1.

6: Matter Has Wavelike Properties

The next real advance in understanding the atom came from an unlikely quarter - a student prince in Paris. Prince Louis
de Broglie was a member of an illustrious family, prominent in politics and the military since the 1600's. Louis began
his university studies with history, but his elder brother Maurice studied x-rays in his own laboratory, and Louis became
interested in physics. After World War I, de Broglie focused his attention on Einstein's two major achievements, the
theory of special relativity and the quantization of light waves. He wondered if there could be some connection between
them. Perhaps the quantum of radiation really should be thought of as a particle. De Broglie suggested that if waves
(photons) could behave as particles, as demonstrated by the photoelectric effect, then the converse, namely that particles
could behave as waves, should be true. He associated a wavelength λλ to a particle with momentum pp using Planck's
constant as the constant of proportionality:
λ=hp(1.6.1)(1.6.1)λ=hp

which is called the de Broglie wavelength. The fact that particles can behave as waves but also as particles, depending
on which experiment you perform on them, is known as the particle-wave duality.

DERIVING THE DE BROGLIE WAVELENGTH

From the discussion of the photoelectric effect, we have the first part of the particle-wave duality, namely, that
electromagnetic waves can behave like particles. These particles are known as photons, and they move at the speed of
light. Any particle that moves at or near the speed of light has kinetic energy given by Einstein's special theory of
relatively. In general, a particle of mass mm and momentum pp has an energy
E=p2c2+m2c4−−−−−−−−−−√(1.6.2)(1.6.2)E=p2c2+m2c4

Note that if p=0p=0, this reduces to the famous rest-energy expression E=mc2E=mc2. However, photons are massless
particles (technically rest-massless) that always have a finite momentum pp. In this case, Equation 1.6.21.6.2 becomes
E=pc.E=pc.

From Planck's hypothesis, one quantum of electromagnetic radiation has energy E=hνE=hν. Thus, equating these two
expressions for the kinetic energy of a photon, we have
hν=hcλ=pchν=hcλ=pc

Solving for the wavelength λλ gives Equation 1.6.11.6.1:

λ=hp=hmv(1.6.3)(1.6.3)λ=hp=hmv

where vv is the velocity of the particle. Hence, de Broglie argued that if particles can behave as waves, then a relationship
like this, which pertains particularly to waves, should also apply to particles.

Equation 1.6.31.6.3 allows us to associate a wavelength λλ to a particle with momentum pp. As the momentum
increases, the wavelength decreases. In both cases, this means the energy becomes larger. i.e., short
wavelengths and high momenta correspond to high energies.

It is a common feature of quantum mechanics that particles and waves with short wavelengths correspond
to high energies and vice versa.
Having decided that the photon might well be a particle with a rest mass, even if very small, it dawned on de Broglie
that in other respects it might not be too different from other particles, especially the very light electron. In particular,
maybe the electron also had an associated wave. The obvious objection was that if the electron was wavelike, why had
no diffraction or interference effects been observed? But there was an answer. If de Broglie's relation between
momentum and wavelength also held for electrons, the wavelength was sufficiently short that these effects would be
easy to miss. As de Broglie himself pointed out, the wave nature of light is not very evident in everyday life. As the next
section will demonstrate, the validity of de Broglie’s proposal was confirmed by electron diffraction experiments of G.P.
Thomson in 1926 and of C. Davisson and L. H. Germer in 1927. In these experiments it was found that electrons were
scattered from atoms in a crystal and that these scattered electrons produced an interference pattern. These diffraction
patterns are characteristic of wave-like behavior and are exhibited by both electrons (i.e., matter) and electromagnetic
radiation (i.e., light).
EXAMPLE 1.6.11.6.1: ELECTRON WAVES
Calculate the de Broglie wavelength for an electron with a kinetic energy of 1000 eV.
Solution
To calculate the de Broglie wavelength (Equation 1.6.31.6.3), the momentum of the particle must be established and
requires knowledge of both the mass and velocity of the particle. The mass of an
electron is 9.109383×10−28g9.109383×10−28g and the velocity is obtained from the given kinetic energy of 1000 eV:
KE=mv22=p22m=1000eVKE=mv22=p22m=1000eV
Solve for momentum
p=2mKE−−−−−−√p=2mKE
convert to SI units
p=(1000eV)(1.6×10−19J1eV)(2)(9.109383×10−31kg)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− ⎷
p=(1000eV)(1.6×10−19J1eV)(2)(9.109383×10−31kg)
expanding definition of joule into base SI units and cancel
p=(3.1×10−16kg⋅m2/s2)(9.109383×10−31kg)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=2.9×10−40m2/s2−
−−−−−−−−−−−−−√=1.7×10−23kg⋅m/sp=(3.1×10−16kg⋅m2/s2)(9.109383×10−31kg)=2.9×10−40m2/s2=1.7×10−23kg⋅
m/s
Now substitute the momentum into the equation for de Broglie's wavelength (Equation 1.6.11.6.1) with Planck
constant (h = 6.626069×10−34J⋅s6.626069×10−34J⋅s). After expanding expanding unites in Planks constant
λ=hp=6.626069×10−34kg⋅m2/s⋅s1.7×10−23kg⋅m/s=3.87×10−11mor38.9pmλ=hp=6.626069×10−34kg⋅m2/s⋅s1.7×10−
23kg⋅m/s=3.87×10−11mor38.9pm
For perspective, the hydrogen atom is about 120 pm across.
EXERCISE 1.6.11.6.1: BASEBALL WAVES
Calculate the de Broglie wavelength for a fast ball thrown at 100 miles per hour and weighing 4 ounces. Comment on
whether the wave properties of baseballs could be experimentally observed.
Answer
This was the prince's Ph.D. thesis, presented in 1924. His thesis advisor was somewhat taken aback, and was not sure if
this was sound work. He asked de Broglie for an extra copy of the thesis, which he sent to Einstein. Einstein wrote
shortly afterwards: "I believe it is a first feeble ray of light on this worst of our physics enigmas" and the prince got his
Ph.D.

