Quantum Phenomena Chapter 3. Photoelectric effect When light shines on the metal surface, a current was detected on the ammeter. Can you explain why?

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Presentation transcript:

Quantum Phenomena Chapter 3

Photoelectric effect When light shines on the metal surface, a current was detected on the ammeter. Can you explain why?

Photoelectric effect When light shines on a metal surface, electrons are found to be emitted from the surface. This effect is called the photoelectric effect and it occurs in many materials, but is most easily observed with metals.

When the photocell is in the dark, the ammeter reads zero. When light of sufficiently high frequency illuminates the plate, the ammeter indicates a current flowing in the circuit. The completion of the circuit is because electrons were ejected by the incoming radiation, flowing from P to C, thus closing the circuit.

The Photoelectric Effect Hertz ( ) noticed that certain metallic surfaces can produce an electric current when exposed to light of very short wavelength, such as UV Became known as the photoelectric effect because it involved both light and electricity The ejected electrons are known as photoelectrons

Light Is it a wave or particles?

Wave theory If it's waves, the energy contained in one of those waves should depend only on its amplitude--that is, on the intensity of the light. Other factors, like the frequency, should make no difference. So, for example, red light and ultraviolet light of the same intensity should knock out the same number of electrons, and the maximum kinetic energy of both sets of electrons should also be the same. Decrease the intensity, and you should get fewer electrons, flying out more slowly; if the light is too faint, you shouldn't get any electrons at all, no matter what frequency you're using.

Particle Theory Planck suggested that energy could only come in discrete lumps, even if the lumps were very small. Based on Planck's work, Einstein proposed that light also delivers its energy in chunks; light would then consist of little particles, or quanta, called photons, each with an energy of Planck's constant times its frequency. In that case, the frequency of the light would make a difference in the photoelectric effect

Higher-frequency photons have more energy, so electrons should come flying out faster; thus, switching to light with the same intensity but a higher frequency should increase the maximum kinetic energy of the emitted electrons. If the frequency is the same but increase the intensity, more electrons should come out (because there are more photons to hit them), but they won't come out any faster, because each individual photon still has the same energy. And if the frequency is too low, then none of the photons will have enough energy to knock an electron out of an atom. So if really low-frequency light is used, no electrons can be emitted, no matter how high the intensity is. Whereas if you use a high frequency, you should still knock out some electrons even if the intensity is very low.

Therefore, with a few simple measurements, the photoelectric effect would seem to be able to tell us whether light is in fact made up of particles or waves. In , R.A. Millikan did a series of extremely careful experiments involving the photoelectric effect. He found that all of his results agreed exactly with Einstein's predictions about photons, not with the wave theory. Einstein actually won the Nobel Prize for his work on the photoelectric effect, not for his more famous theory of relativity.

Some experimental results, like this one, seem to prove beyond all possible doubt that light consists of particles; others insist, just as irrefutably, that it's waves. (Young’s double slit) We can only conclude that light is somehow both a wave and a particle--or that it's something else we can't quite visualize, which appears to us as one or the other depending on how we look at it.

Experiments Show… Increasing the intensity of the light increases the number of electrons emitted per second And For light below a certain Threshold Frequency, f o, no electrons are emitted, even by very intense light

And Above f o the maximum KE of the electrons increases with frequency, but is not affected by intensity. Even very dim light gives some electrons with high KE

Wave Theory The wave theory cannot explain the threshold frequency Nor how low amplitude waves can cause high KE electrons

Wave Theory If light exists as continuous waves, then electrons should absorb energy continuously, allowing them to be emitted as soon as they absorb enough energy This means that there should be no threshold frequency, since any light should eventually allow electrons to be ejected

Einstein’s Explanation Suggested that the energy carried by light comes in the form of energy bundles, called quanta The amount of energy in each light quantum, or photon, depends only on the frequency of the light, according to Planck’s equation E = hf

and more The intensity of a beam of light is determined by the number of photons it contains He further suggested that an electron near the surface of a metal could absorb the energy from the incident photon of light If the light frequency is high enough, the photon can contain enough energy for the electron to escape the metal surface

and more Any excess energy becomes the KE of the electron So the energy of the photon could be expressed in terms of the escape energy and the KE of the electron So E = hf = W o + 1 / 2 m e v max 2

Explanation W o is the energy required to remove the electron, this is known as the work function of the electron given the symbol Φ It can be calculated using Planck’s equation as follows W o = hf o Where f o is the threshold frequency 1 / 2 m e v max 2 is the KE of the electron

Milikan’s experiment on photoelectric emission Inside an evacuated glass bulb, a plate of an alkali metal, such as lithium, sodium, or potassium, was illuminated by monochromatic light at various frequencies. Millikan measured the voltage required to prevent the induced current. The graph of incident light frequency vs. voltage was a straight line. Einstein's equation was verified.

