6. Atomic and Nuclear Physics Chapter 6.4 Interactions of matter with energy

Heinrich Hertz first observed this photoelectric effect in This, too, was one of those handful of phenomena that Classical Physics could not explain. Hertz had observed that, under the right conditions, when light is shined on a metal, electrons are released. The Photoelectric effect Photosurface Photoelectrons The photoelectric effect consists on the emission of electrons from a metallic surface by absorption of light (electromagnetic radiation).

An apparatus to investigate the photoelectric effect was set by Millikan and it allowed him to determine the charge of the electron. The Photoelectric effect When light falls on the surface, the electrons are removed from the metal’s atoms and move towards the positive cathode completing the circuit and thus creating a current.

Whether the photoelectric effect occurs or not depends only on:  The nature of the photosurface The Photoelectric effect  The frequency of the radiation

When the intensity of the light source increases so does the current.  Current and intensity are directly proportional A high current can be due to:  Electrons with high speed  Large number of electrons being emitted To determine what exactly happens we need to be able to determine the energy of the emitted electrons. This is done by connecting a battery between the photosurface and the collecting plate. The Photoelectric effect

When the battery supplies a p.d. the charge of the collecting plate will be negative. This means that the negatively charged electrons can be stopped if a sufficiently negative p.d. is applied to the electrodes. The Photoelectric effect G collecting plate evacuated tube light photosurface electron

The Photoelectric effect The electrons leave the photosurface with a certain amount of kinetic energy – E K. To stop the electrons we must supply a potential difference (called stopping voltage) so that: eV s = E K

The Photoelectric effect The stopping voltage stays the same no matter what the intensity of the light source is. This means that:  The intensity of light affects the number of electrons emitted but not their energy  The energy of the electrons depends on the nature of light: the larger the frequency, the larger the energy of the emitted electrons and thus the larger the stopping voltage

Critical of threshold frequency The two graphs represent the E K of the electrons versus frequency These graphs tell us that:  there is a minimum frequency f c, called critical or threshold frequency, such that no electrons are emitted.  if the frequency of the light source is less than f c then the photoelectric effect does not occur  the threshold frequency only depends on the nature of the photosurface

The Photoelectric effect - Observations 1.The intensity of the incident light does not affect the energy of the emitted electrons (only their number) 2.The electron energy depends on the frequency of the incident light, and there is a certain minimum frequency below which no electrons are emitted. 3.Electrons are emitted with no time delay – instantaneous effect.

The Photoelectric effect Problem: According to Classical Physics, the electron should be able to absorb the energy from light waves and accumulate it until it is enough to be emitted. Solution: Einstein suggested that light could be considered particles of light, photons, packets of energy and momentum or quanta. The energy of such quantum is give by: E = h f where: f is the frequency of the e-m radiation h =6.63x J (constant known as Planck constant)

The Photoelectric effect When a photon hits a photosurface, an electron will absorb that energy. However, part of that energy will be used to pull the electron from the nucleus. That energy is called the work function and represented by Ф. The remaining energy will be the kinetic energy of the free electron. So, E k = hf - Ф

The Photoelectric effect Recalling that E K = eV s So, eV s = hf – Ф that is:

The Photoelectric effect When a photon hits a photosurface, 3 things can happen:  The energy of the photon is not enough to remove the electron  nothing happens  The energy of the photon is just enough to remove the electron: the photon’s energy equals the ionization energy  the electron leaves the atom without any E k  The energy of the photon is larger than the ionization energy  the electron leaves the atom with E k

Exercise: 1.What is the work function for the photosurface (in joules)? 2.What is the energy of the “green” photoelectron? 3.What is the speed of the “green” photoelectron? 4.What is the energy of the “blue” photoelectron? 5.What is the speed of the “blue” photoelectron?

Exercise: 1.What is the work function for the photosurface (in joules)? 2.What is the energy of the “green” photoelectron? 3.What is the speed of the “green” photoelectron? 4.What is the energy of the “blue” photoelectron? 5.What is the speed of the “blue” photoelectron?

Light: wave or particle The photon has an energy given by E = hf But if it is considered a particle it also carries momentum p = m v According to Einstein E = m c 2 ↔ m = E /c 2 So, p = (E /c 2 ) c ↔ p = E / c p = hf /c p = h /λ

Light: wave or particle Light can behave as a particle and the photoelectric effect is evidence for that fact. But if we do Young’s double slit experiment so that make photons of light go through the slits one at each time, the photon will produce an interference pattern. Somehow, even when light behaves like a particle it conserves its wave properties. So, we talk about wave-particle duality.

De Bröglie’s wavelength In 1923, Louis de Bröglie suggested that if light can behave as a particle then particles could have a wave associated to them. Louis de Broglie The wave-particle duality or duality of matter can be applied to matter and energy. All particles have a wave associated to them so that λ = h / p Big particles have a wavelength so small that it can’t be measured. But small particles, like electrons, would have a wavelength that is possible to be measured.

The electron as a wave To have an obstacle with this size we must look at the structure of crystals. The typical distance between atoms in a crystal is of the order of m. When electrons are made to pass through crystals, they do diffract thus proving its wave nature. To prove that the electron behaves like a wave it must have wave properties, such diffraction. To make an electron diffract around an obstacle of size d, its wavelength λ must be comparable to or bigger that d. An electron of mass 9.1x kg and speed of 10 5 m/s will have a wavelength λ = 7.2x10 -9 m.

Davisson and Germer experiment In this experiment, electrons of kinetic energy 54eV were directed at a surface ok nickel where a single crystal had been grown and were scattered by it. Using the Bragg formula and the known separation of the crystal atoms allowed the determination of the wavelength which has then seen to agree with the De Broglie formula. Structural analysis by electron diffraction