Deriving the de Broglie Wavelength

The de Broglie wavelength is the wavelength, λλ, associated with a object and is related to its momentum and mass.
Introduction

In 1923, Louis de Broglie, a French physicist, proposed a hypothesis to explain the theory of the atomic structure. By
using a series of substitution de Broglie hypothesizes particles to hold properties of waves. Within a few years, de
Broglie's hypothesis was tested by scientists shooting electrons and rays of lights through slits. What scientists
discovered was the electron stream acted the same was as light proving de Broglie correct.
Deriving the de Broglie Wavelength
De Broglie derived his equation using well established theories through the following series of substitutions:
De Broglie first used Einstein's famous equation relating matter and energy:
E=mc2(1)(1)E=mc2
with
 EE = energy,
 mm = mass,
 cc = speed of light
Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation:
E=hν(2)(2)E=hν
with
 EE = energy,
 hh = Plank's constant (6.62607 x 10-34 J s),
 νν= frequency
Since de Broglie believed particles and wave have the same traits, he hypothesized that the two energies would be equal:
mc2=hν(3)(3)mc2=hν
Because real particles do not travel at the speed of light, De Broglie submitted velocity (vv) for the speed of light (cc).
mv2=hν(4)(4)mv2=hν
Through the equation λλ, de Broglie substituted v/λv/λ for νν and arrived at the final expression that relates wavelength
and particle with speed.
mv2=hvλ(5)(5)mv2=hvλ
Hence
λ=hvmv2=hmv(6)(6)λ=hvmv2=hmv
A majority of Wave-Particle Duality problems are simple plug and chug via Equation 66 with some variation of
canceling out units
EXAMPLE 11
Find the de Broglie wavelength for an electron moving at the speed of 5.0×106m/s5.0×106m/s (mass of an electron
is 9.1×10−31kg9.1×10−31kg).
SOLUTION
λ=hp=hmv=6.63×10−34J⋅s(9.1×10−31kg)(5.0×106m/s)=1.46×10−10m(7)(7)λ=hp=hmv=6.63×10−34J⋅s(9.1×10−31kg
)(5.0×106m/s)=1.46×10−10m

Although de Broglie was credited for his hypothesis, he had no actual experimental evidence for his conjecture. In 1927,
Clinton J. Davisson and Lester H. Germer shot electron particles onto onto a nickel crystal. What they saw was the
diffraction of the electron similar to waves diffraction against crystals (x-rays). In the same year, an English physicist,
George P. Thomson fired electrons towards thin metal foil providing him with the same results as Davisson and Germer.

