AS LEVEL PHYSICS: ELECTRONS AND PHOTONS Quantum Physics : The Photoelectric Effect By the end of this presentation you should …. Appreciate that the photoelectric effect provides evidence for the particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence for a wave nature. Show an appreciation of the particulate nature of electromagnetic radiation i.e a photon model. Recall E = h f Describe the photoelectric effect and explain it in terms of photon energy and work function energy Explain the significance of the threshold Recall and explain the significance of hf = hf 0 + KE max Explain why the maximum kinetic energy of photoelectrons is independent of intensity, and why the photoelectric current is proportional to the intensity of the incident radiation.

It was Einstein who, in 1905, took Planck's quantum hypothesis and applied it the photoelectric effect and showed how the consideration of the structure of matter has having quantized energy levels accounted precisely for these observations. This was a tremendous boost to the integrity of the quantum concept. In its interesting to note that it was this concept for which Einstein was awarded the Nobel Prize, and not for his work in relativity - something he developed the same year and for which he now is more well-known.

In an experimental arrangement similar to that of the cathode ray tube, scientists had observed that when light impinged upon the surface of the cathode, electrons could be ejected. What was particularly troubling was the effect observed when the wavelength, the intensity of the light which struck the surface or the surface itself, were changed. The observations made when the wavelength was changed is illustrated in the next slide

atoms electrons METAL The electrons were ejected at a certain wavelength or frequency. If the frequency was decreased, this could result in no electrons being released at all, no matter how long the light shone on the metal!

If light of a certain wavelength is shone on a particular metal, which is connected to a circuit, a current will flow in the circuit for as long as the light is on. There is no time delay, and the current stops when the light is switched off!

In reality, the metal (anode) is contained within an evacuated tube. The current is monitored at higher and higher frequencies (lower and lower wavelengths of light). Once a current flows, the intensity of the light is then increased in order to observe the effect on the current. Current (number of e - ejected by cathode) high intensity low intensity Frequency of incident light

The remarkable aspects of the photoelectric effect when it was first observed were: The electrons were emitted immediately - no time lag! Red light on sodium will not cause the ejection of electrons, no matter what the intensity! A weak violet light (low intensity) will eject only a few electrons, but more electrons are emitted with a more intense ( or brighter) light source. Their maximum k.e. remained unaffected! Their maximum kinetic energies are greater for light of higher frequencies!

Explaining the photoelectric effect. + Electrons are held by electrostatic attraction in an orbit around the positive nucleus. Imagine this force as a taut rubber band “binding” the electron to the nucleus. To release the electron, energy equivalent to its “binding energy” must be provided to break the “elastic band”. The quantum of energy carried by a photon, E is… E = hf (where h = Planck’s constant and f = frequency) So the higher the frequency of radiation, the more energy it carries

+ Considering E = hf, X-ray photons would then carry more energy, since the frequency is higher. So when X-rays are incident on the same metal (sodium) more than enough energy is supplied to release the electron, so it is emitted with a higher kinetic energy.. Increasing the intensity of the X-ray beam would increase the number of photons falling on each square centimetre so the number of electrons emitted from the surface of the metal will increase and the photo-current will increase. Current high intensity low intensity Frequency of incident light

+ Considering E = hf, photons of red light would then carry less energy, since the frequency is lower. So when red light is incident on the same metal (sodium) not enough energy is supplied to release the electron. There is no way of storing the energy, so it does not matter how long the red light shines on the metal, no electrons will be emitted from the surface of the sodium metal. Increasing the brightness of the red light simply increases the number of photons falling on each square centimetre. There still will not be any electrons emitted, no matter what the intensity. E = w 0 + k.e. = hf 0 + ½ mv 2 Where w 0 = the work function (the energy required to free the e - ) And f 0 = the threshold frequency, Below this no e - are emitted

Consider the photoelectrons can be detected by a simple circuit with a microammeter in it. The greater the reading on the ammeter, the higher the number of electrons that are being picked up. If a voltage opposing the flow of electrons is set up, the emitted electrons (-), can be stopped from reaching the anode (+). This is called the stopping voltage. + _

