A –Level Physics: Waves and Quanta: Polarisation and the Photoelectric Effect

Objectives: Additional skills gained: Practical Planning 70. be able to use the equation intensity of radiation 92. understand that the absorption of a photon can result in the emission of a photoelectron 93. understand the terms threshold frequency and work function and be able to use the equation 94. be able to use the electronvolt (eV) to express small energies 95. understand how the photoelectric effect provides evidence for the particle nature of electromagnetic radiation. Additional skills gained: Practical Planning Integrating GCSE content

Starter: Explanations Using standing waves as examples; explain constructive and destructive interference (6 marks) Destructive interference: When two waves of identical frequency interact when they are out of phase by a phase difference of 180degrees or pi rads (or a path difference of half a wavelength), they will interfere destructively. If completely in antiphase, this will result in a resultant amplitude of 0. In a standing wave this would be shown as a node. Constructive: If two waves interact in phase (2pi rads or 360degrees) or with a path difference of full wavelengths, they will interfere constructively producing a resultant amplitude that is the combination of the individual amplitudes. This is shown as an antinode on a standing wave. You have 10mins

Polarisation What does the word ‘polarising’ mean? “restricting the vibrations of a transverse wave wholly or partially to a specific direction” Draw what you believe an EM wave looks like Delta What would it look like ‘side on’? Challenge

What commonly used object is made to have this function and why? When waves come from a luminous source the waves will reach us in various directions so can be considered to look like ‘E’ shown on the diagram below Copy the diagram down and then explain what is happening. Use the words ‘restricting’ and ‘polarisation’ What commonly used object is made to have this function and why? Polarisation is when the incoming light is restricted into a single direction by a filter made of small slits. Sunglasses!

Have a try! Grab one of the filters from the box and put it up to see through it to the window. Turn it, does anything change? Put a second one behind it and slowly turn it, what happens? Why is this happening? Delta

Exam Practice: 1 Delta

Time for a history lesson Grab one of the text books and turn to page 274. Read with us through the history of wave particle duality 10m

Draw this set-up in your notes and then we will add the charges! Gold-Leaf Experiment Draw this set-up in your notes and then we will add the charges! Delta

Gold-Leaf Experiment In this experiment, the metal cap (and as such, the stem and gold leaf) were given a negative charge This caused the gold leaf to ______ because _________________ Different colours of light were then shone onto the metal cap Red light had no effect, but ultraviolet caused the gold leaf to go flat again Why did ultraviolet have this effect? [2 marks] Delta Why is it important that the gold leaf be situated in a vacuum?

Photoelectric Effect One ‘packet of energy’ (photon) “the emission or release of electrons from a negatively charged metal when exposed to high frequency radiation” The energy of the incoming light, E = hf Where h= Plank’s constant and f= frequency One ‘packet of energy’ (photon) would allow the emission of one electron (photoelectron) Delta

Exam Practice: 2 5b) Explain why the following observations may be understood by using a photon model of light, rather than a wave model. Light above a certain frequency causes the emission of electrons from the surface of a metal. This emission occurs instantaneously. Light below a certain frequency will not result in the emission of electrons however long it illuminates the surface.(5 marks) ............................................................................................................................................. Delta

What would happen if the work function was greater than the ‘hf’? The electrons in different metals require different amounts of energy to escape. This is the work function of the metal. This is shown by the symbol ‘phi’ So: The energy coming in will be used to overcome work and the left over would be the kinetic energy of the particle! What would happen if the work function was greater than the ‘hf’? The electron would lack enough energy to escape. I could increase the frequency of the incoming light If I kept the type of metal constant, what could I do to increase the kinetic energy of the electrons emitted?

Kinetic Energy You could increase the intensity of incoming light by moving it closer to the metal. (Intensity=power/area). What effect would this have on the energy of the electron emitted? So what would it affect? Increase intensity= increase NUMBER of electrons emitted Increase frequency= increase the KINETIC ENERGY of the electrons Delta

eV Calculating Work Function REMINDER! What was an ‘electronvolt’? REMINDER! It is “the amount of energy an electron gains after being accelerated by a p.d. of 1V.” eV NOTE: to convert joules into electronvolts, divide joules by the charge of an electron, which is about 1.602×10−19 The electron would lack enough energy to escape. I could increase the frequency of the incoming light Charge on a single electron Voltage

Exam Practice: 3 An electron is accelerated from rest through a potential difference of 5.0 kV. The kinetic energy gained by the electron is     A     8.0 × 10−16 J       B     8.0 × 10−19 J       C     3.2 × 10−20 J       D     3.2 × 10−23 J A

Phototube A phototube consists of a anode that is radiated to release electrons that would hit the cathode and read as current Using a variable resistor, a p.d can be created across the system in the reverse direction. This would result in work being done against the flow from anode to cathode eV

Phototube The p.d is increased until no current is recorded. thus the Work done=K.Emax. This is the stopping voltage (Vs) We do this to find out the kinetic energy of the emitted electrons! eVs = 1 2 mv2

Phototube So with our new value of kinetic energy (Ek) we go back to our original equation! Rearrange this equation into ‘y=mx+c’ DON’T FORGET! When you’re dealing with the KE, you use joules!

Phototube work function= incoming energy So on a graph, the gradient would represent plank’s constant and the y-intercept would represent a negative value of the WORK FUNCTION! (phi) The threshold frequency is the mimumum frequency of incident radiation that would cause an electron to be emitted. It would be when the work function= incoming energy

Exam Practice: 4   3. When electromagnetic radiation is incident on a metal plate, electrons may be emitted. (a)  State what is meant by threshold frequency. (1) ............................................................................................................................................. (b)  Calculate the threshold frequency for a metal with a work function of 2.28 eV. Charge on an electron=1.6x10-19C and Plank’s constant = 6.63×10−34Js (3) Threshold frequency = .................................... Delta CLICK FOR ANSWER

Exam Practice: 5 Ultraviolet radiation of wavelength 2.00 × 10−7 m is shone onto a zinc plate.   Calculate the maximum speed of the electrons emitted from the plate.      work function of zinc = 6.88 × 10−19 J (4) Maximum speed of electrons = ........................................... Delta CLICK FOR ANSWER

Independent Study .