Heisenberg's Uncertainty Principle

Heisenberg’s Uncertainty Principle is one of the most celebrated results of quantum mechanics and states that one (often,
but not always) cannot know all things about a particle (as it is defined by it’s wave function) at the same time. This
principle is mathematically manifested as non-commuting operators.
Introduction
Heisenberg's Uncertainty Principle states that there is inherent uncertainty in the act of measuring a variable of a particle.
Commonly applied to the position and momentum of a particle, the principle states that the more precisely the position
is known the more uncertain the momentum is and vice versa. This is contrary to classical Newtonian physics which
holds all variables of particles to be measurable to an arbitrary uncertainty given good enough equipment. The
Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that defines why a scientist cannot
measure multiple quantum variables simultaneously. Until the dawn of quantum mechanics, it was held as a fact that all
variables of an object could be known to exact precision simultaneously for a given moment. Newtonian physics placed
no limits on how better procedures and techniques could reduce measurement uncertainty so that it was conceivable that
with proper care and accuracy all information could be defined. Heisenberg made the bold proposition that there is a
lower limit to this precision making our knowledge of a particle inherently uncertain.

More specifically, if one knows the precise momentum of the particle, it is impossible to know the precise position, and
vice versa. This relationship also applies to energy and time, in that one cannot measure the precise energy of a system
in a finite amount of time. Uncertainties in the products of “conjugate pairs” (momentum/position) and (energy/time)
were defined by Heisenberg as having a minimum value corresponding to Planck’s constant divided by 4π4π. More
clearly:

ΔpΔx≥h4π(1)(1)ΔpΔx≥h4π
ΔtΔE≥h4π(2)(2)ΔtΔE≥h4π

Where ΔΔ refers to the uncertainty in that variable and h is Planck's constant.

Aside from the mathematical definitions, one can make sense of this by imagining that the more carefully one tries to
measure position, the more disruption there is to the system, resulting in changes in momentum. For example compare
the effect that measuring the position has on the momentum of an electron versus a tennis ball. Let’s say to measure
these objects, light is required in the form of photon particles. These photon particles have a measurable mass and
velocity, and come into contact with the electron and tennis ball in order to achieve a value in their position. As two
objects collide with their respective momenta (p=m*v), they impart theses momenta onto each other. When the photon
contacts the electron, a portion of its momentum is transferred and the electron will now move relative to this value
depending on the ratio of their mass. The larger tennis ball when measured will have a transfer of momentum from the
photons as well, but the effect will be lessened because its mass is several orders of magnitude larger than the photon.
To give a more practical description, picture a tank and a bicycle colliding with one another, the tank portraying the
tennis ball and the bicycle that of the photon. The sheer mass of the tank although it may be traveling at a much slower
speed will increase its momentum much higher than that of the bicycle in effect forcing the bicycle in the opposite
direction. The final result of measuring an object’s position leads to a change in its momentum and vice versa.
All Quantum behavior follows this principle and it is important in determining spectral line widths, as the uncertainty in
energy of a system corresponds to a line width seen in regions of the light spectrum explored in Spectroscopy.
What does it mean?
It is hard to imagine not being able to know exactly where a particle is at a given moment. It seems intuitive that if a
particle exists in space, then we can point to where it is; however, the Heisenberg Uncertainty Principle clearly shows
otherwise. This is because of the wave-like nature of a particle. A particle is spread out over space so that there simply
is not a precise location that it occupies, but instead occupies a range of positions. Similarly, the momentum cannot be
precisely known since a particle consists of a packet of waves, each of which have their own momentum so that at best
it can be said that a particle has a range of momentum.
Figure 11: A wave packet in space
Let's consider if quantum variables could be measured exactly. A wave that has a perfectly measurable position is
collapsed onto a single point with an indefinite wavelength and therefore indefinite momentum according to de Broglie's
equation. Similarly, a wave with a perfectly measurable momentum has a wavelength that oscillates over all space
infinitely and therefore has an indefinite position.
You could do the same thought experiment with energy and time. To precisely measure a wave's energy would take an
infinite amount of time while measuring a wave's exact instance in space would require to be collapsed onto a single
moment which would have indefinite energy.

