3.12.2.4 The Discovery Of Photoelectricty

Question 1

What is a black body?

Answer 1

An object that absorbs and emits all possible wavelengths of radiation

Question 2

What did wave theory predict in terms of UV radiation from a black body?

Answer 2

As the wavelength of radiation decreases, the intensity of the radiation increases, leading to a prediction of an infinite amount of ultraviolet radiation being emitted

Question 3

What is the UV catastrophe?

Answer 3

The widely accepted wave theory predicted an impossible amount of UV radiation and it could not be used to explain experimental measurements

Question 4

How could the UV catastrophe be resolved?

Answer 4

By Planck’s interpretation of EM waves

Question 5

What was Planck’s interpretation of EM waves?

Answer 5

EM waves travel in discrete packets called quanta/photons which have an energy directly proportional to their frequency (E = hf)

Question 6

What are the reasons for why the photoelectric effect couldn’t be explained by wave theory?

Answer 6

1. Wave theory suggests that any frequency of light should be able to cause photoelectric emission as the energy absorbed by each electron will gradually increase with each incoming wave, so can’t explain threshold frequency
2. The photoelectric effect is immediate, contradicting wave theory which suggests time is needed for the energy supplied to the electrons to reach the work function
3. Increasing the intensity of the light does not increase the speed of photoelectric emission as would be suggested by wave theory, but instead increases the number of photoelectrons released per second
4. Photoelectrons are released with a range of kinetic energies

Question 7

What was Einstein’s explanation of photoelectricity that could be used to explain the points wave theory couldn’t?

Answer 7

1. When a photon interacts with an electron, all of its energy is transferred to it and an electron can only interact with a single photon. If this energy is above the work function, a photoelectron is emitted, if this energy is below the work function, the electron remains in place. As the energy of a photon is directly proportional to frequency (E=hf), the threshold frequency is the frequency at which the photon energy is equal to the work function of the metal​.
2. The photon energy is transferred to the electron immediately when they interact, so photoelectrons are emitted immediately.
3. Intensity is equal to the number of photons released per second​. A higher intensity increases the number of photoelectrons emitted because ​more photons interact with electrons per second​.
4. All electrons receive the ​same amount of energy​ from a photon of light, but electrons deeper in the ​metal lose energy through collisions​ when leaving the metal, so have a lower kinetic energy.

Question 8

What happens to the kinetic energy of photoelectrons if the metal surface is positively charged and why?

Answer 8

The ​kinetic energy of the photoelectrons decreases as they must do work against the electrostatic force of attraction towards the surface​
As the potential on the surface is increased, the number of photoelectrons released decreases, because the number of photoelectrons with a high enough kinetic energy to be emitted decreases