560 likes | 746 Views
Quantum Nature of Matter & Energy. Is light a wave or a particle?. Wave Properties. -Polarization -Diffraction -Interference.
E N D
Quantum Nature of Matter & Energy Is light a wave or a particle?
Wave Properties -Polarization -Diffraction -Interference
Max Planck studied how blackbodies emit radiation (EM waves).A blackbody is an object that absorbs & does not reflect any light incident on it. As Temp goes up, power (intensity) goes up.Frequency goes up too!
T vs. Intensity of a blackbody Classical Physics predicts E becomes infinite as l approaches 0.
As T , molc. vibration. Planck proposed that the E is emitted by vibrating molecules is quantized—E could only take on certain values. Energy come in little "chunks" of the frequency multiplied by a constant now called Planck's constant, h:
Matter absorbs & emits energy in discrete units called quanta or photons. Plank’s formula gives the amount of energy based on the frequency of waves. E = hf. h is Plank’s constant 6.63 x 10-34 Js. E is energy in Joules f is frequency of radiation
Planck’s theory’s predictions agreed with experimental observations.
Ex 1. Each photon of a certain color light has an energy of 2.5 eV. What is the frequency of and color of the light?
Solution:E = hff = E/h convert eV to Joules.(2.5 eV)(1.6 x 10-19J/eV) = 6 x 1014 s-1 or Hz 6.626 x 10-34 J s
When you need to find # joules in a wave/photon use E = hf.OrE = hc/l.
The fact that EM waves comes in discrete packets seems to indicate E is made of particles. • More evidence for the particle nature of light was later to come from work with the photoelectric effect.
Light as E particles (photons, quanta) • E = hf (packets) • Photoelectric effect.
Photoelectric Effect It's been known (previously) that when EM waves shine on a metal surface, the surface can emit e-. You can start a current in a circuit just by shining a light on a metal plate. Materials that emit e- in this way are called photoemissive. The ejected e- are called photoelectrons.
Light is a type of EM wave, & waves carry energy. If a wave of light hits an e- in one of the atoms, it may transfer enough energy to knock the electron out of the atom.
Photoelectric – supports Planck idea. Observations Emission depends on f – not amplitude/intensity
Each metal, has a threshold frequency Light frequencies below the fo eject no e-, no matter how intense or bright the light.
Light frequencies above the fo eject electrons, no matter how low the intensity or how dim.
Einstein proposed that EM radiation delivers E in chunks, or quanta, he called photons, each photon has an Energy of hf. Higher-frequency photons have more E, so they can knock out e- & make the e- come flying out faster with greater KE.
Increasing the intensity/amplitude, increases the rate of e- emission or the current; the # of e-, but each e- won’t gain any extra energy. Only f increases E of photoelectrons.If the frequency is too low, none of the photons will have enough energy to knock an electron out of an atom.
All EM waves can be described as photons. The energy carried by photons is: Ephoton = hf or Ephoton = hc/l. (for photon traveling at speed of light).This energy is absorbed by photo-emissive materials.
The min. f to free e- is fo.The min E to free an e- is the work function Wo, or F.Metals have low F.
max KE of photo e- vs. f for metal. As f of EM wave increases, KE increases, slope = h. F (work function), is minimum energy needed to eject e-. Work function
The threshold frequency fo, is the minimum frequency needed to eject e- so the minimum energy needed, work function in J is: Wo = F = hfo. Any photon E left over after the work function, goes into KE of e-.
Since a photon’s E goes to work function & KE, The total E of photoelectrons is:Etot = Fo + KE.The maximum KE of ejected e- is:KEelc = hfphoton – F. h is Plank’s constantF. is the work functionKEe- = hf – hfo.
Ex 2: Photoelectric Effect:Light having f = 1 x 1015 hz shines on a sodium surface. The photoelectrons have a maximum KE of 1.86 eV.Find the threshold frequency for sodium.
KEmax = hf – WoWo is the minimum energy needed to eject e-. Since E = hf, the minimum energy occurs at the threshold frequency. We can say KEmax = hf – hfo where hfo is the Wo. Since we want to find fo, then rearrangefo = (hfphoton – KEmax)/(h)
change eV to Joules: (1.86 eV) (1.6 x 10-19 J/eV) = 2.85 x 10-19 Jfo = (hfphoton – KEmax)/(h)(6.63 x 10-34 Js)(1 x 1015 hz) - (2.85 x 10-19 J) (6.63 x 10-34 Js)fo = 5.5 x 1014 Hz.Below this frequency no electrons will be ejected.
In 1913-1914, R.A. Millikan did a series of extremely careful experiments involving the photoelectric effect. He found that all of his results agreed exactly with Einstein's predictions about photons, not with the wave theory. Einstein actually won the Nobel Prize for his work on the photoelectric effect, not for his more famous theory of relativity.
(IB) Stopping Potential Milliken’s experiment measured the KE of photo e- by applying a stopping potential – a voltage that brought the velocity of e- to 0.
Stopping Potential overcomes KE of e-. The energy of the voltage = qV where q is the charge on one e-. When e- velocity = 0 then eV (qV)= KEmax. Then graph is made.
Some experimental results, like this one, seem to prove that light consists of particles; others insist, that it's waves. We can only conclude that light is somehow both a wave and a particle--or that it's something else we can't quite visualize, which appears to us as one or the other depending on how we look at it.
Hist of Quantum pt 1 15 min http://www.youtube.com/watch?v=zBTbqOgdfEY
http://www.youtube.com/watch?v=B7pACq_xWyw&feature=related Watch this. Good summery of light. 9:45 min.
IB Read Hamper pg 239 – 242 • Do take home qz packet on photoelectric effect.
Hwk Text 23 – 1 and 23 – 2Do page 833 all andpage 856 # 2, 4, 7, 10, 11
Einstein realized that matter contains energy. There is an equivalence of mass & energy.Energy is stored in the nucleus of atoms.The energy stored any mass obeys Einstein’s equation: E = energy in J.E = mc2. m = mass kg c = vel of light
Ex 2: How much energy is produced when 2.5 kg of matter are completely converted to energy?How much energy is that in eV?
E = mc2.=(2.5 kg )(3x108 m/s)2. = 2.25 x 1017 Jin eV(2.25 x 1017 J)(1 eV / 1.6 x 10 –19 J) = 1.4 x 1036 eV.
Atomic Mass Units: amu or u Mass of atoms very small so they are measured in amu or u. Since mass is equivalent to energy, 1 amu = 931 MeV or 931 x 106 eV.
Ex 3: One universal atomic mass unit is equivalent to an energy of 931 MeV. Calculate the mass in kg of one universal mass unit.Hint: Use E = mc2 where energy is known in eV.
Don’t forget to convert MeV to eV.(1 u) x (931 MeV/u) x (106eV/MeV) x (1.6 x 10 –19 J / eV) =1.49 x 1010 J E = mc2 so m = E/c2.(1.49 x 1010 J) / (3x108 m/s)2 = 1.66 x 10 –27 kg
The mass units are based on the mass of a proton or 1H. (A hydrogen nucleus)
Particle Properties of Waves extend to conservation of energy and momentum. Photons may give up all or part of their energy in collisions, but the sum of the momentums and energy before must equal the sum after.