Photoelectric Effect—
Post-Lab Summary
Photon Energy
• Energy of Photons is dependent on the
frequency of the light.
• Reminders:
– Speed of light (c) = 3.00 x 108 m·s-1
𝑐 = 𝑓𝜆
Energy of a Photon:
𝐸 = ℎ𝑓
Practice Problem
• A beam of blue light (wavelength =
455. nm) is emitted from a light
source.
– What is the frequency
of this light?
8
𝑐
3.00𝑥10
14
𝑓= =
=
6.59𝑥10
𝐻𝑧
−7
𝜆 4.55𝑥10
– What is the energy of the blue photons?
𝐸 = ℎ𝑓 = 6.626𝑥10−34 6.59𝑥1014
𝐸 = 4.37𝑥10−19 𝐽
1 𝑒𝑉
−19
𝐸 = 4.37𝑥10 𝐽
= 2.73 𝑒𝑉
−19
1.60𝑥10 𝐽
What usually happens as energy
is absorbed?
• Classical Explanation:
– a substance will continue to gain energy every
time energy is transferred to it
– For example, if a child on a swing is given a
push every time he reaches the peak of his
swing, he will continually gain energy from each
subsequent push, and will steadily swing higher
and higher.
– If one push isn’t enough to make him reach a
certain height, eventually he’ll get there. (It
might just take a while)
Does this work for electrons in a
metal when energy is added?
• In a word, NO.
• Electrons must gain a certain amount of
energy in order to break completely free
from the surrounding atoms.
• This amount of energy must come at once,
not over a period of time
Energy gained by electrons
• When photons hit a metallic surface, the
energy from the photon is absorbed by an
electron
• IF there isn’t enough energy in the single
photon to make the electron break free, it
won’t. Simple as that.
• If there, is, then it will. Simple as that.
Threshold Frequency
(Cutoff frequency)
• The frequency that carries the
minimum amount of energy
necessary for an electron to absorb
in order for it to be emitted from the
surface of a metal
Work Function (ϕ)
• The energy, in eV, that must be
absorbed by an electron in order for it
to be emitted from the surface of a
metal
• Work Function is dependent on the
type of metallic surface. Not its size,
or temperature or anything else—just
the type.
Practice Problem #2
• The work function for a piece of
Aluminum is 4.08 eV.
• How much energy, in Joules, must
an electron absorb in order to be
emitted from the surface of
aluminum?
• What frequency of light would allow
this to occur?
• Where in the spectrum is this light?
What about intensity of light?
• Does intensity affect the amount of energy
an electron absorbs?
• NO! Each photon of light can eject a single
electron. So if there is a higher intensity,
that means there are more photons. More
photons = more electrons…but not more
energy per electron.
What if a photon has more energy
than the electron needs?
• Additional energy, over and above
the work function, that is supplied by
the photon will be converted into the
kinetic energy of the electron
Graphical Representation
Work Function
• What does the
slope represent?
Graphical Representation
Work Function
𝑬𝒌,𝒎𝒂𝒙 = 𝒉𝒇 − 𝝓
Stopping Voltage
• When a piece of metal is
subjected to light and
produces a photocurrent,
the stopping voltage is the
amount of potential that
must be added in order to
just stop the forward motion
of the electrons.
• The voltage that just brings
the current down to zero (0)
Stopping Voltage
• What must be done to stop an electron?
– A force must be applied; work must be done
• What must change when the electron
comes to a stop?
– Kinetic energy of the electron reaches 0
• How do we represent that
mathematically?
𝑾 = 𝒒𝑽
𝑬𝒌 = 𝒆 ∙ 𝑽

Assume that a 180 W lightbulb gives
off 2.80% of its energy as visible light. How
many photons of visible light are given off
in 2.90 minutes? (Use an average visible
wavelength of 550 nm.)

A metal with a work function of 2.30 eV is
illuminated with a beam of monochromatic
light. It is found that the stopping potential
for the emitted electrons is 2.30 V. What is
the wavelength of the light in nanometers?


The work function of a photoelectric material
is 3.5 eV. Assume the material is illuminated
with monochromatic light (λ =250 nm).
(a) What is the stopping potential of the
emitted photoelectrons?
(b) What is the cutoff frequency?