

n hE

mL=

These are the only values of energy allowed for the particle in the box.

If the energy of the electrons drops from a higher value to a lower value, a photon having

the lost energy is emitted, as in the Bohr atom.

Example 11.3.1

An electron is trapped in a one-dimensional box 0.7 nm wide.

a) What are the four lowest values of energy that this electron can have?

The energies are

( )( )( )

2342 2

19

1 22 31 9

6.626 1011.23 10 0.767

8 8 9.11 10 0.7 10

J shE J eV

mL kg m

−

−

− −

× ⋅= = = × =

× ×

2 2

2

2 2

22 0.767 3.070

8

hE eV eV

mL= = ⋅ =

2 22

3 2

33 0.767 6.907

8

hE eV eV

mL= = ⋅ =

2 22

4 2

44 0.767 12.279

8

hE eV eV

mL= = ⋅ =

b) What is the wavelength of the photon emitted if the electron passes from the third

to the first energy level?

The energy of the photon is

6.907 0.767

6.139

photon i fE E E

eV eV

eV

= −

= −

=

Therefore, the wavelength of the photon is



1240202

6.139

hc eVnmnm

E eVλ = = =

The graphs of ψ² for the first three levels are as follow.


These graphs show the probability of finding the particle at a specific location in the box

if its position is measured. For example, when the particle is at the third level, there are

three locations where there is a good chance of finding the particle while it is impossible

to find the particle at x = L/3 and x = 2L/3.

Finite Potential Well

In a finite potential well, the particle is trapped between x = 0 and x = L, but it can get out

of the region if its energy is sufficiently high. Let’s take a specific example to illustrate.

Let’s assume that the potential energy is 0 eV between x = 0 and x = 0.7 nm, and 10 eV for

x < 0 and x > 0.7 nm.

Thus, if the energy of the particle is less than 10 eV, it is trapped between x = 0 and

x = 0.7 nm (remember that objects cannot go to the places where the potential energy is

greater than the mechanical energy according to classical physics). However, the particle

can be anywhere if its energy is greater than 10 eV (and thus be out of the box).

The solutions of the Schrödinger equation are completely different for each section if the

energy of the particle is smaller than 10 eV. Then, the solutions are:

- A sine wave between x = 0 and x = 0.7 nm.

- Exponential functions for x < 0 and x > 0.7 nm.



As the function and the derivative of the wave function must be continuous, there are only

a few possible solutions for which the sine function and the exponential functions fit. The

possible solutions, in this case, are

tccc.iesl.forth.gr/education/local/quantum/vqm/figs/5fwf.gif

In addition, there is an infinite number of solutions with energies higher than 10 eV since

the particle can have any energy if E > 10 eV. The following observations can be made:

1- The number of energy levels below 10 eV is limited. If the well gets deeper,

then there are more energy levels.

2- The energy is not quantified if the energy of the particle is greater than the

depth of the well. The energy of the particle can take any value if it is greater

than 10 eV.

3- The energy levels are slightly lower than those obtained with the infinite

potential well (see example 10.3.1) because the sine function is slightly

stretched. If the wavelength is increased, the momentum and the energy of the

particle decreases.

4- Curious thing: the wave goes a little inside the prohibited zones for the four

levels below 10 eV. This means that there is a certain probability of finding the

particle in the regions x < 0 and x > 0.7 nm, even if this is completely

impossible according to classical mechanics.

With quantum mechanics, it was quickly understood that the atomic nucleus model used at

the time was not good. Then, it was thought that the atomic nucleus was composed of

protons and electrons. With quantum mechanics, a fairly rudimentary model of the nucleus



in which the particles are enclosed in a finite well can be made. In this case, the box has a

width of the order of 10-15 m and the height of the potential well is about 10 MeV. It was

then realized that there is no solution where the electron can be trapped in the potential

well. With an infinite potential well, the energy of the first level is about 3000 MeV, which

is more than the energy required to get out of the box. This difference is so great that, even

with a more realistic potential well, it was calculated that it is impossible to have electrons

trapped inside an atomic nucleus. This result forced them to assume, in the 1920s, that a

neutral particle with almost the same mass as the proton exist (the neutron), whose

existence was proven in 1932 by James Chadwick.

Barrier Crossing

It was noticed in the previous section that a particle has a certain probability of being in

places forbidden by classical mechanics. An even more surprising result is obtained when

quantum mechanics is applied in the case of a barrier. Here is a situation representing this.

