“The temperature of a lava flow can be approximated by merely observing its colour. The result agrees nicely with the measured temperatures of lava flows at about 1,000 to 1,200 °C.”

• In the late 19th Century, physicists began to study light emitted from hot solids and gases

• They studied non-reflecting solids that appeared black at room temperature

Why do black objects appear black?

•When a solid object is heated, it emits EMR with a range of frequencies. (The atoms are vibrating – as predicted by Maxwell’s equations.)

• It followed that objects at higher temp’s should emit EMR with higher frequencies.

frequency of wave animation

• Classical theory predicted that the intensity of the radiation would increase as the square of the frequency but this did not fit with observations

It’s an ultraviolet

catastrophe!

As the temp. increased: the max brightness (intensity) increased and more EMR was emitted as shorter wavelengths (higher freq.)

However, the frequencies reached a peak, then dropped.

Black Body Animation

•This contradiction was called the ultraviolet catastrophe

•Max Planck developed a theory that proposed that the vibrating atoms could only have, emit, or absorb certain discrete amounts of energy (or whole number multiples)

•He called the bundles quanta (singular: quantum)

•Planck’s theory was revolutionary because it proposed that an object cannot vibrate with any amount of energy – only specific amounts.

•Whereas, classical theory said energy could be sent out in continuous streams of waves

… may be thought of as a wave packet kinda like:

A quantum of light is known now as a photon

The energy of a quantum (photon) of light was proportional to its frequency: Planck's Law

hfE E

nerg

y pe

r pho

ton

Freq

uenc

y (H

z)

Pla

nk’s

Con

stan

t

= 6

.63

x 10

-34 J

s

• Planck’s equation fit all observations of blackbody radiation

• Since c = f, Planck’s Law can be written

hc

E

Example 1Determine the energy of a photon of light (in J) with a frequency of 2.56 x 1014 Hz.

E = 1.70 x 10-19 J

• The energy of a photon is very small, a more convenient unit is the “electron volt” (eV)

the amount of energy an electron gains or loses as it moves through a potential difference of 1 volt

1 eV = 1.60 x 10-19 Jh = 4.14 x 10-15 eVs

Example 2Determine the energy of a photon in eV, of light with a frequency of 4.88 x 1014 Hz.

E = 2.02 eV

Example 3Determine the energy in eV of a photon emitted by a laser in a retail scanner (λ = 632 nm).

E = 1.97 eV

Example 4The same retail scanner expends 1.0 mW of power. How many photons are emitted per second?

E = 3.17 x1015 photons / second

CBC: Quirks and Q

uarks: “The New Solar Pow

er

CBC: Quirks and Q

uarks: “The New Solar Pow

er”

• Hertz noticed that certain metals lost negative charges when exposed to ultraviolet light

• The term PHOTOELECTRIC EFFECT was given to this phenomenon and the electrons given off were called photoelectrons

• Many experiments were done to measure the effects of the incident radiation on the energy and number of the photoelectrons

As the voltage increases, the current through the tube decreases to zero, this is known as the stopping voltage and is related to the maximum kinetic energy of the photoelectrons.

stopk qVE max

animation

Must be in Joules!

Example #1Photoelectrons emitted from a piece of selenium have a maximum kinetic energy of 7.2 eV. Calculate the stopping voltage of the photoelectrons.

Vstop = 7.2 V

Experiments on lots of different metals and different frequencies of incident light gave results which could not be completely explained by classical physics:

1) Electrons are emitted from the metal only if the incident frequency is above a certain threshold frequency (fo).

2) Intensity (brightness of the light) had no effect on fo. No matter how bright the light, if it is below fo, no electrons are emitted.

3) Each metal has its own value of fo.

4) As the frequency of the incident radiation increased, the kinetic energy of the photoelectrons increased

5) The photoelectrons are emitted immediately (classical physics predicted time delays of weeks for very faint light)

6) The brightness of the light increased the amount of photoelectrons, but did not change the kinetic energy

• Einstein proposed that each photon gave all of its energy to just one electron.•The amount of energy was given by Planck’s equation• The electron required a certain amount of energy to be freed from its atom.

called the work function

• Electrons emitted from the surface of a metal would have the maximum amount of kinetic energy limited by the work function

WEhf k max

Energy of Energy of incident incident photonphoton

Energy of Energy of photoelectronphotoelectron

Work Work functionfunction

outin EE

ohfW

You Loser!You Loser!

incident frequency (Hz)

kin

etic

en

erg

y o

f p

ho

toel

ectr

on

s (e

V)

Kinetic energy of photoelectrons as a function of incident frequency

Metal Metal Metal1 2 3

• The slope of the lines is Planck’s constant, x-intercept is fo, the y-intercept is the work

function of the metal

Millikan used the photoelectric effect to accurately determine Planck’s constant and to verify Einstein’s work

Practice Problem #1Determine the work function of a metal which emits photoelectrons only when the wavelength of the incident light is equal to or greater than 500 nm.

W = 2.48 eV

Get it? … it’s a “work

function!”

Practice Problem #2

Light with a frequency of 7.00 x1015 Hz strikes a crystal inside of a photovoltaic cell which has a work function of 1.50 eV. Determine the kinetic energy of the photoelectrons.

Ekmax = 27.5 eV

Practice Problem #3

Determine the maximum speed of the photoelectrons in the last example.

v = 3.11 x106 m/s

Remember:If it has mass, it’s

moving slower than 3.00 x 108 m/s!