Consequences
The Heisenberg Principle has large bearing on practiced science and how experiments are designed. Consider measuring
the momentum or position of a particle. To create a measurement, an interaction with the particle must occur that will
alter it's other variables. For example, in order to measure the position of an electron there must be a collision between
the electron and another particle such as a photon. This will impart some of the second particle's momentum onto the
electron being measured and thereby altering it. A more accurate measurement of the electron's position would require
a particle with a smaller wavelength, and therefore be more energetic, but then this would alter the momentum even
more during collision. An experiment designed to determine momentum would have a similar effect on position.
Consequently, experiments can only gather information about a single variable at a time with any amount of accuracy.
Problems
1. The uncertainty in the momentum ΔpΔp of a football thrown by Tom Brady during the superbowl traveling
at 40m/s40m/s is 1×10−6 1×10−6 of its momentum. What is its uncertainty in position ΔxΔx? Mass=0.40kg
2. You notice there is 2 mL of water traveling on the football at the same speed and ΔpΔp. Calculate its ΔxΔx.
3. An electron in that molecule of water traveling at the same speed has the same ΔpΔp. Calculate its ΔxΔx if the
mass of an electron is 9.1×10−31kg 9.1×10−31kg.
4. Comment on the differences in the uncertainty of momentum between the ball, water, and electron. How does
the mass effect this value?
5. Taking into account all of the information presented above, can you state a situation in which the Heisenberg
Uncertainty Principle has little effect on measuring the momentum and position of one object, but dominates
for that of another when both objects are part of the same system?
Answers
1.
p=mv=(0.40kg)(40m/s)=16kgmsp=mv=(0.40kg)(40m/s)=16kgms
Δp=p(1×10−6)=16kgms(1×10−6)=16×10−6kgmsΔp=p(1×10−6)=16kgms(1×10−6)=16×10−6kgms
ΔpΔxΔx≥h4π≥h4πΔp≥6.626×10−34Js4π(16×10−6kgms)≥3.3×10−30mΔpΔx≥h4πΔx≥h4πΔp≥6.626×10−34Js4π(16×1
0−6kgms)≥3.3×10−30m
Note that 1J=1kgms1J=1kgms.
2.
The volume is not the property that matters, but the mass. So convert to mass with density.
(2mL)(1g1mL) density of water(1kg1,000g)=2×10−3kg(2mL)(1g1mL)⏟density of
water(1kg1,000g)=2×10−3kg
p=mv=(2×10−3kg)(40m/s)=8×10−2kgmsp=mv=(2×10−3kg)(40m/s)=8×10−2kgms
Δp=p(1×10−6)=(8×10−2kgms)(1×10−6)=8×10−8kgmsΔp=p(1×10−6)=(8×10−2kgms)(1×10−6)=8×10−8kgms
ΔpΔxΔx≥h4π≥h4πΔp≥6.626×10−34Js4π(8×10−8kgms)≥6.6×10−28m(3)(3)ΔpΔx≥h4πΔx≥h4πΔp≥6.626×10−34Js4π(8
×10−8kgms)≥6.6×10−28m

3. p=mv=(9.1×10−31kg)(40m/s)=3.6×10−29kgmsp=mv=(9.1×10−31kg)(40m/s)=3.6×10−29kgms
Δp=p×10−6=3.6×10−29kgms×1×10−6=3.6×10−35kgmsΔp=p×10−6=3.6×10−29kgms×1×10−6=3.6×10−35kgms
ΔpΔxΔx≥h4π≥h4πΔp≥6.626×10−34J×s4π3.6×10−35kgms≥1.5mΔpΔx≥h4πΔx≥h4πΔp≥6.626×10−34J×s4π3.6×10−35
kgms≥1.5m
4. The mass of the football is 4×10−1kg 4×10−1kg, the water is 2×10−3kg 2×10−3kg, and the electron
is 9.1×10−31kg 9.1×10−31kg. The mass of the water is 2 orders of magnitude smaller than that of the football, and the
resulting position uncertainty is 2 orders of magnitude larger. Between the electron and water there is a difference of 28
orders of magnitude for both mass and ΔxΔx. There is a direct correlation of inverse proportionality between
the ΔxΔx and ΔpΔp as described by the Heisenberg Uncertainty Principle, and the much smaller electron has a larger
position of uncertainty of 1.5 m compared to the larger football's of 3.3×10−30m 3.3×10−30m

5. One example that can be used is a glass of water in a cup holder inside a moving car. This glass of water has multiple
water molecules each consisting of electrons. The water in the glass is a macroscopic object and can be viewed with the
naked eye. The electrons however occupy the same space as the water, but cannot be seen and therefore must be
measured microscopically. As stated above in the introduction, the effect of measuring a tiny particle causes a change
in its momentum and time in space, but this is not the case for the larger object. Thus, the uncertainty principle has much
greater bearing on the electrons rather than the macroscopic water.