An electron with an energy of 5 eV is in an area where the potential energy is zero. It is

heading towards a fairly thin (0.1 nm) section where the potential energy is 10 eV. As the

potential energy of this area (10 eV) is larger than the total energy of the particle (5 eV),

the particle should not be able to enter this region according to classical mechanics and,

therefore, should not be able to cross this barrier. But the quantum mechanical solution is

different. As before, the solution is a sine function for the areas where the energy of the

particle is larger than the potential energy and an exponential function for the area where

the energy of the particle is smaller than the potential energy. The following figure

represents the solution.




In the barrier, the exponential function decreases rapidly, but it does not vanish when it

reaches the other side. As the wave function must be continuous, the sine function has a

non-vanishing amplitude across the barrier. Since the wave function is not zero to the right

of the barrier, there is a certain probability of finding the particle across the potential barrier

while it should have been impossible to find it there according to classical mechanics. In

this example, the probability to cross the barrier is 33.3% (see the formula a bit further).

This video shows the wave encountering a barrier.

http://www.youtube.com/watch?v=_3wFXHwRP4s

This probability of crossing areas that would have been impossible to cross according to

classical mechanics is called quantum tunnelling. This probability is given by (formula

given only as a curiosity).

( )( )

12

200 0

0

1 sinh 24

U LT m U E

E E U

−

= + − − �

One might wonder if this probability to cross a barrier is important for everyday life objects.

For example, let’s assume that a 100 g object having a 4 J energy is trying to cross a 1 mm

wide region where the potential energy is 5 J. In this case, the probability of crossing the

barrier is one chance in 1 000 000 000 000 000 000 000 000 000 00010 . In other words, it’s pretty impossible.

Do not try to walk through walls by quantum tunnelling…

Quantum tunnelling is the basis for the operation of tunnelling microscopes (really!). This

apparatus with unmatched precision was invented in 1981. A very sharp tip (there is only

a single atom at the end) passes over a surface. The tip is loaded with electrons, and the

surface is positively charged. The electrons are, therefore, attracted to the surface but they

cannot go to this surface according to classical mechanics because they don’t have enough

energy to cross the space between the tip and surface. However, with quantum tunnelling,

some electrons are able to cross this

space and reach the surface. If the

space between the tip and the surface

decreases, more electrons can cross

the gap. The number of electrons

crossing is fairly easy to measure with

the electric current passing through

the tip. If the current is larger, the

surface is closer to the tip because

more electrons are then able to pass to

the other surface. By scanning the

surface, one can determine the shape

of the surface. fr.wikipedia.org/wiki/Microscope_à_effet_tunnel



The precision is quite remarkable as the following picture shows. This image represents a

silicon surface. The green dots are silicon atoms. A missing atom in the structure can be

noticed at the bottom right of the picture.

www.aip.org/history/einstein/atoms.htm

Since the beginning of the 1990s, individual atoms can even be moved one at a time with

some tunnelling microscopes. This image, made in April 1990, is the result of the first

manipulation of atoms with the microscope at IBM. Atoms have been moved to write the

letters IBM on a metal surface.

www-03.ibm.com/ibm/history/exhibits/vintage/vintage_4506VV1003.html



Harmonic Oscillator

The potential energy of a particle in harmonic oscillation is ½kx². If a particle subjected to

such a potential energy has a mechanical energy E, then, according to classical mechanics,

it does a harmonic oscillation between points x' and x" and its speed is the largest at x = 0.

In addition, it is impossible, always according to classical mechanics, for the particle to go

in the region x > x" or in the region x < x' (since U is greater than E in these regions).

Eisberg, Resnick, Quantum physics, John Wiley, 1985

To find out what is happening in this case according to quantum mechanics, the

Schrödinger equation with this potential energy must be solved.

2

2122 2

2( ) 0

d mE kx

dx

ψψ+ − =

�

It may seem easy at first glance, but this is a mathematical challenge since the solutions

are Hermite polynomials (that you probably don’t know). For example, the graph of the

wave for the thirteenth energy level is




Even if the details of these solutions are not known, some elements show that these

solutions are consistent with what was learned up to now.

1. The amplitude of the wave increases as the potential energy increases. This is quite

normal because the kinetic energy decreases when the potential energy increases.

Thus, the particle moves at a lower speed and spends more time there than in places

where the potential energy is low. If it spends more time at this place, then the

probability of finding the particle at this place is larger and so the wave has a greater

amplitude.

2. The wavelength increases as the potential energy increases. When the potential

energy increases, the kinetic energy must decrease and the momentum must also

decrease. As the wavelength is h/p, the wavelength must also increase as U

increases.

