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

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/h
ome/animations3/modernPhysics/photoele
ctricEffect.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 Einstein
1879 – 1955
Momentum
• Previously learned
p = 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 Planck
1858 - 1947
=
Ephoton = energy of a photon
measured in J or eV
max KE of
ejected
electrons
fo
frequency
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
Photon
KEmax
Ephoton = hf
Wo = hfo
e- e- e- e- e- ePhotoelectrons
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.