1
$\begingroup$

In the explanation of photoelectric effect it is written that intensity and frequency of radiation have different results i.e. higher intensity means greater number of emitted photoelectrons and higher frequency means greater kinetic energy of emitted photoelectrons.

But what we have learned about intensity is that it is the energy released per unit area per unit time.

So if we say about the intensity of light in terms of photons it would look like this :

I = nhf/area/time.

If we take area to be 1 unit and time to be 1 second then the equation will look like this :

I = nhf

The equations above show that with fixed intensity if frequency is increased then number of photons or say photoelectrons decreases and vice versa. But according to Einstein's theory number of photoelectrons is affected only by the intensity of light not its frequency.

What's the cause of this contradiction ?? Is it okay to define intensity only as number of photons ?? In that case there should be no units of intensity since it is just the number of photons .

Am I wrong somewhere ?

I am just a beginner so please cooperate.

$\endgroup$
  • $\begingroup$ Intensity is power per unit area I guess? $\endgroup$ – Jack Rod Aug 22 '20 at 2:56
  • $\begingroup$ @Yuvraj And power is energy per unit time 😂 $\endgroup$ – A student Aug 22 '20 at 3:58
2
$\begingroup$

What you just discovered is that if you keep the intensity of a light source constant and increased the frequency, the number of photons per unit area must decrease. What Einstein told us is that intensity alone is not enough to tell us whether an electron will be ejected from a solid. Increasing the intensity and keeping the frequency constant only increase the number of ejected electrons, and if there are no electrons ejected then nothing happen, none start coming out. However changing the frequency of of the light (even when intensity is kept constant) cause the electrons come out with higher energies.

$\endgroup$
  • $\begingroup$ Is my above mentioned intuition wrong somewhere as it seems contradictory with that of Einstein's theory ?? $\endgroup$ – A student Aug 21 '20 at 17:14
  • $\begingroup$ I don't fully understand what you mean. Increasing the intensity by keeping the number of photons constant and increasing the frequency has a different effect than increasing the intensity by keeping the frequency constant and increasing the number of photons. $\endgroup$ – A. Jahin Aug 21 '20 at 17:17
  • $\begingroup$ I meant that if intensity is increased ( say two times) and at the same time the frequency is decreased ( say half) then the number of photons should increase (by four times) . But Einstein told that frequency doesn't affect the number of photons. What is the cause of this contradiction ?? $\endgroup$ – A student Aug 22 '20 at 4:08
  • $\begingroup$ There are two situations here. (1) The energy of a photon after you half the frequency is lower than the work function (the minimum energy to eject electrons). In this case the number of electrons drop to zero. (2) The energy of the photons is still bigger than the work function, in this case more electrons will come out but with lower energies. $\endgroup$ – A. Jahin Aug 22 '20 at 12:05
  • 1
    $\begingroup$ Well yes. But also notice that (1) the energies of the electrons will decrease by more than half.. And (2) if you are measuring current, just increasing the number of electrons ejected will not mean a direct increase in the current, since the current also depends on the kinetic energy of the ejected electrons. $\endgroup$ – A. Jahin Aug 22 '20 at 12:12
1
$\begingroup$

Intensity, in a manner of speaking, is how bright the source of light is. Brighter being more intense.

What Einstein postulated (inspired by Planck) was that light was made up of packets of energy $\epsilon$ (proportional to frequency but we’ll get to that later). So Einstein said if the total energy of a monochromatic light source that you observe per unit area per unit time (intensity) is $E$, then it is made up of $n$ photons. Where n is given by: $$n=\frac{E}{\epsilon}$$

Note that $\epsilon$ is a property of individual packets of light where as intensity $E$ is only defined for a collection of such packets.

Each packet can knock out one electron from the metal provided it has energy $\epsilon$ greater than the binding energy of the electron historically known as the work function $\phi_0$. The ejected electrons then have the kinetic energy $\epsilon - \phi_0$

Now if we were to increase the brightness of our source, we would be increasing the intensity $E$. Since $\epsilon$ is still the same (as it is the same source), our equation above says that we’ll have a bigger $n$. This means more electrons will be ejected.

Now coming to your question:

I = nhf

In both the equations we see that intensity and frequency are related. But in Einstein's theory both have different results.

It is further seen that the individual energy packets $\epsilon$ are related to frequency as $\epsilon=hf$.

Now, say we have a light source with intensity $E=100$ and the work function $\phi_0=1$. Consider the following two cases:

  • $\epsilon=10$
  • $\epsilon=50$

Both cases there’ll be ejection of electrons. However the number or electrons ejected aren’t the same. In the first case there will be $n=10$ electrons each with kinetic energy of $9$ units. In the second case, only $n=2$ electrons will be ejected albeit with a much higher kinetic energy of $49$ units.

Now let us double the brightness in the same experimental setup. So we have $E=200$ and $\phi_0=1$. Same two cases of:

  • $\epsilon=10$
  • $\epsilon=50$

Again in both cases there’ll be ejection of electrons. In the first case there will be $n=20$ electrons each with kinetic energy of $9$ units, same as before. In the second case, only $n=4$ electrons will be ejected albeit with a much higher kinetic energy of $49$ units, same as before.

What we observe is that the kinetic energy of each ejected electron is unchanged with changing intensity. However the number of ejected electrons has doubled with doubling intensity.

$\endgroup$
  • $\begingroup$ what you did is just an example of Einstein's theory. But what's wrong with my intuition ?? $\endgroup$ – A student Aug 22 '20 at 2:34
  • $\begingroup$ Also did you mean that by increasing frequency number of photons decreased ?? This is fundamentally the same as what I wrote above. $\endgroup$ – A student Aug 22 '20 at 2:39
  • $\begingroup$ And again n=E/ϵ . So shouldn't E be proportional to n as well as ϵ (which is proportional to frequency )?? So aren't intensity and frequency interlinked ?? $\endgroup$ – A student Aug 22 '20 at 2:41
  • $\begingroup$ Also in Einstein's theory it is given that increasing frequency has no effects on the number of photoelectrons and it is affected by intensity only but in your example we saw that it decreased . Isn't it a contradiction ?? $\endgroup$ – A student Aug 22 '20 at 2:55
  • $\begingroup$ @Ankit what you did is just an example of Einstein's theory. But what's wrong with my intuition ?? no what I just described is what happens in nature. $\endgroup$ – Superfast Jellyfish Aug 22 '20 at 4:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.