These two remarks do not only apply to a particle undergoing a harmonic oscillation. They

are valid regardless of the way the potential energy changes.

Despite the complexity of the solutions, the possible energy values are surprisingly quite

simple. They are given by

Energy Levels for a Particle in Harmonic Oscillations

1

2n

E n hf

= +

The energy levels are, therefore, all evenly spaced in energy. The gap between the levels

is always hf.

A Look Back at Planck’s Hypothesis

The atoms in a solid are subject to forces such that every atom

is doing a harmonic oscillation (when the amplitudes are

small). When the energy of an atom drops by one level, the

energy of the emitted photon is hf. If the energy of the atom

drops by 2 levels, the emitted photon has energy 2hf. If the

energy of the atom drops by three levels, the energy of the

emitted photon is 3hf, and so on. It is easy to see that the

energy of the emitted photons must be nhf, where n is an

integer. Therefore, hot objects can only emit photons whose

energy is nhf. This is exactly what Planck assumed in 1900 to

explain the radiation of hot objects. This is an incredible luck

to have energy levels evenly spaced like this for a particle

undergoing a harmonic oscillation. If the energy differences

had been irregular, the energy of the photons would have been



very different from nhf and it is unlikely that Planck would have guessed the energy

formula and solve the hot body problem.

Energy at Absolute Zero

When the temperature of a body becomes very low, the atoms go down to the lowest energy

levels, so that they all are at the level lowest at 0 K. However, it can be seen that, even on

the first level n = 0, the energy is not zero (E = ½hf). So, even at 0 K, there is some energy

left and the atoms are still oscillating.

The Hydrogen Atom

To know the wave function for the electron in a hydrogen atom, the formula for the

potential energy of the electron when it is subjected to the force of electric attraction of the

nucleus (U = –ke²/r) must be used in the Schrödinger equation. Again, the solution is quite

difficult to obtain, especially since the Schrödinger equation in 3 dimensions must be used.

This time, the solutions are Bessel functions, which you probably don’t know either.

Graphically, the solutions shown in the next figure are obtained. (Those are the wave

functions squared giving the probability densities.)




You certainly recognize the different orbitals of the hydrogen atom seen in chemistry. Even

if the solutions of the wave functions are relatively complicated, the possible energies are

given by quite a simple formula.

Energy Levels in the Hydrogen Atom

2

13,61n

eVE

n

−=

Oddly enough, these are the same levels of energy that Bohr had obtained in 1913 with a

much less refined theory. With these energy levels, the wavelengths of the spectral lines

can be calculated. In fact, this formula predicts wavelengths that are close to the



experimental results but that do not exactly agree with them. However, they do agree

exactly if the following corrections are made to the theory.

1. Relativity. The relativistic kinetic energy formula gives slightly different values

than the Newtonian kinetic energy formula. This correction was made by Paul Dirac

in 1928 when he formulated the relativistic equivalent of the Schrödinger equation.

This equation also predicted the existence of the spin of the electron and of

antimatter (discovered in 1932).

2. Mass of the nucleus. The wave function of the electron is not centred on the

nucleus but on the centre of mass of the atom. This slightly changes the force

between the nucleus and the electron and, consequently, the energies of the levels.

3. Spin-orbit interaction. The electron in the orbitals around the nucleus generates a

magnetic field. This magnetic field acts on the electron since the electron is like a

small magnet (the spin measures the intensity of this magnet). This slightly changes

the force on the electron and, consequently, the energies of the levels.

4. Spin-spin interaction. The nucleus also has a spin that creates a magnetic field that

acts on the electron. This slightly changes the values of energy levels.

5. Vacuum polarization. In a vacuum, particles and antiparticles continually appear

to disappear almost immediately. During the very brief life of these particles, they

exert a force on the nucleus and the electron, and this changes the energy levels

very slightly.

With all those small corrections, the energy levels slightly differ from those given by the

above formula. The energies of the transitions between these corrected levels then

correspond exactly to the energies of the photons in the hydrogen spectrum. In fact,

quantum electrodynamics (used to make these calculations) is the most accurate known

scientific theory. For some measurements, the first 10 significant digits of the experimental

and theoretical values agree. As a curiosity, here is the formula for the energy level in

hydrogen with all the corrections.