If the emf of the power supply is initially zero, a photocurrent will flow. As the supply is turned up, the emitter becomes more positive (because it is connected to the positive terminal of the supply). So electrons leaving it are attracted back towards it. If they leave with enough energy they can overcome this attraction and cross to the collector. There will then be a reading on the ammeter. If they don’t have enough energy, they can’t cross the gap. By increasing the emf of the supply you can find the pd at which no electrons are able to cross the gap. Even those with the maximum energy can’t do it. At that point, the energy needed to cross the gap = maximum Ek of the electrons.

The experiment may be conducted with a variety of target metals, with different frequencies of incident radiation and at different intensities. Look carefully at the graphs on the following slides, and see if you can answer the questions.

This shows the results when the metal is changed. Look at one of the lines. Why does it slope upwards? Why don’t the lines pass through the origin? Why do the results with different metals give different lines, don’t they both emit the same thing- electrons? f 0 for A f 0 for B w 0 = hf 0 As the frequency of the incident light increases, the photon energy increases too. Electrons are then emitted with higher and higher K.E. so a greater potential is required to stop them. This is because electrons are not emitted below the threshold frequency. Atoms of different metals have different sizes, or crystal structure and particles within each atom. The binding energy of electrons of different atoms are bound to vary. So the threshold frequency of different metals will be different.

GRAPH 1 :Here is a plot of the measured stopping potentials obtained for several light frequencies for two different metals. 1.The slope of this plot is the same in both cases and is equal to Planck's constant h. 2.The x-intercept corresponds to the lowest frequency light that is able to eject electrons for each metal (threshold frequency). 3.This frequency times Planck's constant is the work function of the metal, particular to each metal, and represents the amount of energy necessary for an electron to "climb" out of bulk metal materials.

Which colour has the lowest frequency? How will the energy of a “red” photon compare to the energy of a “blue” photon? In which case will the electron have more KE after “breaking free”? Consider the light falling on a certain metal plate. All three colours have the same intensity. RED Remember E = hf so, a red photon will have less energy than a blue photon. There will be more energy left over from the blue photon, so that electron will move with a higher KE, and will be harder to stop. Photocurrent (A) Retarding voltage (V) Increasing light frequency Stopping potential

If each light has the same intensity, why is the photocurrent slightly different in each case? Consider the light falling on a certain metal plate. All three colours have the same intensity. Intensity = power / area If you consider a time of 1s, then Intensity = energy / area If the energy falling on the metal surface is the same in each case, but a photon of red light carries less energy than a photon of blue light; so more “red photons” must be falling on the metal target in order to deliver the same amount of energy per second. Since each photon will release one electron, more charge flows; so a higher photocurrent is recorded for red light. Photocurrent (A) Retarding voltage (V) Increasing light frequency

GRAPH 2 : This graph shows the typical results of an experiment. 1.For a given light frequency and a particular metal for the cathode, the photocurrent is constant until a large enough retarding potential prevents the ejected electrons from reaching the anode. 2.This potential is called the stopping potential and provides just enough energy (e x V s ) to oppose the kinetic energy of the ejected electrons. 3.When a higher energy light source is used (greater frequency, shorter wavelength), the ejected electrons possess a greater kinetic energy and so the stopping potential is correspondingly larger.

This time we use the same colour of light, the same target metal, but we use a brighter and brighter light. What is the difference between a light of low intensity, and one of high intensity? Why does a more intense light cause a larger current to flow? How come the stopping voltage is the same even when the light is brighter? A more intense light source produces more photons per unit area. More photons are falling on each square cm. So more electrons can be freed from the metal. The colour of the light has nothing to do with its brightness. Although more photons hit the metal, each one still carries the same amount of energy.

GRAPH 3 : This graph shows how the photocurrent increases when the light intensity increases but the wavelength is held constant. The stopping potential is the same for all intensities, suggesting that the kinetic energy of the ejected electrons is the same and hence independent of light intensity.