Aim: How can we apply mathematics to the photoelectric effect?

In the photoelectric effect, how do you increase:

The number of ejected electrons?

The KE of the ejected electrons? Increase intensity of the wave

Increase frequency of the wave

• http://www.stmary.ws/highschool/physics/home/animations3/modernPhysics/photoelectricEffect.html

Wave-Particle Duality

• According to Einstein, light has particle characteristics

• Light travels as a photon

Photon – A “bundle” or “packet” of energy

Has zero rest mass but has momentum and energy

Albert Einstein1879 – 1955

Momentum

• Previously learnedp = mv

• You need mass to have momentum

• Photons have no mass but have momentum

• Contradicts classical physics!

The Planck Hypothesis

• In 1900, Max Planck proposed that energy could exist only in discrete quanta which were proportional to the frequency.

Max Planck1858 - 1947

=

Ephoton = energy of a photon

measured in J or eV

frequency

max KE of ejected

electrons

fo

Slope = h

Also….

How much energy does a photon of yellow light have?

Ephoton = hf

Ephoton = (6.63 x 10-34 J·s)(5.20 x 1014 Hz)

Ephoton = 3.45 x 10-19 J

How many electrons will be ejected if the threshold frequency is 6.20 x 1014 Hz?

None – If the frequency is below the threshold frequency, no electrons get ejected

Minimum Energy

• The minimum frequency gives the minimum energy

• Minimum frequency = threshold frequency (fo)

• Minimum energy = work function (Wo)

Can you figure out the formula for work function?

Wo = hfo

What is the work function of zinc if the threshold frequency is 9.6 x 1014 Hz?

Wo = hfo

Wo = (6.63 x 10-34 J·s)(9.6 x 1014 Hz)

Wo = 6.4 x 10-19 J

KE Model for the Ejected Photoelectrons

e- e- e- e- e- e-

Photon

Photoelectrons

Ephoton = hf

Wo = hfo

KEmax

Formula

KEmax = Ephoton – Wo

KEmax = hf – hfo

KEmax = h(f – fo)

Light with a frequency of 4.5 x 1015 Hz strikes zinc whose work function is 6.4 x 10-19 J. What is the maximum kinetic energy of the ejected electrons?

KEmax = hf – Wo

KEmax = (6.63 x 10-34 J·s)(4.5 x 1015 Hz) – 6.4 x 10-19 J

KEmax = 3.0 x 10-18 – 6.4 x 10-19

KEmax = 2.34 x 10-18 J

The work function of chromium is 4.6 eV. If a photon with 5.0 eV of energy strikes chromium, what is the maximum kinetic energy of the ejected electrons?

KEmax = Ephoton – Wo

KEmax = 5.0 eV – 4.6 eV

KEmax = 0.4 eV

Convert this to Joules.