Physics 25 Chapter 29 Dr. Alward

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1 Physics 25 Chapter 29 Dr. Alward Photons and Matter Waves Planck s Constant: h = 6.63 x J-s E = hf E = hc/λ 1

2 Example A: Red light of wavelength λ = 720 nm consists of a stream of photons of what energy (in ev)? E = hc/λ = 6.63 x x 3.0 x 10 8 /720 x 10-9 = 2.76 x J 1.0 electron-volt (ev) = 1.6 x J E = 1.73 ev (See faster way at the right.) The 1243 Equation hc = (6.63x10-34 J-s)(3.0x10 8 m/s) = 1.99 x J-m = 1.99 x J-m/1.6 x J/eV = x 10-6 ev-m = (1.243 x 10-6 ev-m) / 1.0 x 10-9 m/nm = 1243 ev-nm E = 1243 ev-nm / λ In order to obtain E in ev, λ must be in nm, which will cancel the nm units. The example at the left is solved below: E = 1243 ev-nm/720 nm = 1.73 ev Example B: A laser pen using red light of wavelength λ = 730 nm is operating at an output power of 40.0 milli-watts. How many photons are emitted each second? E = 1243/730 = 1.70 ev 40.0 x 10-3 J/s /1.6 x J/eV = 2.50 x ev/s (2.50 x ev/s)/1.70 ev/photon = 1.47 x photons/s 2

3 The Photoelectric Effect Electrons at the surface of metal require the least amount of photon energy to escape: that amount of energy is called the work function, and its symbol is W. In the example below, the surface is gold, whose work function is 5.10 ev. After delivering all of the photon s energy to the surface electron, the electron will escape if it has any energy left over after using 5.10 ev of it to leave the surface. What is left over is the kinetic energy of the photo-emitted electron: Maximum Kinetic Energy = Photon Energy - Work Function K = hf - W Electrons lying not at the surface, but deeper inside the metal, such as the electron at the far right above, require more than 5.10 ev to escape. In the case of that electron, it uses 1.62 ev to fight it way to the surface, and then 5.10 ev more to escape. Example A: The mass of an electron is 9.11 x kg. What is the maximum speed of electrons emitted when 160 nm light is shined on gold? In the diagram above, we see that photoemitted electrons at the surface have kinetic energy of 2.67 ev. Example B: In the example at the far right in the diagram above, an electron deep within the metal receives the photon energy. How much energy (call it α) did the electron use to fight its way to the surface? ½ (9.11 x ) v 2 = 2.67 (1.6 x ) v = 9.7 x 10 5 m/s α = 1.05 α = 1.62 ev 3

4 Photon Momentum In colliding with the proton above, the light wave exerts a force on it, according to the first right-hand rule. The EM wave interacting with an object exerts a force on it, just as would a particle having momentum exert a force on an object when it collides with it. What is the momentum of a photon? Without proof, we state here that each of the photons in the stream has a momentum: p = E/c = hf/c = h/(c/f) = h/λ A wave-like entity (light) has a particle-like attribute (momentum): p = h/λ 4

5 de Broglie Waves (also called Matter Waves ) Twenty-six year-old Louis-Victor-Pierre- Raymond de Broglie was a 26-year-old French physicist who made groundbreaking contributions to quantum theory. In his 1924 PhD thesis he postulated the wave nature of electrons and suggested that matter has wavelike properties. de Broglie noted electromagnetic waves have a particle-like property (momentum p), as expressed through the equation obtained previously, p =h/λ, where λ is the wavelength of the EM wave. He then guessed that perhaps the reverse is true: that particles have a wave-like property--wavelength, as expressed by the same equation, twisted around: Example: What is the matter-wavelength of a 1000-kg automobile traveling at 20 m/s? p = mv = 2.0 x 10 4 kg-m/s λ = h/p = 6.63 x / 2.0 x 10 4 = 3.32 x m This about five-hundred times smaller than the Planck length, 1.6 x meter, which is thought to represent the theoretical limit of the smallest possible object. Nothing is this small, so it would be impossible for this matter wave to interfere with, or be disrupted by, any object. Thus, the wavelength nature of this automobile would forever be hidden from us, and is therefore of no consequence. Only matter waves of very small moving objects--such as electrons, are of any consequence. λ = h/p p = momentum of the object λ = matter-wavelength (Also called de Broglie wavelength) 5

6 Example: Calculate the de Broglie wavelength of the electron in a hydrogen atom orbiting in a circular path of radius r = 2.12 x m. Circumference = 2πr = 2π (2.12 x ) = 1.33 x 10-9 m F = ma ke 2 /r 2 = mv 2 /r = (p 2 /m)/r p 2 = (mke 2 /r) = (9.11 x )(9 x 10 9 )(1.6 x ) 2 /2.12 x = 9.90 x p = 9.95 x kg-m/s λ = 6.63 x /9.95 x = 6.66 x m For the orbiting hydrogen electron above, explore below the issue of constructive versus destructive interference of waves: How many 6.66 x m matter-wavelengths fit around the 1.33 x 10-9 m circumference? Answer: 1.33 x 10-9 / 6.66 x = 2 Two de Broglie wavelengths fit neatly around the circular path, allowing the wave to constructively interfere with itself, creating a standing wave. If this were not the case, we would have the impossible situation of the electron self-destructing as it orbited the hydrogen nucleus. 6