Why is the photoelectric effect cited as an example of a particle-like nature of photons?
The photon's not physically knocking off the electron, right?
It's supplying energy to break the bond, hence the work function of a metal.
It's like placing a popcorn kernel over a flame. It pops because it's had energy supplied, not because the fire punched it.
Am I missing something?
The energy is delivered to the metal in discrete packets.
If you think of light as a wave, than one may expect that a low-energy colour of light should be able to (eventually) liberate electrons from the metal if you wait long enough (as more and more energy is deposited into the metal). Because electrons are not liberated below a certain energy threshold (color of light which happens to be the work function of the metal), the conclusion is that the energy is given to the metal in discrete packets (particles).
There are two approaches that give different results. Classical wave theory of light fails to explain some phenomenons of photoelectric effect but the quantum theory, which assumes particle nature of light, explains them fruitfully.
Photons do not give their energy in parts, they either will give all the energy or none at all. The energy given to the electrons should be more than or equal to the work function of the metal, if the electron is to be ejected. The excess energy will become the electron's kinetic energy.
Failures of classical wave theory of light to explain laws of photoelectric effect :
(1) . $\space$ According to the classical wave theory if intensity of the wave is increase the oscillating electric field vector $E$ of the light increases in amplitude $(Recall\space I \propto A^2 )$. The force applied to the electron in the metal will be $eE$ due to the falling radiation and this force will increase on increasing the intensity of light.
This suggests that the kinetic energy of the emitted photoelectrons should also increase due to the larger force applied in emitting them out as the light beam is made more intense.
However, observations show that the maximum kinetic energy is independent of light's intensity (at constant frequency).
$KE_{max}=h\nu - h\nu_{\circ}$,
As this relation suggests, maximum kinetic energy depends only on the frequency of the falling radiation.
(2). $\space$ According to the wave theory, the photoelectric effect should take place at any frequency of light, provided that the light is intense enough to supply the energy needed to eject the photoelectrons.
However, observation show that the photoelectric effect does not occur if the frequency of the falling radiation is less than the threshold frequency, even if the light is highly intense.
(3). $\space$ In the classical theory, the light energy is uniformly distributed over the wave front. So, when the light falls on the metal, the energy of the incident light will not entirely go to a particular electron in the metal but will be distributed uniformly to other electrons also.
So, the electrons will take some time to accumulate enough energy to escape from the metal surface. Hence, there should be a measurable time lag between the impinging of the light on the surface and the ejection of photoelectron.
However, no time lag is observed. Photoelectrons are ejected the instant the light falls on the metal.
Quantum theory's explanations :
(1). $\space$ Let N be the number of photons falling on metal plate per unit area per unit time then,
$N=\frac{IA}{h\nu}$
Now, if we double the intensity, we double the number of photons which consequently doubles the photocurrent. This is done without changing the energy of the individual photon.
Hence, Kinetic energy does not increase when you double the intensity of light (keeping energy of individual photon constant).
(2). The second problem can be explained if we consider the case $KE_{max}=0$. In this case we have,
$Work function\space (\phi)=h\nu_{\circ}$.
This tells us that the photon has just enough to eject the photoelectrons and no extra energy appear as kinetic energy.
If frequency is reduced below threshold frequency $(\nu_{\circ})$, no photoelectrons will be ejected no matter how intense the radiation may be.
Remember that, the intensity of radiation can be increased by either increasing the number of photons in it or by increasing the energy of each photon.
(3). $\space$ The third conundrum follows from the photon theory because the required energy is supplied in a concentrated bundles of energy (i.e photons).
Energy is not spread uniformly over the beam's cross section as in wave theory.
The ejection of photoelectrons is instantaneous. The time difference between incidence of light and emission of photoelectrons is very very small, may be even less than $10^{-9}s$.
$Conclusion$ : In photoelectric effect light exhibits its particle nature. +1 for photon.
For more info on intensity , check out some of my answers given in these links : Are number of photons in an incident radiation proportional to its intensity?
