Download presentation
Presentation is loading. Please wait.
1
Light and Matter Review
Some learnings…..
2
Light EMR travels at the speed of light and is a transverse wave.
Maxwell proved this mathematically, and explained how EMR is a self-propagating wave of perpendicular electric and magnetic fields. This eliminated the ether model. Visible light is between nm, and follows ROYGBIV, with red the longest. EMR ranges from smallest wavelength to longest as: cosmic, gamma, X-ray, UV, Visible, IR, micro, FM/TV/Radar, AM and AC waves. All EMR obeys the Universal Wave Equation v = ƒλ
3
Light Which EMR are longer than visible light?
Arrange the EMR spectrum from lowest to highest frequency.
4
X-Rays X-rays are created when high energy electrons collide with a target releasing X-rays. The accelerating voltage determines the energy of the X-ray. Photons of light are chunks of light where E α ƒ Find the wavelength of the X-ray created by electrons with 30.0 MeV of energy.
5
interference Waves overlap to create interference: CI if the amplitude increases (gets brighter) and DI if it decreases. A node is DI, and anti-node is CI. A stable pattern exists when the sources are coherent (constant in frequency) creating nodal lines and anti-nodal lines. 𝜆= 𝑑𝑥 𝑛𝐿 and 𝜆= 𝑑𝑠𝑖𝑛𝜃 𝑛 for CI and replace n with n + ½ for DI Look for key words: bright fringe (CI), absence of light (DI), in phase (CI)
6
example A 3rd order nodal line is located 2.00 cm from the central max and 21.0 cm from the midpoint of the 2 sources. If the 2 sources are 3.00 mm apart, what is the wavelength of the wave?
7
polarization An arrangement of crystals that acts like millions of tiny slits, allowing light waves to pass through in one plane only.. (affects the electric field). A polarizer by itself will eliminate 50% of the light passing through it. A second polarizer can be used to control how much light is transmitted after the first one (≤ 50%) Obeys Malus’ Law: 𝐼 2 = 𝐼 1 𝑐𝑜𝑠 2 𝜃 and I1 = ½ I0 Find the percentage of transmitted light for two polarizers at an angle of 75o to each other.
8
Photoelectric effect Einstein took work from Hertz, Millikan and Planck to explain this. Conservation of energy is assumed to follow Ek(max) = hƒ – W Light shines on a metal and the photon energy does work (binding energy) to free electrons and the remaining energy goes to the electrons as Ek. Increasing brightness only increases the current, not the energy. The work function is constant for a metal and corresponds to a threshold frequency: W = hƒo The stopping voltage measures the Ek of the photoelectrons. It’s the voltage needed to stop the electrons, so the current is zero. Ek = qVstop
9
PE example Find the velocity of emitted photoelectrons from a metal with binding energy of eV if the incident light has a frequency of 2.00 x 1014 Hz. The work function for a metal is 2.0 x J. What is the threshold frequency and the stopping voltage for 500 nm light?
10
Compton Effect and de Broglie wavelength
Compton explained how photons interact with electrons using momentum. A photon in a collision acts like a particle, but has no mass. An X-ray, hits an electron (which excites the electron out of the atom, and another electron falls to emit spectra) and the photon loses energy and momentum to the electron. The momentum of a photon follows p= ℎ λ De Broglie stated that if photons have momentum, then particles could have wavlengths! The de Broglie wavelength is found as 𝜆= ℎ 𝑚𝑣 Application: scanning electron microscope.
11
Examples A 200 nm X-ray hits an electron and scatters with a wavelength of 300 nm. What is the velocity of the electron? (Assume vi = 0) Find the deBroglie wavelength of an alpha particle accelerated by 200 V.
12
relativity Einstein postulated that the speed of light is constant for all inertial frames of reference. The effects of the math revealed that mass, length and time all change at relativistic velocities (>20% the speed of light). The mass effect affects the Ek formula and is modified as Ek = (m – mo)c2 The mass formula is: 𝑚= 𝑚 𝑜 1− 𝑣 2 𝑐 and ∆t = 𝑡 𝑜 1− 𝑣 2 𝑐 and L = Lo 1− 𝑣 2 𝑐 2 The binding energy in a nucleus comes from some mass being converted into energy and follows E = mc2.
13
Examples Find the kinetic energy of an alpha particle moving at 0.88c.
Find the binding energy for U-235 (Z = 92) given the mass of the atom is x kg.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.