These notes were typed in association with Physics for use with the IB Diploma Programme by Michael Dickinson For further reading and explanation see: Physics, Tsokos (purple): Ch 6.4 Physics, Giancoli (mountain): Ch 27 OPTION B THE WAVE NATURE OF MATTER

This is a continuation of , Photoelectric Effect First off a little review and of what we were last talking about. The photoelectric effect is seen when light is shown onto a piece of metal and knocks off electrons. The light has to have a sufficent frequency, called threshold frequency(f 0 ), to be able to do this. As the light’s frequency increases the energy the freed electron will have will increase as well. This can be seen in the following graph.

– Describe the photoelectric effect – Describe the concept of the photon, and use it to explain the photoelectric effect – Describe and explain an experiment to test the Einstein model. At low frequencies the photon energy is low and electrons are not emitted. Work Function Φ – the minimum amount of energy of photons incident on a surface required to cause photoelectric emission. Φ = hf 0 From, E = hf we can say… IB Equations hf = Φ + E K(max) hf = hf 0 + eV

– Describe the photoelectric effect – Describe the concept of the photon, and use it to explain the photoelectric effect – Describe and explain an experiment to test the Einstein model. All this can be arranged in y = mx + b form… eV s = hf – hf 0 y is eV s or E K(max) m is planck’s constnat or h b is hf 0 or Φ IB Definition h – planck’s constant Is 6.63 x Js

Now time for a little overview. Check out this video. It’s just an overview. 36A5A90DF 36A5A90DF If the link doesn’t work just copy past it into the address bar.

Practice time! Get your pencil and paper ready. Also be ready to pause the video and give everyone a chance to work out the problem BEFORE he gives the explanation. You only need to watch the first half for now. You’ll watch the second half in a little bit. 36A5A90DF 36A5A90DF Second set of practice. This video gives and over and then a series of questions. There is no pause button so BE READY. es/photoeffect%20explanation.swf D d

Example Problem 1 The apparatus shown below is used to measure the stopping potential V s for photoelectrons emitted from a metal surface. V s is measured for different frequencies of light f, incident on the surface. The broken line on the graph, labeled “R”, shows the results obtained when the metal plate is zinc. The zinc plate is then replaced with another metal having a higher work function. Which line on the graph would best represent the results obtained in this case?

Example Problem 1 Solution: The gradient of the graph for any metal will be equal to Planck’s constant therefore it cannot be Q. The negative y intercept will give the work function, therefore a line with the same gradient as R, giving a greater negative y intercept than R will be the correct answer, which is line S!

Example Problem 2 The work function of a metal may be defined as a)The minimum frequency of the incident electromagnetic radiation required to cause photoelectric emission. b)The minimum wavelength of the incident electromagnetic radiation required to cause photoelectric emission. c)The minimum energy of photons, incident on a surface, required to cause photoelectric emission. d)The minimum energy required to take an electron from the interior to the surface to cause photoelectric emission.

Example Problem 2 The work function of a metal may be defined as a)The minimum frequency of the incident electromagnetic radiation required to cause photoelectric emission. b)The minimum wavelength of the incident electromagnetic radiation required to cause photoelectric emission. c)The minimum energy of photons, incident on a surface, required to cause photoelectric emission. d)The minimum energy required to take an electron from the interior to the surface to cause photoelectric emission. SOLUTION: The work function of a metal may be defined as the minimum energy of photons incident on a surface, required to cause photoelectric emission. Answer C

Now we move onto the wave nature of matter Go ahead and finish the second half of the following video. A5A90DF

Describe the de Broglie Hypothesis and the concept of matter waves Louis de Broglie had the idea that all objects have a wave associated with them. He suggested that an object of mass, m, traveling with velocity, v, would have a wavelength, λ. λ = h / p Where h is Planck’s constant and p is the momentum(mv) of the object. IB EQUATION p = h / λ

Describe the de Broglie Hypothesis and the concept of matter waves De Broglie hypothesized that each electron orbiting a nucleus is a standing wave. This was like the standing waves on a plucked guitar string, except that the electron wave is circular unlike the linear guitar string. The problem is that the effect of this idea is the wavelength of everyday objects is far to small to see. Remember that Planck’s constant is tiny and the momentum(mv) will be huge in comparison giving an even tinier wavelength. Also remember that diffraction only occurs noticeably when the slit (think back to Thomas Young’s double slit experiment) width and the wavelength are similar size.

Describe the de Broglie Hypothesis and the concept of matter waves Wave-Particle Duality The photoelectric effect and Einstein’s explanation of this effect shows that light behaves as a particle. The properties of reflection, refraction, diffraction and interference shows that light behaves as a wave. Is one correct and the other incorrect? No!?!? BOTH theories are correct. Light sometimes behaves like a wave and sometimes like a particle This two sided nature of light is call wave-particle duality.

Describe the de Broglie Hypothesis and the concept of matter waves Example Problem 4 Calculate the de Broglie wavelength for a baseball of mass 0.45kg thrown with a speed of 12m/s. Answer: 1.23x m A wavelength way to small for the wave properties to be noticed.

Outline an experiment to verify the de Broglie hypothesis. See previous video A5A90DF

Solve problems involving matter waves. Example Problem 5 Calculate the de Broglie wavelength for an electron accelerate through a potential difference of 50V. Answer: 1.73 x m

Solve problems involving matter waves. Remember the equation the equations: E K(max) = E elec ½ mv 2 = eV s Re-arrange it so it looks like v = √ 2eV/m