An Experiment to test Einstein’s Model Light Plate Vacuum d.c. supply (variable and reversible) G V A

The Principle The material is illuminated with light of a known frequency f Emitted electrons reach plate A The galvanometer detects a current in the circuit The KE max of the emitted electrons is found be applying just enough opposing voltage to stop them reaching A

Principle 2 This is called the stopping voltage V s The galvanometer falls to zero therefore e V s = 1 / 2 m e v max 2 and therefore e V s = hf - hf o

How would a graph of V s against f look like? VsVs

The De Broglie Hypothesis As the photoelectric effect demonstrated the particle nature of light The next question was whether matter, made up of particles, might have a wave nature First addressed by Prince Louis Victor de Broglie

What did he do? He started with the equation for mass- energy equivalence E = mc 2 Where m is the “mass” of the photon c is the speed of light or a photon

And then? From E=mc 2 E = hf = hc / = mc 2 mc = h / Therefore p = h /

What does it mean? The greater the momentum The shorter the wavelength Thus a fast moving particle such as an electron should produce very short waves of the order of X-rays

Electron Diffraction 1926 Davisson and Germer beamed electrons at a single nickel crystal They discovered the electrons reflected very strongly at certain special angles But not at others Recognised the pattern formed by the electrons reflecting and interfering as a diffraction pattern

Diffraction? They used the equation for light passing through a diffraction grating to calculate the wavelength of the electrons using n = d sin 

And De Broglie’s They compared this to the wavelength predicted by de Broglie’s hypothesis To calculate the velocity they started with the KE KE = 1 / 2 mv 2 = eV (Where V is the p.d. through which the electron is accelerated) Thus the velocity =  (2eV / m)

And... Their momentum is given by p = mv = m  (2eV / m) =  (2eV m) The equation for de Broglie wavelength is therefore = h / p = h /  (2eVm)

Example If the accelerating p.d. is 54 volts what is the wavelength? = h /  2eVm = ____6.63 x _______________  (2 x 1.6 x x 54 x 9.11 x ) = 1.67 x m

The Conclusion The wavelength calculated using the light equation was the same as that using de Broglie’s equation This suggests that the electrons themselves are diffracted in the same way as one would expect the de Broglie waves to be Therefore electrons have wave properties!

Confirmation 1927 G.P. Thomson confirmed with his own experiment bombarding gold leaf with an electron beam (G.P. Thomson was the only son of J.J.Thomson)

Spectroscopy Atomic spectra are generally observed by exposing substances to a source of continuous energy, which excites the electrons of the substance. Spectroscopy is the science of using spectral lines to figure out what something is made of. Each type of atom gives off a unique set of colors. The colored lines (or Spectral Lines) are a kind of "signature" for the atoms.

Atomic spectra

Flame Testing Flame testing is a simple experimental procedure based on visible light emission. It is particularly useful in identifying certain metallic elements. With this method, a bunsen burner is used to provide the energy to excite the electrons in the atoms, and then one observes a particular color of light as the electrons return to lower energy levels. This color, as well as the corresponding line spectrum, are characteristic of the metal.

Various substances are held in the flame of a Bunsen burner. The light emitted from the heated elements was separated into spectra using a prism For example, elemental sodium will turn a flame a characteristic yellow colour, while elemental copper emits a characteristic blue ‑ green colour. One of the simplest forms of spectroscopy is the flame test, which illustrates emission of light.

Separation of light by a prism

Origin of spectra

Emission spectra are produced by thin gases in which the atoms do not experience many collisions (because of the low density). The emission lines correspond to photons of discrete energies that are emitted when excited atomic states in the gas make transitions back to lower-lying levels.

An absorption spectrum occurs when light passes through a cold, dilute gas and atoms in the gas absorb at characteristic frequencies; The electrons in the gas absorb the exact frequencies that when excited they re-emitted. These vacancies show up as "black lines" in the solid’s spectrum

All other spectroscopic techniques rely on supplying some form of energy to a substance, and studying the absorption and subsequent emission lines produced. A prism or a diffraction grating is generally used to observe the spectrum of the substance, which will exhibit characteristic lines. The separation of colours is very much like that seen in a rainbow, except that the spectrum will generally have very sharply defined lines at wavelengths characteristic to the element or compound being observed.

On the left is a helium spectral tube excited by means of a 5000 volt transformer. On the right of the image are the spectral lines through a 600 line/mm diffraction grating.

Nitrogen Spectrum Argon spectrum Iodine Spectrum Mercury spectrum