( ) ( ) ( )( )( )

( )( ) ( )

220

2 2 3 2

0

5 2

3

1 1 113,61 31

1/ 2 4 8 1/ 2 1

1 1,

1/ 2 1/ 24

s B

e

j j l l s seV nE g

jn n n a l l l

m c k n lj ln

µαµ

π

απ

+ − + − + −= + − +

+ + +

+ ± + +

Here is the diagram of the energy levels with these corrections.



backreaction.blogspot.ca/2007/12/hydrogen-spectrum-and-its-fine.html

Electron Microscope

Nowadays, the wave aspect of matter is used in electron microscopes. In these devices,

electrons rather than light are used to form an image. Otherwise, the principle is the same

as for a conventional microscope. Electric fields are used to play the role of lenses in these

microscopes. Very accurate images can be obtained because electrons with a wavelength

smaller than 1 nm are used. The resolution of a microscope being roughly equal to the

wavelength, very small details can be seen. As the lenses for electrons are not perfect, the

maximum resolution is of the order of 5 to 10 nm, which is much better than a microscope

operating in visible light whose maximum resolution of approximately 200 nm. Here is an

image produced by an electron microscope.



www.boston.com/bigpicture/2008/11/peering_into_the_micro_world.html

Wave Packet

The fact that matter is a wave implies certain restrictions on the accuracy of some

measurements. If a wave has a very specific wavelength, then the wave is sinusoidal with

a constant amplitude.

However, this wave is infinite, and its amplitude is constant. This means that it is possible

to find the particle anywhere in the universe. To obtain a particle that is localized in a

certain place, the wave must look like this.

For this wave, the amplitude as a function of the position is



Then, the particle can only be found in the region whose length is ∆x.

If the position of the particle is measured, any value of x in this interval can be obtained.

So, there is some uncertainty on the position of the particle which does not depend on the

device used. This uncertainty comes from the wave nature of matter.

However, to get a wave with this shape, there is only one way: several sine functions must

be added together.

hyperphysics.phy-astr.gsu.edu/hbase/waves/wpack.html

Actually, an infinite number of sine function must be added to form such a wave packet.

All the waves of wavelengths lying in a ∆λ wavelength interval must be added together,

and the amplitude of these waves must vary with the wavelength according to the following

graph.

If the interval ∆λ is larger, the wave packet is shorter (∆x is smaller).



www.ipod.org.uk/reality/reality_quantum_intro.asp

If there are many wavelengths at the same time in the wave, then the particle has many

momenta at the same time since the wavelength and momentum are related by de Broglie

formula.

h

pλ =

Therefore, if the momentum is measured, one of those possible momenta will be obtained.

So, there is some uncertainty on the momentum that will be measured. This uncertainty

does not come from a lack of precision of the measuring device but from the fact that the

particle has several momenta at the same time. It is an uncertainty due to the wave nature

of matter.

Advanced calculations (using Fourier integrals…) show that the following relation exists.

x p h∆ ∆ ≈

If ∆p is large, ∆x is small. This actually agrees with what was said earlier: if ∆p is large,

then several momenta are present at the same time. This means that many wavelengths are

present at the same time, and the wave packet is smaller.

An experiment can never give results with uncertainties smaller than that. So, for any

experiment, the following relation must hold.

Heisenberg Uncertainty Principle (With p and x)

x p h∆ ∆ ≥



Example 11.4.1

An electron is confined in a region 10 nm wide. What is the uncertainty of its momentum?

The uncertainty is

34

9

26

6.626 10

10 10

6.626 10kgm

s

p x h

hp

x

Jsp

m

p

−

−

−

∆ ∆ =

∆ =∆

×∆ =

×

∆ = ×

This could mean, for example, that the momentum of the particle can range from

10- 24 kgm/s to 1.06626 x 10-24 kgm/s (the gap between these two is ∆p). If the

momentum of such a particle is measured, any value of p between these two extreme

values can be obtained. If the momentum of several of these particles is measured,

many different results between these two values would be obtained, even if the

conditions are exactly the same.

Example 11.4.2

What is the minimum possible uncertainty on the position of a baseball (m = 145 g)

travelling at 40 m/s if the speed is known with an uncertainty of 0.01%?

The minimum uncertainty is

p x h

m v x h

∆ ∆ ≥

∆ ∆ ≥

( ) ( )

34

30

6.626 10

0.145 0.004

1.142 10

ms

hx

m v

Jsx

kg

x m

−

−

∆ ≥∆

×∆ ≥

×

∆ ≥ ×



The uncertainty is so small for a macroscopic object that it is imperceptible. The

uncertainty principle does not really limit the measurement of the position and the

speed of a baseball.

