1
PHOTOELECTRIC EFFECT
Light in
Electrons
out
Metal Surface
DETECT NUMBER OF ELECTRONS AND
THEIR KINETIC ENERGY
OBSERVATIONS
# e
–
# e
–
ν
I
KE
ν
KE
I
ν
0
ν
0
ν
>
ν
0
ν
>
ν
0
PHOTON
– a “particle” of light
ENERGY OF A PHOTON
E = h
ν
where h is the Planck’s constant:
h = 6.626 x 10
-34
J s
LIGHT AS A PARTICLE
LIGHT AS A PARTICLE
ENERGY AND WAVELENGTH
E = h
ν
and
c =
λ
ν
, then
E = (h c) /
λ
APPLICATION
SAMPLE CALCULATION
A laser emits red light with a wavelength of
633 nm, what is the energy of the photons being
released at this wavelength?
APPLICATION
SAMPLE CALCULATION
Given: wavelength = 633 nm
We know:
E = h
ν
and
c =
λ
ν
so:
E = (h c) /
λ

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
2
APPLICATION
SAMPLE CALCULATION
E = (h c) /
λ
= (6.626 x 10
-34
J s) x (3 x 10
8
m s
-1
)
633 x 10
-9
m
=
3.14 x 10
-19
J
SAMPLE CALCULATION – 2
The particular laser describe before has an
output of 1 mW (1 x 10
-3
J s
-1
).
How many
photons are emitted per second?
no. of photons = Output (power) / E
=
(1 x 10
-3
J s
-1
)
(3.14 x 10-
19
J / photon)
= 3.18 x 10
15
photons / s
APPLICATION
ENERGY OF A PHOTON
E = h
ν
LIGHT AS A PARTICLE
# e
–
ν
ν
0
THRESHOLD FREQUENCY
Energy greater than E
ν
0
is required to eject
an electron
E
ν
0
= h
ν
0
BINDING ENERGY
–
the force that atoms
exert to hold onto their electrons
So, part of the energy of the photon is absorbed
to overcome the binding energy.
As
ν
is increased and
ν
>
ν
0
part of the energy is
still absorbed to overcome the binding energy

This is the end of the preview.
Sign up
to
access the rest of the document.