17.2.3 Experimental Evidence

The photon theory sounds really sexy. But does it have the support of experimental observations?

Emission Rate

Firstly, can the photon theory explain why higher intensity results in higher rate of emission? The picture in our head is that of photons raining down at the metal. Of all the photons arriving at the metal, only a tiny fraction of the photons are absorbed by electrons[1]. Nevertheless, higher intensity means more photons arriving at the metal per unit time, which means more electrons being “hit” by photons per unit time, which means more photoelectrons liberated per unit time. Yay!

KEmax

Next, can the photon theory explain why increasing the intensity of light does not result in higher KEmax? Well, the answer is clear from the photoelectric equation

\displaystyle K{{E}_{{\max }}}=hf-\Phi

Increasing the intensity of light merely increases the number of photons arriving per unit time. It does not increase the amount of energy packed in each individual photon hf. So while there are more liberations per unit time, the energy constraint for each liberation has not changed. The maximum KE is still being limited by the same hf. Hence the same KEmax. Yay!

What about the observation that increasing the frequency of light results in higher KEmax? Well, light of a higher frequency contains photons which individually packs more energy. The energy constraint for each liberation has been lifted. One just has to look at the photoelectric equation again. The least tightly bound electrons still require the same amount of work function F, but with a higher hf, they now escape with a higher KEmax! Yay!

Threshold Frequency

Can the photon theory explain why photoelectric effect cannot occur below the threshold frequency, regardless of the light intensity? Let’s see.

With a bit of thought, one should realize that the threshold frequency f0 must be the frequency at which the photon energy is exactly equal to the work function of the metal.

\displaystyle h{{f}_{0}}=\Phi

At this frequency, a photon packs just sufficient energy to knock out the least tightly bound electron with zero KE.

\displaystyle K{{E}_{{\max }}}=h{{f}_{0}}-\Phi =0

It does not matter how many the photons are raining on the metal per unit time. Below f0, since each photon does not pack sufficient energy to liberate even the least tightly bound electron in the lattice, not a single electron is liberated. Hence no photoelectric effect! Yay!

Now the alert students will ask, if one photon does not have sufficient energy, why can’t the electron accumulate the energy of a few photons? Well, it turns out that the electron is unable to hold on to any surplus energy for long: if it does not escape after absorbing the energy of one photon, it gives up the energy immediately and start from square one again. Besides, the chance of an electron being hit by two photons simultaneously at the same time is practically zero (except at extremely high intensity situations). Hence, one punch one kill is the rule of the game.

Zero Latency

Can the photon theory also explain why photoelectric effect is instantaneous and has zero latency?

Unlike a wave which delivers energy continuously to many electrons over a duration of time, a photon delivers its entire packet of energy to one electron instantaneously. The moment one electron absorbs the energy of one photon, one photoelectron is liberated (if the photon energy is at least equal to the work function). This explains why a steady photoelectric current (albeit a small one) is registered the moment the light (no matter how low the intensity) arrives on the metal. Yay!

Concept Test

3406

3412

Comics

One-Shot-One-Kill


[1] The so-called yield rate is seldom more than 0.1 %. Well, the absorption rate can’t be too high since most metal surfaces are highly reflective and shiny.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s