Quite a similar reasoning can be done with the energy of a wave and the duration of the

wave. The result is

E t h∆ ∆ ≈

An experiment can never give results with uncertainties smaller than that. So, for any

experiment, the following relation must hold.

Heisenberg Uncertainty Principle (With E and t)

E t h∆ ∆ ≥

Example 11.4.3

An electron is in an excited state whose energy is 10 eV. The electron remains on that level

for 10-8 seconds and then goes down to the fundamental level whose energy is 1 eV. What

is the uncertainty of the energy of the excited level?

The uncertainty is

34

8

26

7

6.626 10

10

6.626 10

4.136 10

E t h

hE

t

JsE

s

E J

E eV

−

−

−

−

∆ ∆ ≈

∆ ≈∆

×∆ ≈

∆ ≈ ×

∆ ≈ ×

This means that the energy of the excited level can take any value between

9.9999997932 eV and 10.0000002068 eV (the difference between these values is

4.136 x 10-7 eV). This uncertainty can be seen by measuring the energy of the photon

emitted when the electron goes back down to the fundamental level. There is no

uncertainty on the energy of the fundamental level since the electron remains at this

level for a long time. Thus, the energy of the emitted photon can take any value

between 8.9999997932 eV and 9.0000002068 eV. If the energy of the photon is

measured for several of these atoms, a distribution of energy that looks like this is

obtained.



This kind of uncertainty can be seen in the

measurement of the mass of very short-lived

particles. Here is the graph obtained when the

mass of several Z bosons, who live for just

3 x 10-25 sec (which gives an uncertainty of

13 GeV for the energy), is measured.

www.etp.physik.uni-muenchen.de/opal/opal_en.html




It was said earlier that the square of the amplitude of the wave function gives the probability

of finding the particle at a specific location. In fact, quantum mechanics can be used to

calculate much more than just the position of a particle. For example, it can be used to

predict the result of the measurement of the spin of a particle. However, in all cases, the

interpretation remains the same: quantum mechanics only give the probability of measuring

a specific value. Here is an example to illustrate this concept. An electron is heading

towards a detector measuring the spin of the particle. With an electron, there are only two

possible outcomes: the spin can be either upwards (+) or downwards (–). With quantum

mechanics, the probability to measure + or – can be calculated depending on the situation.

The result could be, for example, 70% chance of measuring + and 30% chance of

measuring –.

However, the Copenhagen interpretation goes way further than that. In this interpretation,

the spin of the particle is not simply + or – before the measurement is made, it is in a state

where the two possible outcomes exist at the same time. In our example, the spin of the

particle is 70% + and 30% –. It is only when the measurement is made that the spin of the

particle become either + or –. Before the measurement, the spin of the particle was both +

and – at the same time, and, after the measurement, it is only + if + was measured or is

only – if – was measured. This is called the wave function collapse.

Many physicists, including Einstein and Schrödinger, disliked this interpretation. The latter

invented in 1935 a tough experiment to illustrate the ridiculousness of this situation:

Schrödinger’s cat. A cat is locked up in a box with a randomly triggered device (poison)

that can kill the cat. Suppose that after 2 hours, there is a 60% chance that the mechanism

has been triggered and that the cat is dead. In this case, according to the Copenhagen

interpretation, the cat is in a mixed

state (60% dead, 40% living) and the fate of

the poor feline will only be decided when

the box is open to look at the cat. After the

observation, the cat will be either only dead

or only alive.

For Einstein and Schrödinger, this

superposition of states did not make any

sense. For them, the cat cannot be dead and

alive at the same time before the box is open. He is either dead or living at any moment in

the box, even if nobody observed it. If only the probability that the cat is dead or alive after

a while can be calculated, it is because not enough information is known. Therefore, this

probability only represents the ignorance of the observers, and there is no real superposition

of states. For Einstein and Schrödinger, when the spin of a particle is measured and + is

en.wikipedia.org/wiki/Schrödinger’s_cat



obtained, it means that the spin was + before the measurement. If – is obtained, then the

spin was – before the measurement. The fact that quantum mechanics only enable us to

calculate probabilities simply means that it is an incomplete theory. According to them, a

more refined theory which ought to make more precise predictions must exist, a theory that

would allow us to predict the results of the measurement, to know the spin of the particle

before measuring it, rather than just giving probabilities.

The EPR Paradox

Another situation, imagined by Einstein, Podolsky, and Rosen (EPR) in 1935, was devised

to show that the Copenhagen interpretation cannot be right. In this experiment, a particle

with zero spin decays into two particles heading in opposite directions. If each of these

particles have a spin, the spins of the two particles should have opposing signs according

to the laws of conservation used in particle physics. This means that if the spin of a particle

is measured and + is obtained, then the spin must be – for the other particle.

