1. THE COMPTON EFFECT Energy and momentum are conserved when a photon collides with an electron.

2. KE=1/2mv 2 E2E2 E 1 E 1 = E 2 + KE (electron) X-ray

3. EXAMPLE A ) A Photon having an energy of 3.8eV collides with an electron at rest. The electron has a gain in kinetic energy of 1.2eV, what will be the energy of the emitted photon in eV and joules? E 1 = E 2 + KE (electron) 3.8eV = E 2 + 1.2eV E 2 = 2.6eV E 2 = 4.16x10 -19 J *(1 eV = 1.6 x 10 -19 J)*

4. B) What is the frequency of the scattered light wave and what color is it? E 2 =hf 4.16x10 -19 J= 6.63 x 10 -34 Js (f) f = 6.23 x 10 14 Hz color = blue

5. c) What is the velocity of the electron? KE= 1/2mv 2 1.2eV(1.6 x 10 -19 J/eV)= ½ (9.11x10 -31 kg)v 2 1.92 x 10 -19 J = 4.56x10 -31 (v 2 ) 4.2 x 10 11 = v 2 v= 6.49 x 10 5 m/s

6. DeBroglie Wavelength If light has particle properties then particles must have wave properties λ= h = h mv ρ λ- wavelength h- Planks constant m- mass v-velocity ρ- momentum

7. EXAMPLE 1 What is the DeBroglie wavelength of a 0.25kg ball thrown at 20m/s? λ = h = 6.63 x 10 -34 Js mv (0.25kg)(20m/s) λ =1.33 x 10 -34 m *This is too small to observe.*

8. EXAMPLE 2 What is the DeBroglie wavelength of an electron traveling at 1x10 6 m/s? λ = h = 6.63 x 10 -34 Js mv (9.11x10 -31 kg)(1x10 6 m/s) λ= 7.29 x 10 -10 m * Suitable for interference and diffraction and can be observed*

9. EXAMPLE 3 What is the momentum of a proton having a DeBroglie wavelength of 2.81x10 -11 m? λ=h/ρ 2.81x10 -11 m = 6.63 x10 -34 Js ρ ρ = 2.36x10 -23 kgm/s

10. Heisenberg Uncertainty Principle “To study the nature and motion of a moving electrons by bombarding them with photons would change the motion and position of the electron and lead to uncertainty” You can’t accurately measure both position and momentum of a moving electron at any given time.