Heisenberg's Uncertainty Principle

Heisenberg’s Uncertainty Principle is one of the most celebrated results of quantum mechanics and states that one (often,
but not always) cannot know all things about a particle (as it is defined by it’s wave function) at the same time. This
principle is mathematically manifested as non-commuting operators.
Introduction
Heisenberg's Uncertainty Principle states that there is inherent uncertainty in the act of measuring a variable of a particle.
Commonly applied to the position and momentum of a particle, the principle states that the more precisely the position
is known the more uncertain the momentum is and vice versa. This is contrary to classical Newtonian physics which
holds all variables of particles to be measurable to an arbitrary uncertainty given good enough equipment. The
Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that defines why a scientist cannot
measure multiple quantum variables simultaneously. Until the dawn of quantum mechanics, it was held as a fact that all
variables of an object could be known to exact precision simultaneously for a given moment. Newtonian physics placed
no limits on how better procedures and techniques could reduce measurement uncertainty so that it was conceivable that
with proper care and accuracy all information could be defined. Heisenberg made the bold proposition that there is a
lower limit to this precision making our knowledge of a particle inherently uncertain.
More specifically, if one knows the precise momentum of the particle, it is impossible to know the precise position, and
vice versa. This relationship also applies to energy and time, in that one cannot measure the precise energy of a system
in a finite amount of time. Uncertainties in the products of “conjugate pairs” (momentum/position) and (energy/time)
were defined by Heisenberg as having a minimum value corresponding to Planck’s constant divided by 4π4π. More
clearly:

ΔpΔx≥h4π(1)(1)ΔpΔx≥h4π
ΔtΔE≥h4π(2)(2)ΔtΔE≥h4π
Where ΔΔ refers to the uncertainty in that variable and h is Planck's constant.

Aside from the mathematical definitions, one can make sense of this by imagining that the more carefully one tries to
measure position, the more disruption there is to the system, resulting in changes in momentum. For example compare
the effect that measuring the position has on the momentum of an electron versus a tennis ball. Let’s say to measure
these objects, light is required in the form of photon particles. These photon particles have a measurable mass and
velocity, and come into contact with the electron and tennis ball in order to achieve a value in their position. As two
objects collide with their respective momenta (p=m*v), they impart theses momenta onto each other. When the photon
contacts the electron, a portion of its momentum is transferred and the electron will now move relative to this value
depending on the ratio of their mass. The larger tennis ball when measured will have a transfer of momentum from the
photons as well, but the effect will be lessened because its mass is several orders of magnitude larger than the photon.
To give a more practical description, picture a tank and a bicycle colliding with one another, the tank portraying the
tennis ball and the bicycle that of the photon. The sheer mass of the tank although it may be traveling at a much slower
speed will increase its momentum much higher than that of the bicycle in effect forcing the bicycle in the opposite
direction. The final result of measuring an object’s position leads to a change in its momentum and vice versa.

All Quantum behavior follows this principle and it is important in determining spectral line widths, as the uncertainty in
energy of a system corresponds to a line width seen in regions of the light spectrum explored in Spectroscopy.
What does it mean?
It is hard to imagine not being able to know exactly where a particle is at a given moment. It seems intuitive that if a
particle exists in space, then we can point to where it is; however, the Heisenberg Uncertainty Principle clearly shows
otherwise. This is because of the wave-like nature of a particle. A particle is spread out over space so that there simply
is not a precise location that it occupies, but instead occupies a range of positions. Similarly, the momentum cannot be
precisely known since a particle consists of a packet of waves, each of which have their own momentum so that at best
it can be said that a particle has a range of momentum.
Figure 11: A wave packet in space
Let's consider if quantum variables could be measured exactly. A wave that has a perfectly measurable position is
collapsed onto a single point with an indefinite wavelength and therefore indefinite momentum according to de Broglie's
equation. Similarly, a wave with a perfectly measurable momentum has a wavelength that oscillates over all space
infinitely and therefore has an indefinite position.
You could do the same thought experiment with energy and time. To precisely measure a wave's energy would take an
infinite amount of time while measuring a wave's exact instance in space would require to be collapsed onto a single
moment which would have indefinite energy.
Consequences
The Heisenberg Principle has large bearing on practiced science and how experiments are designed. Consider measuring
the momentum or position of a particle. To create a measurement, an interaction with the particle must occur that will
alter it's other variables. For example, in order to measure the position of an electron there must be a collision between
the electron and another particle such as a photon. This will impart some of the second particle's momentum onto the
electron being measured and thereby altering it. A more accurate measurement of the electron's position would require
a particle with a smaller wavelength, and therefore be more energetic, but then this would alter the momentum even
more during collision. An experiment designed to determine momentum would have a similar effect on position.
Consequently, experiments can only gather information about a single variable at a time with any amount of accuracy.