According to the Copenhagen interpretation, each particle is in a superimposed state (half

+, half –) before the measurement of the spin. When one of the spins is measured, the wave

function collapse, thus forcing the other particle to have a spin with the opposite sign.

Suppose this is done with the setup shown in the next figure and that – was measured on

the sensor to the right.

www.clker.com/clipart-15121.html

Then, we know that the result of the measurement of the spin on the sensor to the left will

give +. But then, how can the particle travelling towards the left knows that it must change

form a mixed + and – state to the + only state? The answer seems to be simple at first

glance: the measurement of the spin of the particle on the right sensor caused the collapse



of the wave function, thereby changing the state of the particle to the left from a mixed

state to the + only state. However, this collapse cannot be instantaneous since nothing can

travel faster than light according to Einstein’s relativity. At best, the collapse spreads at the

speed of light.

This is where things get complicated. What will happen if the spin of the particle to the left

was measured only a short time after the measurement of the spin of the particle to the

right? In fact, the time between the measurements can be so small that a beam of light

would not have enough time to move from one particle to the other during this time.

Therefore, the collapse of the wave function does not have enough time to reach the particle

to the left. Then, how can the second particle, which was in a mixed + and – state, know

that the measurement must be + if this information doesn’t have enough time to move from

one particle to another? And yet, if this experiment is done, the result of the measurement

will always be +!

Actually, the situation is even weirder if the point of view of another observer is taken

according to relativity. Remember that the time at which events occur changes according

to the observers. Here, there are even observers who will say that the spin of the particle

on the left was measured before the spin of the particle to the right. For them, it is the

measure on the particle to the left that caused the collapse of the wave function which then

force the particle to the right to have a spin –, and this, even before the information had

enough time to reach this particle!

For Einstein, the serious issues raised by this thought experiment showed that the

Copenhagen interpretation could not be correct because it implied that the collapse of the

wave function has to spread faster than the speed of light, which leads to all kinds of

paradoxes. Einstein gave a different interpretation of this situation. As soon as the particles

are emitted, they each have a specific value of spin that is measured a bit later. As they

already have the spins that will be measured and there is no superposition of states at all,

there is no need to have a collapse of the wave function. Spins are opposed because, right

from the start, the particles had these opposite values of spin.

At this point, you surely agree with Einstein…

Bell’s Theorem

For a few years, there were several confrontations between Bohr and Einstein in which the

latter presented different scenarios to show that the Copenhagen interpretation had to be

false. Each time, Bohr was able to find a way to explain the situation using the Copenhagen

interpretation. For example, Bohr claimed that the collapse of the wave function actually

travels faster than light for the EPR paradox, and that this is possible since no information

can be transmitted by this process. A (friendly) war between the two clans continued until

exhaustion, and none had provided the decisive argument in favour of its interpretation.



The situation changed in 1964 when John Stewart Bell discovered a way to determine

experimentally which interpretation was right (unfortunately, Bohr and Einstein were both

dead by then). It can be done with an experiment very similar to the one described for the

EPR paradox, except that the two spins are measured along different axes. Bell determined

that the results of this experiment should respect certain inequalities (Bell’s inequalities) if

Einstein’s interpretation is correct while those inequalities would not be true with the

Copenhagen interpretation. The details are a little complicated but what really matters here

is that they finally had a way to experimentally determine who was right. The experiment

was conducted for the first time in the early 1970s and again several times later, getting

more accurate every time. These experiments proved without a shadow of a doubt that

Einstein’s interpretation cannot be correct. The Copenhagen interpretation is correct. An

object is really in a mixed state before measurements are made and there is a collapse of

the wave function when a measurement is made!

Going back to what has been said about the EPR paradox, then this means that the collapse

of the wave function actually travels faster than light.

The Hydrogen Atom According to the Copenhagen Interpretation

To further illustrate the Copenhagen interpretation, let’s have a look at the current atomic

model of the hydrogen atom. You will certainly be surprised to learn that very few people

really know this current atomic model, and this even includes many chemists. Many people

will say that in the hydrogen atom, one electron revolves around the nucleus. Yet, there is

no such thing in the actual atomic model. Of course, there is a nucleus and an electron

around the nucleus, but there is no mention of the motion of the electron around the nucleus.

