Module 4: C13 - Quantum Physics

Question 1

What are Photons

Answer 1

Einstein proposed that, although light exhibited wave-like properties, it travelled in particles called photons.

He explained that photons contain discrete ‘energy packets’ called quanta, and that the energy of an individual quantum depends on the frequency of the light.

A photon is a ‘packet’ of light energy
or
A photon is a ‘packet’ of electromagnetic energy

Question 2

Photon Definition

Answer 2

A photon is a ‘packet’ of light energy

or

A photon is a ‘packet’ of electromagnetic energy

Question 3

Equation for Energy (involving Planck’s constant)

Answer 3

The energy of each photon is directly proportional to its frequency.

E = ℏf

Energy (J) = Planck’s Constant (Js) x Frequency (Hz)

Question 4

What is Planck’s Constant, ℏ

Answer 4

Planck’s Constant, ℏ:

6.63x10^-34

Question 5

What is an Electronvolt

Answer 5

An electronvolt is equal to the amount of energy transferred to a single electron if it is accelerated through a potential difference of 1 V.

1eV = 1.6x10^-19 J

Question 6

Conversions between 1eV to J and 1J to eV?

Answer 6

1 eV = 1.6 x 10^-19 J

1J = 1/(1.6x10^-19)eV = 6.25 x 10^18 eV

Question 7

Equation relating Wave Speed, Frequency, and Wavelength

Answer 7

Wave Speed = Frequency x Wavelength

c = fλ

Question 8

Example Question:

What is the photon energy, in electronvolts, of red light of wavelength 685 nm?

Answer 8

c = fλ => f = c/λ
f = 3x10^8/6.85x10^-7
f = 4.38x10^14

E=ℏf
E = 4.38x10^14 x 6.63x10^-34
E = 2.90x10^-19

2.90x10^-19 / 1.6x10^-19 = 1.81eV

Question 9

Practice Question:

What is the photon energy of electromagnetic radiation with a wavelength of 5.75x10^-12m.

Answer 9

c = fλ => f = c/λ
3x10^8/5.75x10^12 = 5.22x10^19Hz

E = ℏf
5.22x10^19 x 6.63x10^-34 = 3.46x10^-14 J

3.47x10^-14 / 1.6x10^-19 = 216195.7eV

= 216000eV (3sf)

Question 10

What is the photon energy of electromagnetic radiation with a wavelength of 470nm?

Answer 10

c = fλ => f = c/λ
3x10^8/4.7x10^-7 = 6.38x10^14

E = ℏf
E = 6.63x10^-34 x 6.38x10^14
E = 4.23x10^-19 J

4.23x10^-19 / 1.6x10^-19 = 2.64eV

Question 11

What is the wavelength of electromagnetic radiation with a photon energy of 2.98x10^-24J

Answer 11

E = ℏf => f = E/ℏ
f = 2.98x10^-24 / 6.63x10^-34
f = 4494720965

c = fλ => λ = c/f
λ = 6.68x10^-2

Question 12

What do you call an electron when it moves outwards and inwards between shell(s)

Answer 12

When an electron moves out a shell it’s ‘excited’

When an electron moves in a shell it ‘relaxes’ (and releases a photon (light))

Question 13

Given a set of values for atomic energy levels in a particular element, how it is possible to calculate the wavelengths of radiation it can emit or absorb

Answer 13

The difference between two energy levels gives the energy of the photon corresponding to that jump, and this can be used to find frequency and wavelength.

hf = E1 – E2
hc/λ = E1 – E2

Question 14

How can electrons jump up one or more energy levels? (+ how is an absorption spectrum formed)

Answer 14

Just as an electron can drop between energy levels in an atom, releasing a single photon, it can also jump up one or more energy levels if it absorbs a photon of the right energy.

Only a single photon of the relevant energy can cause this. It is not possible for an electron to ‘store up’ energy from smaller quanta until it has enough to make the jump.

One result of this is that shining a continuous spectrum of light at a transparent material leads to a few discrete frequencies being absorbed, while the rest are transmitted. This forms an absorption spectrum.

Question 15

What is an Emission Line Spectrum

Answer 15

A set of specific frequencies of electromagnetic radiation, visible as bright lines in spectroscopy, emitted by exited atoms as their electrons make transitions between higher and lower energy states, losing the corresponding amount of energy in the form of photons as they do so - every element has a characteristic line spectrum.

