Zinc Negatively charge Zinc Negatively charge a sheet of zinc. 2. Shine long wavelength radio waves on the zinc. 3. Zinc remains negatively charged. 4. Increase intensity of radio waves… 5. Zinc remains negatively charged. 6. Change wavelength of EM radiation 7. Zinc discharges!

What’s happening? E-M radiation is causing the electrons to leave the metal, making it discharge. These are called electrons photoelectrons Negatively charge Zinc e e e e e e E-M waves deliver energy If they deliver enough energy to a particular electron, that electron could use the energy to leave the metal surface. Why? Any wave will deliver energy ! If you shine any radiation onto the metal for long enough eventually enough energy will be delivered to allow electrons to leave. Also, if you increase the intensity of the radiation, it should occur sooner. Any wave will deliver energy ! If you shine any radiation onto the metal for long enough eventually enough energy will be delivered to allow electrons to leave. Also, if you increase the intensity of the radiation, it should occur sooner. WAVE THEORY E-M radiation does not always behave like a wave - a smooth continuous stream of energy being delivered to a point. QUANTUM THEORY We can only explain the photoelectric effect if E-M radiation is behaving like packets of energy being delivered one by one. We call these lumps quanta or photons. Zinc

What’s the energy of a photon? p E = h f f = c / λ HIGH ENERGY !LOW ENERGY ! h = Planck's constant f = the frequency of the radiation

Each photon only interacts with one electron. It delivers its energy to the electron and disappears (because it is a packet of pure energy and nothing else.)  Electron now has extra energy. If it has enough extra energy it can leave the metal atom. So only photons above the threshold frequency, f o, will cause photoelectric emission. More intense radiation simply means that more packets of energy (photons) are delivered each second. But the energy of each packet is unchanged. So if there wasn't enough energy to cause photoelectric emission, making it brighter won't change anything. The Rules! How come more intense (brighter) radiation doesn't cause emission?

Compare energy received and used by the electron at the surface: Energy supplied to the electron = hf (energy from the photon) Energy used by the electron: used as work function (Φ) to escape from the surface or is left with the electron after it has escaped, in which case it is in the form of E k. Energy Supplied = Energy Consumed hf = Φ + E kmax It actually takes energy for an electron to leave the surface of the metal Einstein This energy is the work function, Φ. …Conservation of Energy Principle

hf = Φ + E kmax E kmax : only electrons on the surface (Φ= 0) will have the maximum kinetic energy on leaving. Electrons from deep inside which make it to the surface and still have enough energy to escape will have used some energy getting to the surface. These electrons will have less energy left once they are free of the metal and so they will have E < E kmax If E photon = hf o (i.e. the energy of the threshold frequency photons), only the electrons on the surface will only have enough energy to overcome the work function and no more. So once they have escaped they will have no energy left (E k = zero)  The photoelectric equation becomes: hf 0 = Φ e e e e e e hf = hf 0 + Φ hf = hf 0 + ½ mv 2

A Metal Plate (Emitter) Metal Plate (Collector) If the EMF is > 0: the emitter becomes more positive  Electrons leaving it are attracted back towards it. e e The emitter gives out electrons: cathode. If EMF of power supply = 0: Flow of electrons across the gap sets up an electromotive force (EMF) between the plates that causes a current to flow around the rest of the circuit. 0 V Only really energetic electrons can make it across- if they don't have enough energy, they can't cross the gap. By increasing the EMF of the supply you can find the potential difference at which no electrons are able to cross the gap. At that point, the energy needed to cross the gap = maximum E k of the electrons. Work done moving a charge through a potential difference is W = QV, Q=e eV stop = ½ m v 2

VSVS Frequency eV stop = ½ mv 2 eV stop = h(f-f 0 ) V stop = (h/e)(f-f 0 ) f0f0 Nothing happens until you get photons with energy > hf 0 After that, increasing the frequency of the photons increases the maximum kinetic energy of the photoelectrons, so stopping them takes a larger value of stopping potential. Gradient is h/e Stopping Potential

Current-voltage for high intensity and low intensity light What happens when we change the light intensity?

I EMF VSVS Voltage to turn around most energetic electron (stopping potential)

I EMF VSVS Voltage to turn around most energetic electron (stopping potential) Fewer electrons pop off metal Current decreases. Current proportional to light intensity. Low High Same KE electrons popping off metal. So same “stopping potential”.

What happens to the initial KE of the electrons as the frequency of light changes? (Light intensity is constant) 0 Frequency of light Initial KE

0 Frequency Initial KE 0 Frequency Initial KE 0 Frequency Initial KE 0 Frequency Initial KE AB C D E. something different

0 Frequency of light Initial KE As the frequency of light increases (shorter λ)  KE of electrons being popped off increases. (it is a linear relationship) There is a minimum frequency below which the light cannot kick out electrons… Even if wait a long time