Problems

1. The uncertainty in the momentum ΔpΔp of a football thrown by Tom Brady during the superbowl traveling
at 40m/s40m/s is 1×10−6 1×10−6 of its momentum. What is its uncertainty in position ΔxΔx? Mass=0.40kg
2. You notice there is 2 mL of water traveling on the football at the same speed and ΔpΔp. Calculate its ΔxΔx.
3. An electron in that molecule of water traveling at the same speed has the same ΔpΔp. Calculate its ΔxΔx if the
mass of an electron is 9.1×10−31kg 9.1×10−31kg.
4. Comment on the differences in the uncertainty of momentum between the ball, water, and electron. How does
the mass effect this value?
5. Taking into account all of the information presented above, can you state a situation in which the Heisenberg
Uncertainty Principle has little effect on measuring the momentum and position of one object, but dominates
for that of another when both objects are part of the same system?
Answers
p=mv=(0.40kg)(40m/s)=16kgmsp=mv=(0.40kg)(40m/s)=16kgms
Δp=p(1×10−6)=16kgms(1×10−6)=16×10−6kgmsΔp=p(1×10−6)=16kgms(1×10−6)=16×10−6kgms
ΔpΔxΔx≥h4π≥h4πΔp≥6.626×10−34Js4π(16×10−6kgms)≥3.3×10−30mΔpΔx≥h4πΔx≥h4πΔp≥6.626×10−34Js4π(16×1
0−6kgms)≥3.3×10−30m

Note that 1J=1kgms1J=1kgms.

2. The volume is not the property that matters, but the mass. So convert to mass with density.
(2mL)(1g1mL) density of water(1kg1,000g)=2×10−3kg(2mL)(1g1mL)⏟density of
water(1kg1,000g)=2×10−3kg
p=mv=(2×10−3kg)(40m/s)=8×10−2kgmsp=mv=(2×10−3kg)(40m/s)=8×10−2kgms
Δp=p(1×10−6)=(8×10−2kgms)(1×10−6)=8×10−8kgmsΔp=p(1×10−6)=(8×10−2kgms)(1×10−6)=8×10−8kgms
ΔpΔxΔx≥h4π≥h4πΔp≥6.626×10−34Js4π(8×10−8kgms)≥6.6×10−28m(3)(3)ΔpΔx≥h4πΔx≥h4πΔp≥6.626×10−34Js4π(8
×10−8kgms)≥6.6×10−28m
p=mv=(9.1×10−31kg)(40m/s)=3.6×10−29kgmsp=mv=(9.1×10−31kg)(40m/s)=3.6×10−29kgms
Δp=p×10−6=3.6×10−29kgms×1×10−6=3.6×10−35kgmsΔp=p×10−6=3.6×10−29kgms×1×10−6=3.6×10−35kgms
ΔpΔxΔx≥h4π≥h4πΔp≥6.626×10−34J×s4π3.6×10−35kgms≥1.5mΔpΔx≥h4πΔx≥h4πΔp≥6.626×10−34J×s4π3.6×10−35
kgms≥1.5m
4. The mass of the football is 4×10−1kg 4×10−1kg, the water is 2×10−3kg 2×10−3kg, and the electron
is 9.1×10−31kg 9.1×10−31kg. The mass of the water is 2 orders of magnitude smaller than that of the football, and the
resulting position uncertainty is 2 orders of magnitude larger. Between the electron and water there is a difference of 28
orders of magnitude for both mass and ΔxΔx. There is a direct correlation of inverse proportionality between
the ΔxΔx and ΔpΔp as described by the Heisenberg Uncertainty Principle, and the much smaller electron has a larger
position of uncertainty of 1.5 m compared to the larger football's of 3.3×10−30m 3.3×10−30m
5.One example that can be used is a glass of water in a cup holder inside a moving car. This glass of water has multiple
water molecules each consisting of electrons. The water in the glass is a macroscopic object and can be viewed with the
naked eye. The electrons however occupy the same space as the water, but cannot be seen and therefore must be
measured microscopically. As stated above in the introduction, the effect of measuring a tiny particle causes a change
in its momentum and time in space, but this is not the case for the larger object. Thus, the uncertainty principle has much
greater bearing on the electrons rather than the macroscopic water.