Question 16

What is an Absorption Line Spectrum

Answer 16

A set of specific frequencies of electromagnetic radiation, visible as dark lines in an otherwise continuous spectrum on spectroscopy. They are absorbed by atoms as their electrons are excited between energy states by absorbing the corresponding amount of energy in the form of photons - every element has a characteristic line spectrum.

Question 17

How was Photoelectricity First Discovered?

Answer 17

Photoelectricity was first discovered in 1887 by Heinrich Hertz during investigations into radio waves using a ‘spark gap’.

Radio waves are produced when a high voltage is supplied across two electrodes causing a spark in the gap.

Hertz found that if ultraviolet light was shone on the electrodes, the sparks were much stronger and thicker.

Question 18

How does a Photocell work?

Answer 18

When light is shone on the cathode of the photocell, electrons are released. They are attracted to the anode, causing a current to flow.

Question 19

What is the Photoelectric Effect

Answer 19

In 1887 Heinrich Hertz reported that when he shone UV radiation onto zinc, electrons were emitted from the surface on the metal.

This is the photoelectric effect. The emitted electrons are sometimes called photoelectrons. They are normal electrons, but their name describes their origin - emitted through the photoelectric effect.

Question 20

How can we used the Gold-Leaf Electroscope to demonstrate how electrical charges repel each other?

Answer 20

Briefly touching the top plate with negative electrode from a high-voltage power supply will charge the electroscope. Excess electrons are deposited onto the plate and stem of the electroscope. Any charge developed on the plate at the top of the electroscope spreads to the stem and the gold leaf. As both the stem and gold leaf have the same charge, they repel each other, and the leaf lifts away from the stem. If a clean piece of zinc is placed on top of a negatively charged electroscope and UV radiation shines on the zinc surface, then the gold leaf slowly falls back towards the stem. This shows that the electroscope has gradually lost its negative charge, because the incident radiation (in the case UV) has caused the free electrons to be emitted from the zinc. These electrons are known as photoelectrons.

Question 21

What are the 3 Key Observations from the Photoelectric Effect

Answer 21

When different frequencies of incident radiation were investigated in more detail, scientists at the time made 3 key observations.

1. Photoelectrons were emitted only if the incident radiation was above a certain frequency (called the threshold frequency f0) for each metal. No matter how intense the incident radiation (how bright the light), not a single electron would be emitted if the frequency was less than the threshold frequency.
2. If the incident radiation was above the threshold frequency, emission of photoelectrons was instantaneous.
3. If the incident radiation was above the threshold frequency, increasing the intensity of the radiation did not increase the maximum kinetic energy of the photoelectrons. Instead more electrons were emitted. The only way to increase the maximum kinetic energy was to increase the frequency of the incident radiation.

Question 22

What needs to happen for Photoelectron to be emitted

Answer 22

As the energy of the photon is dependant on it’s frequency (E =hf), if the frequency of the photon is too low, the intensity of the light does not matter, as a single photon deliver its energy to a single surface electron in a one-to-one interaction.

If a photon does not carry enough energy on its own to free an electron, the number of photons makes no difference. However, when the frequency of the light is above the threshold frequency f0 for the metal, then each individual photon has enough energy to free a single surface electron and so photoelectrons are emitted.

Question 23

Why is there no time delay between when a photon hits a surface and when a photoelectron is emitted

Answer 23

This also explained why there was no time delay. As long as the incident radiation has frequency greater than or equal to the threshold frequency, as soon as the photons hit the surface of the metal, photoelectrons are emitted. Electrons cannot accumulate energy from multiple photons. Only one-to-one interactions are possible between photons and electrons.

Question 24

What is Work Function, Φ, and why is it different depending on an electron’s position

Answer 24

Einstein was also able to explain the third observation. Depending on their position relative to the positive ions in the metal, electrons would require different amounts of energy to free them. Einstein defined a constant for each metal, called the work function . This is the minimum energy required to free an electron from the surface of the metal.

Question 25

What affect does increasing the Intensity of radiation on a surface have on the rate of photoelectrons being emitted?

Answer 25

Increasing the intensity of the radiation means more photons per second hit the metal surface. As each photon interacts one-to-one with a single surface electron, as long as the radiation has frequency above the threshold frequency for the metal, more photons per second means a greater rate of photoelectrons emitted from the metal.