The only thing there is in quantum mechanics are the orbitals and these orbitals only

represent the places where there is a strong chance of finding the electron.

However, if the position of the electron is not

measured, the electron is in a state that

corresponds to the superposition of all possible

outcomes of the measurement. This means that

the electron is everywhere at the same time in the

orbital. The electron is at several places at the

same time! If the position of the electron in the

orbital is measured, a certain position is obtained.

This means that the wave function of the electron

has then collapsed to pass from everywhere at the

same time in the orbital to a specific location.

The current atomic model says nothing more

about atoms. According to this model, the

electrons are everywhere at the same time in the orbital. No path, no orbit, no rotation

around the nucleus (even if the electron has a certain angular momentum in the orbital). If

www.ipod.org.uk/reality/reality_quantum_intro.asp



there are orbitals with the same energy, like the three orbitals 2p, then the electron is

everywhere at the same time in all three orbitals. If there is a molecule that can have two

configurations with the same energy (such as benzene), then the molecule is not in a

configuration or the other, it has both configurations at the same time. The benzene

molecule will take one configuration or the other only if a measurement is made.

Decoherence

The Copenhagen interpretation brings some conceptual difficulties. For example, what

happens if there is a fly with Schrödinger’s cat in the box? Will the observation of the cat

by the fly causes the collapse of the wave function and prevents the cat from being in a

superposition of states? Some would say no by asserting that only conscious human beings

can cause the collapse of the wave function! Apart from the problem of deciding which

living beings are conscious, or even deciding what consciousness is, this conception brings

us to ask ourselves what would happen if all the conscious beings of the universe were to

disappear. Would the universe then be sentenced to stay forever in a superposition of many

states?

Since the 1980s, this problem seems to be solved with the decoherence theory. According

to this theory, no conscious beings are required. In fact, as soon as there is interaction with

the environment, the wave function collapse. If the interaction is stronger, then collapse is

faster. The collapse of the wave function, therefore, does not only occur during a

measurement but can also happen spontaneously if the system interacts with the

environment. One could, therefore, conclude that a simple photon locked up with the cat

in the box causes the collapse of the wave function. And even if there is no photon, the

interactions of the atoms of the cat among themselves will cause the collapse of the wave

function. The very large number of atoms in a cat implies that it cannot be in a superposition

of dead and alive states for more than 10-23 seconds. There is, therefore, no chance of seeing

a real cat in a superposition of dead and alive states. However, simpler systems can remain

in a superposition of state for a much longer time, thereby allowing physicists to observe

systems in superimposed states experimentally. This was done by Serge Haroche team in

1996 (Nobel Prize 2012), among others.

The theory is still unable to determine what will happen to the cat. The cat can’t be observed

in a superposition of dead and alive states, but the ultimate fate of the cat (dead or alive)

cannot be known. The theory is unable to predict the outcome of this experiment; it can

only give the probability of each outcome.

Everett’s Many Worlds Interpretation

In 1957, a surprising interpretation was given by Hugh Everett, then a student at Princeton

University. He was trying to give an explanation to the fact that, among all possible

outcomes of a measurement, only one is observed. What happened to all the other possible

states?



Everett’s answer is surprising: each time an observation of a phenomenon is made, the

universe splits into several universes in which each possible outcome is observed.

Let’s take an example to clarify this situation. Suppose the spin of a particle is measured

and quantum mechanics tell us that there is a 50% probability of measuring + and a 50%

probability of measuring –. When the measurement is made, the universe splits in two

universes: a universe in which the measured value is + and another universe in which

measured value is –. In each of these worlds, the observer wonders what happened to the

other possibility that disappeared during the measurement. In this interpretation, the two

possibilities did not disappear. They both occurred but in different universes!

The cat experiment can then be reinterpreted. If the cat in a superposition of dead and alive

states is observed, then, according to Everett, the universe splits into two universes. In one

universe, the cat is dead, and in the other universe, the cat is alive.



De Broglie Wavelength

h

pλ =


The square of the amplitude (ψ²) at a location is proportional to the

probability of finding the particle at this location.

Possible Wavelengths for a Particle Confined in a One-Dimensional Box

2n

L

nλ =

Possible Energies for a Particle Confined in a One-Dimensional Box

2 2

28n

n hE

mL=

Energy Levels for a Particle in a Harmonic Oscillation

1

2n

E n hf

= +

Energy Levels in the Hydrogen Atom

2

13,61n

eVE

n

−=

Heisenberg Uncertainty Principle (With p and x)

x p h∆ ∆ ≥

Heisenberg Uncertainty Principle (With E and t)

E t h∆ ∆ ≥



For these exercises, use the following masses.

