2. Wave Particle Duality Quantum Physics Lesson 3

3. Today’s Objectives Explain what is meant by wave-particle duality. Explain what is meant by wave-particle duality. Describe the main points of de Broglie’s hypothesis that matter particles also have a wave-like nature. Describe the main points of de Broglie’s hypothesis that matter particles also have a wave-like nature. State and use the equation λ = h/p = h/mv State and use the equation λ = h/p = h/mv Describe evidence for de Broglie’s hypothesis. Describe evidence for de Broglie’s hypothesis.

4. Wave particle duality We have seen…………….. We have seen…………….. Photons : Quanta (particles) of light Electrons: Being diffracted. A property of waves

5. Prince Louis de Broglie 1892-1987 Electrons should not be considered simply as particles, but that frequency must be assigned to them also. Electrons should not be considered simply as particles, but that frequency must be assigned to them also. (1929, Nobel Prize Speech)

6. De Broglie (1924) Suggested that particles such as electrons might show wave properties. Suggested that particles such as electrons might show wave properties. He summised that the de Broglie wavelength, λ was given by: He summised that the de Broglie wavelength, λ was given by: m = mass v = velocity of the particle

7. Note that:- This is a matter wave equation not electromagnetic wave The de Broglie wavelength can be altered by changing the velocity of the particle.

8. In words...

9. The diffraction tube

10. Summary of Experiment Beam of electrons directed at a thin metal foil. Beam of electrons directed at a thin metal foil. Rows of atoms cause the electron beam to be diffracted in certain directions only. Rows of atoms cause the electron beam to be diffracted in certain directions only. We observe rings due to electrons being diffracted by the same amount from grains of different orientations, at the same angle to the incident beam. We observe rings due to electrons being diffracted by the same amount from grains of different orientations, at the same angle to the incident beam.

12. What we should see

13. Electron diffraction 1927: Davisson & Gerner confirmed this prediction with experiments using electron beams. 1927: Davisson & Gerner confirmed this prediction with experiments using electron beams. They actually used a nickel target instead of a carbon one (we used) They actually used a nickel target instead of a carbon one (we used) The wavelength they measured agreed with de Broglie The wavelength they measured agreed with de Broglie There is a relationship between the accelerating voltage V and the k.e. of the particles There is a relationship between the accelerating voltage V and the k.e. of the particles

14. Diffraction effects have been shown for Hydrogen atoms Helium atoms Neutrons Neutron diffraction is an excellent way of studying crystal structures.

15. What is the wavelength of a human being, assuming he/she weighs 70 kg, and is running at 25 m/s? De Broglie Wavelength In 1932, De Broglie discovered that all particles with momentum have an associated wavelength.

16. Practice Questions 1.Find the wavelength of an electron of mass 9.00 × 10 -31 kg moving at 3.00 × 10 7 m s -1 1.Find the wavelength of an electron of mass 9.00 × 10 -31 kg moving at 3.00 × 10 7 m s -1 2. Find the wavelength of a cricket ball of mass 0.15 kg moving at 30 m s -1. 2. Find the wavelength of a cricket ball of mass 0.15 kg moving at 30 m s -1. 3. It is also desirable to be able to calculate the wavelength associated with an electron when the accelerating voltage is known. There are 3 steps in the calculation. Calculate the wavelength of an electron accelerated through a potential difference of 10 kV. 3. It is also desirable to be able to calculate the wavelength associated with an electron when the accelerating voltage is known. There are 3 steps in the calculation. Calculate the wavelength of an electron accelerated through a potential difference of 10 kV.

17. Step 1: Kinetic energy Step 1: Kinetic energy E K = eV = 1.6 × 10 -19 × 10000 = 1.6 × 10 -15 J Step 2: Step 2: E K = ½ mv 2 = ½m (mv) 2 = p 2 / 2m, so momentum p = √2mE k = √2 × 9.1 × 10 -31 × 1.6 × 10 -15 = 5.4 × 10 -23 kg m s -1 Step 3: Wavelength Step 3: Wavelength λ = h / p = 6.63 × 10 -34 / 5.4 × 10 -23 = 1.2 × 10 -11 m = 0.012 nm.

18. Slits Laser Screen 1 L1L1 Slit spacing, d Wavelength, Distance to screen, L Fringe spacing, x Screen 2L2L2 d1d1 d2d2

19. End