The rate of emission of photoelectrons is directly proportional to the intensity of the incident radiation. Double the intensity and you double the number of photons per second, leading to a doubling in the number of electrons emitted from the metal per second.

Question 26

How can you determine the Kinetic Energy of an emitted photoelectron.

Answer 26

Using the principle of conservation of energy, Einstein deduced that the kinetic energy of each photoelectron depends on how much energy was left over after the electron was freed from the metal.

At a given frequency, all photons have the same amount of energy, and the metal a specific work function, so there is a maximum value of kinetic energy that any emitted photo electrons can have. Increasing the intensity results in a greater rate of emission, but none of the emitted photoelectrons will move any faster.

Question 27

What is the only way to increase maximum kinetic energy of emitted Photoelectrons?

Answer 27

The only way to increase the maximum kinetic energy of the emitted photoelectrons is to increase the frequency of the radiation. In this case each photon has more energy and so each electron has more kinetic energy after it has been freed from the metal.

Question 28

Work Function Definition

Answer 28

The amount of energy needed to release an electron from a metal is called the “work function” and is given the symbol φ. Generally, work functions are lower for more reactive metals.

Photon Energy = work function + kinetic energy of electron

Question 29

Equation Involving Work Function (with Kinetic Energy)

Answer 29

Photon Energy = work function + kinetic energy of electron

Question 30

Equation for Work Function (involving threshold frequency)

Answer 30

Work Function = Planck’s Constant x Threshold Frequency

Φ = hf0

Question 31

Equation for Energy of an Electron

Answer 31

Energy of Electron = QVs = 1/2mv^2

Question 32

Equation for Energy/Work involving Charge and Voltage

Answer 32

E = QV
or
W = QV

This equation states that the work done on an electron (of charge 1.6x10-19C) as it moves through a potential difference of 1V is given by:

W = QV = 1.6x10^-19 x 1 = 1.6x10^-19 J

Question 33

Calculate the threshold frequency if the work function is 8.12x10^-19 J

Answer 33

8.12x10^-19 / 6.63x10^-34 = 1.22x10^15 Hz

f = 1.22x10^15 Hz

Question 34

What are 4 Conclusions from the Photoelectric Effect

Answer 34

• For a given metal, no photoelectrons are emitted if the radiation has a frequency below the threshold frequency.
• The photoelectrons are emitted with a variety of kinetic energies ranging from zero to a maximum value. This maximum kinetic energy increases with the frequency of the radiation.
• The maximum kinetic energy is unaffected by varying the intensity of the radiation.
• The number of photoelectrons emitted per
  second is proportional to the intensity of the radiation.

Question 35

General Equation relating the energy of each proton, the work function of the metal, and the Max KE of the emitted Photoelectron

Answer 35

hf = Φ + KEmax

Energy of a Single Photon = Minimum energy required to free a single electron from the metal surface + maximum kinetic energy of the emitted electron

Question 36

How is the Energy of each individual photon conserved?

Answer 36

Einstein realised that the energy of each individual photon must be conserved. This energy does 2 things:

• It frees a single electron from the surface of the metal in a one to-one interaction
• Any remainder is transferred into the kinetic energy of the Photoelectron.

Question 37

Worked Example: Emitting Photoelectrons from a metal surface

Photoelectrons with a maximum kinetic energy of 9.34x10^-19J are emitted from a metal with a work function 2.40eV.
Calculate the frequency of the incident radiation.

Answer 37

Step 1: Identify the correct equation to calculate the frequency of the radiation.

hf = Φ + KEmax
Convert the work function into joules.
2.40eV x 1.60x10^-19 = 3.84x10^-19 J

Step 2: Substitute the values into the equation to calculate the energy of a single photon.

hf = 3.84x10^-19 + 9.34x10^-19 = 1.32x10^-18 J
Using this valve we can determine the frequency of the radiation.

f = 1.32x10^18/h = 2.32x10^-18/6.63x10^-34 = 1.99x10^15 Hz
f = 1.99x10^15 Hz

Question 38

Why is maximum kinetic energy used in Einstein’s Photoelectric Effect Equation ‘hf = Φ + KEmax’

Answer 38

Some electrons in the surface of the metal are closer to the positive metal ions than other. Their relative positions affect how much energy is required to free them. The work function is the minimum energy required to free an electron from the metal - most electrons need a
little more energy than the work function to free them.