Electron me = 9.1094 x 10-31 kg

Proton mp = 1.6726 x 10-27 kg

Neutron mn = 1.6749 x 10-27 kg

11.1 De Broglie Waves 11.2 Interpretation of Quantum Mechanics (Part 1)

1. What is the wavelength of a proton travelling at 104 m/s?

2. What is the wavelength of a proton travelling at 2 x 108 m/s?

3. What is the wavelength of an electron if its kinetic energy is 10 eV?

4. An electron has a kinetic energy of 6 eV when it is in a region where U = 0 eV. The

electron is travelling towards a region where U = 2 eV. By how much will the

wavelength of the electron change when it enters the region where U = 2 eV?

5. Young’s experiment is done with electrons whose kinetic energy is 2 eV. The

electrons pass through two slits 0.1 µm apart and the arrival of the electrons is

observed on a screen located 3 m from the slits. The figure shows what is then

observed. What is the distance x in the figure?

web.utk.edu/~cnattras/Phys250Fall2012/modules/module%202/matter_waves.htm



11.3 Some Applications of Quantum Mechanics

6. A neutron is confined in a 10-14 m wide box (infinite potential well).

a) What are the energies of the first two levels (in MeV)?

b) What is the wavelength of the photon emitted when the neutron goes from the

level n = 2 to the level n = 1?

c) What is the wavelength of the neutron when it is at the level n = 1?

7. An electron is confined in a box (infinite potential well). At the level n = 4, the

energy of the electron is 10 eV. What is the width of the box?

8. An electron is at the level n = 1 in a 2 nm wide box (infinite potential well). What

must be the wavelength of the photon that the electron must absorb to bring it up to

the level n = 4?

9. An electron is confined in a 6 nm wide box (infinite potential well). What is the

smallest speed that this electron can have?

10. The energy of the level n = 4 is 24 eV for a particle confined in a box (infinite

potential well). What is the energy of the level n = 3?

11. An electron undergoes a harmonic oscillation with a period of 4 x 10-15 s.

a) What is the smallest energy that this electron can have?

b) What is the wavelength of the photon emitted if the electron goes from the level

n = 3 to the level n = 1?

12. When an electron undergoing a harmonic oscillation goes from the level n = 5 to

the level n = 2, a photon having a 496 nm wavelength is emitted. What is the period

of the oscillations of the electron?

11.4 Heisenberg Uncertainty Principle

13. The momentum of an electron ranges from 2 x 10-23 kgm/s to 2.05 x 10-23 kgm/s.

What is the minimum uncertainty on the position of this electron?



14. The lifetime of an excited state in an atom is 10-8 s. What is the uncertainty of the

photon energy (in eV) emitted during a transition between this excited level and the

fundamental level?

Challenges

(Questions more difficult than the exam questions.)

15. It was said that the probability of finding a particle at a certain location is

proportional to ψ². More precisely, the probability of finding a particle between

x = a and x = b is

2

b

a

dxψ∫

As the sum of all the probabilities of finding the particle at different locations must

total 100%, the following equation must hold

2

all possiblelocations

1dxψ =∫

Now, these ideas will be applied to a particle confined in a box (infinite potential

well) at the first energy level. Calculate the probability of finding an electron

between x = 0 nm and x = 3 nm if the box has a width of 10 nm.

Here’s an integral that can be useful:

( )2 1sin sin 2

2 4

xax ax

a= −∫

16. For a harmonic oscillator, the potential energy is given by U = ½kx². Show that the

function

2Bx

Aeψ −=

is a solution of the Schrödinger equation if the energy of this level is E = ½hf. (A

and B are constants.)



11.1 de Broglie Waves 11.2 Interpretation of Quantum Mechanics (Part 1)

1. 0.0396nm

2. 1.476 x 10-15 m

3. 0.3879 nm

4. 0.1125 nm

5. 10.41 cm

11.3 Some Applications of Quantum Mechanics

6. a) 1st level: 2.045 MeV 2nd level: 8.181 MeV b) 2.021 x 10-4 nm

c) 2 x 10-14 m

7. 0.7757 nm

8. 879 nm

9. 6.062 x 104 m/s

10. 13.5 eV

11. a) 0.517 eV b) 599.6 nm

12. 4.963 x 10-15 s

11.4 Heisenberg Uncertainty Principle

13. 1.325 nm

14. 4.136 x 10-7 eV

Challenges

15. 14.86%

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