Photoelectric Effect*Post-Lab Summary

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
– What is the frequency
of this light?
𝑓= =
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
𝐸 = 4.37𝑥10 𝐽
= 2.73 𝑒𝑉
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
• 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
• 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
• 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
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?