An electron that requires the minimum amount of energy to free it (the work function of the metal) would have the most energy left over from the incident photon. Only a few of the emitted Photoelectrons have this maximum kinetic energy - most have a little but less, and so travel a little slower.

Question 39

What happens when a photon strikes the surface of a metal at the threshold frequency of that metal?

Answer 39

If a photon strikes the surface of the metal at the threshold frequency f0 for the metal then it will only have enough energy to free a surface electron, with none left over to be transferred into kinetic energy of the electron.

In this case Einsteun’d photoelectric effect equation becomes
hf0 = Φ + 0

Question 40

What is the Wave-Particle Duality Model

Answer 40

This wave-particle duality is a model used to describe how all matter has both wave and particle properties. De Broglie, who was warded the 1929 Nobel Prize in Physics for his insight, realised that all particles travel through space as waves. Anything with mass that is moving has wave-like properties. These waves are referred to as matter waves or de Broglie waves.

Question 41

What did De Broglie say the wavelength of a particle depended on (+ equation)

Answer 41

De Broglie deduced that a particle had a wavelength, and it was dependent on only one thing – the momentum of that particle:

λ=h/p

Three years later, this hypothesis was confirmed for electrons with the first observations of electron diffraction.

Question 42

Name the Wave-Like Properties and Particle-Like Properties

Answer 42

Wave-Like Properties:
- Reflection
- Refraction
- Diffraction
- Polarisation

Particle-Like Properties:
- Photoelectric Effect
- Emission Spectra
- Bohr’s Model

Question 43

How was the de Broglie equation created

Answer 43

De Broglie realised that the wavelength λ of a particle is was inversely proportional to its momentum p. As the momentum of a particle increase by a certain factor it’s wavelength reduces by the same factor.

λ ∝ 1/p

Further investigation led to him finding:
λ = h/p

(Where λ is the wavelength in m, h is the Planck constant, and p is the momentum of the particle in kgms^-1)

Question 44

Worked Example: The wavelength of a fast-moving electron

Calculate the wavelength of an electron travelling at 2.00x10^8ms^-1.

Answer 44

Step 1: Identify the correct equation to calculate the wavelength.

λ = h/p

As momentum is p = mv we can substitute this into the previous equation giving:

λ = h/mv

Step 2: Substitute in known values in SI units and calculate the wavelength in metres.

λ = (6.63x10^-34 / 9.11x10^-31 x 2.00x10^8) = 3.64x10^-12m

(Notice how small this wavelength is - less than the diameter of an atom. In order diffract electrons tiny gaps are needed.

Question 45

How are particles wave-like properties change as particles become larger

Answer 45

As particles become larger, their wave properties become harder to observe. The mass of individual protons is much greater than electrons, so at the same speed their momentum is significantly greater and therefore their wavelength is much smaller, and much harder to observe.

Question 46

Example Question:

Find the wavelength of an electron travelling at 4.5x10^7 ms^-1 if it’s mass is 9.1x^-31kg.

Answer 46

λ = h/p
p = 4.5x10^7 x 9.1x10^-31
p = 4.095x10^-23

λ = 6.63x10^-34/4.095x10^-23
λ = 1.62x10^-11 m

Question 47

Example Question:

Find the velocity of a particle if it has a wavelength of 3.6x10^-21m, and a mass of 4.2x10^-10kg.

Answer 47

λ = h/p
p = h/λ
p = 6.63x10^-34 / 3.6x10^-21
p = 1.84x10^-13

p = mv
v = p/m
v = 1.84x10^-13 / 4.2x10^-10
v = 4.38x10^-4 s

Question 48

What conditions can electrons diffract and how do they diffract?

Answer 48

Under certain conditions we can make electrons diffract. They spread out like waves as they pass through a tiny gap and can even form diffraction patterns in the same way as light.

If an electron gun fires electrons at a thin piece of polycrystalline graphite, which has carbon atoms arranged in many different layers, the electrons pass between the individual carbon atoms in the graphite. The gap between the atoms is so small (~10^-10m) the it is similar to the wavelength of the electrons and so the electrons diffract, as waves, and for, a diffraction pattern seen in the end of the tube.

We do not normally notice the wave nature of electrons because we need a tiny gap in order to observe electrons diffracting. For diffraction to occur the size of the gap through which the electrons pass must be similar to their wavelength.