Chapter 13: Quantum Physics

Question 1

1. Define photon.
2. Describe the energy of a photon in terms of its frequency.
3. Describe how the energy of a photon can be expressed in terms of its wavelength.
4. Describe the proportionality between the energy of a photon and its wavelength.

Answer 1

1. A photon is defined as a quantum of electromagnetic energy.
2. The energy of each photon is directly proportional to its frequency. Specifically
   E = h f
   where E is the energy of the photon in J, f is the frequency of the electromagetic radiation in Hz, and h is the Planck constant (6.63 x 10^-34 Js).
3. Combining E = h f with c = f 𝛌 gives the energy of a photon in terms of its wavelength and c (the speed of light):
   E = hc / 𝛌.
4. The energy of a photon is inversely proportional to its wavelength.

Question 2

1. What is the unit used for measuring energies at the quantum scale?
2. Define the energy of 1eV.
3. How is the work done on an electron moving through a p.d. calculated?
4. Calculate the energy of 1eV.

Answer 2

1. Instead of the SI unit for energy (joules, J) the electronvolt (eV) is often used when measuring energies at the quantum scale.
2. The energy of 1eV is defined as the energy transferred to or from an electron when it moves through a potential difference of 1V.
3. The work done on an electron moving through a p.d. is equal to the p.d. x charge on the electron (W = VQ = Ve).
4. The work done on an electron as it moves through a p.d. of 1V is given by
   W = 1V x 1.60 x 10^-19 C = 1.60 x 10^-19 J
   so 1eV is equivalent to 1.60 x 10^-19 J, or 1J is equal to 6.25 x 10¹⁸ eV.

Question 3

Give the typical energy (in J and eV) and the typical wavelength of types of photons.

Answer 3

Question 4

What do LEDs do?

Answer 4

LEDs convert electrical energy into light energy. They emit visible photons when the p.d. across them is above the critical value (the threshold p.d.)

Question 5

Describe and show the I-V characteristic for a typical LED.

Answer 5

• When the p.d. reaches the threshold p.d. the LED lights up and starts emitting photons of a specific wavelength.

Question 6

Describe how the Planck constant can be determined experimentally.

Answer 6

• Setting up a circuit with a battery, voltmeter, LED and variable resistor, the threshold p.d. can be determined.
  
  • A black tube placed over the LED helps to show exactly when the LED lights up.
• The wavelength of the photons emitted by the LED is also found.
  
  • At the threshold p.d., the energy transferred by an electron is approximately equal to the energy of the single photon it emits. Hence
    threshold p.d. x charge on electron ≈ energy of emitted photon
    V e ≈ h f
  • Expressing this in terms of the wavelength of the emitted photon 𝛌 gives
    e V = h c / 𝛌
• By gathering data from a variety of different-wavelength LEDs (threshold p.d. dictates colour of LED), we can plot a graph of V (y-axis) against 1/𝛌 (x-axis). The Planck constant canthen be determined from the gradient of the graph, hc/e.

Question 7

What is the photoelectric effect?

Answer 7

The photoelectric effect describes the emission of electrons (or photoelectrons) from a metal surface when electromagnetic radiation above a threshold frequency is incident on the metal.

Question 8

Describe and explain how the photoelectric effect can be shown.

Answer 8

• A simple demonstration of the effect can be seen with a gold-leaf electroscope.
  
  • A clean piece of sinc is placed on top of the electroscope.
  • Briefly touching the top metal plate of the electroscope with the negative electrode from a high-voltage power supply will charge it.
  • Any charge developed (from excess electrons) on the plate at the top of the electroscope spreads to the stem and gold leaf. As both the stem and the gold leaf have the same charge, they repel each other, and the leaf lifts away from the stem.
  • If EM radiation above the threshold frequency shines onto 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 has caused the free electrons to be emitted from the zinc. These electrons are known as photoelectrons.

Question 9

Describe three key observations that can be made from the photoelectric effect.

Answer 9

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), no electrons would be emitted if the frequency was less that 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 max KE of the photoelectrons. Instead, more electrons were emitted. The only way to increase the max KE was to increase the frequency of the incident radiation.

Question 10

Why can the wave model not be used to explain the observations made from the photoelectric effect?

Answer 10

• These observations cannot be explained using the wave model of EM radiation.
  
  • For example, if the threshold frequency for a particular metal is in the green part of the visible spectrum, bright red light does not cause emission yet very dim blue light does.
  • This does not fit the wave model if which the rate of energy transferred by the radiation is dependent on its intensity (brightness) as it would follow that bright red light has more energy than dim blue light.

Question 11

Explain the observations of the photoelectric effect using the photon model.

Answer 11

• The photon model proposes that electromagnetic radiation is a stream of photons, rather than continuous waves.
• Each electron on the surface of the metal requires a certain amount of energy in order to escape from the metal, and each photon can transfer its exact energy to one surface electron in a one-to-one interaction. The energy of a photon is dependent on its frequency (E=hf) so if the frequency is too low, the intensity of the light (number of photons per second) does not matter.
• This also explains why there was no time delay: as long as the incident radiation is above or equal to the threshold frequency, photoelectrons are emitted as soon as the photons hit the surface of the metal.
• As for the third observation, increasing the intensity of the radiation means more photons per second hit the metal surface so more photoelectrons are emitted. The only way to increase max KE is to increase the energy of the photon so there is more energy left over after it releases the electron. As E=hf, the energy of the photon is increased by increasing frequency of the radiation.

Question 12

Using the principle of conservation of energy, deduce Einstein’s photoelectric effect equation.

Answer 12

Using the principle of conservation of energy, it can be deduced that the energy of each individual photon does two things: it frees a single electron from the surface of the metal and any remainder must transferred to the kinetic energy of the photoelectron.

The work function 𝜙 represents the energy required to free and electron from the surface of a metal. So

energy of a photon = minimum energy required to free a single electron + maximum kinetic energy of the emitted electron

OR

hf = 𝜙 + KE_max

Question 13

Describe what happens if a photon strikes the surface of the metal at the threshold frequency.

Answer 13

The photon will only have enough energy to free a surface electron; hence, KE_max=0.

Question 14

Why is the term ‘maximum kinetic energy’ used?

Answer 14

• This term is used as it refers to the remainder of energy from the incident photon after an electron that requires the minimum amount of energy to be released has been released by it.
• This minimum energy is the work function of the metal. Most electrons need more energy than the work function to free them, dependent on their relative proximity to positive metal ions in the surface of the metal.

Question 15

Describe how the work function and the Planck constant can be determined from the results of a gold-leaf electroscope experiment.

Answer 15

• Plotting the maximum kinetic energy of photoelectrons (y-axis) against the frequency of radiation on the metal surface (x-axis) gives an area of 0 kinetic energy where f<f>0 and no photoelectrons are emitted, followed by a linear pattern.</f>
• Considering the general equation for a straight-line graph (y=mx+c), and rearranging hf = 𝜙 + KE_max to match the generic equation, we get KE_max = hf – 𝜙.
• This shows that the gradient of the graph is the Planck constant h and the y-intercept is equal to –𝜙, where 𝜙 is the work function of the metal.

Question 16

Describe and explain the diffraction of electrons.

Answer 16

Question 17

Discuss the de Broglie equation.

Answer 17