How does Einstein’s photon hypothesis explain all the observational facts? We list here the
explanations for the same six observational facts above:
1. Say the intensity is low, i.e., light is very dim:
there are few photons irradiating
the surface. It is not very likely then that the photon will collide with any atom in
the material, and therefore will not eject any electron.
If the light is very bright,
there are many photons, and a certain fraction thereof collides with atoms and ejects
electrons.
If we now double the intensity, we also double the number of incident
photons, and therefore also double the number of collisions and correspondingly the
number of ejected electrons. (Recall that this experimental result is partially explained
also by classical theory.)
2. When a photon collides with an atom it ionizes it and releases an electron on the
atomic transition time scale. There is no need to “build up” the energy transferred to
the material, and all the energy is imparted at once.
3. For fixed
I
, if we increase
f
the number of photons
n
decreases (for the product
nhf
to remain constant). With fewer photons, there are also fewer ejected electrons.
4. Below the cuto
ff
frequency the photon does not carry enough energy t release the
electron, regardless of how many photons there are. Recall that electrons are released
by a single photon. (The probability of multi–photon interaction is very small.)
5. Increasing the intensity means increasing the number of photons, but the energy im
parted by a single photons is unchanged.
6. When the frequency
f
increases, the photon carries more energy. The work function
remains unchanged, so that the additional energy is converted to a higher kinetic
energy, i.e., higher
K
max
. As we showed above
V
stop
∝
f
with the same proportionality
coe
ffi
cients for all materials (
h/e
), and
k
max
=
eV
stop
.
The Compton e
ff
ect
If light is made of particles, then photons should collide with other particles like particles do.
We therefore analyze the elastic collision of two particles, letting one of them have energy
according to Einstein’s hypothesis (
E
=
hf
), and allow for velocities to be relativistic. We
take the particle to be static before the collision, and the photon moving before the collision
in the
x
–direction. We denote the angle the scattered photon makes with the
x
–axis by
φ
and the angle the scattered particle makes with the same axis by
θ
.
114
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Conservation of energy:
hf
=
hf
+ (
γ

1)
mc
2
where (
γ

1)
mc
2
is the recoil relativistic kinetic energy of the particle the photon collides
with.
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 Spring '09
 LIORBURKO
 Physics, Energy, Photon, Light, Compton, Compton Edge, hf − hf

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