Say a photon of frequency $\nu_1$ strikes the metal, and we have certain kinetic energy which is stopped by $V_{stop1}$. Now for frequency $\nu_2$ for same metal it will have $V_{stop2}$. Now $V_{stop}=\frac{1}{2}mu^2$. So $V_{stop}$ determines the $u$. Now current is defined as rate of flow of charge. So the flow of charge per unit time depends upon the number of electrons (here intensity) and how fast do they move (here kinetic energy or $u$). So for same intensity and different frequency, my logic says that current will differ and vice versa when we change intensity keeping frequency same. This seems to contradict photoelectric effect observation. Can you tell me where I have gone wrong? Thanks in advance


2 Answers 2


A photon that strikes the metal either has enough energy to release a valence electron from its bond with the atoms in the metal, from a certain frequency and up, or its energy is just not enough to eject the electron from the metal. The surplus of energy over the energy needed to eject the electron is used as kinetic energy of the electron and determines the speed with which it escapes the metal.

enter image description here

The current that starts running, however, is determined by the number of electrons that is ejected from the metal, as this creates a potential difference within the metal. The speed at which they move away from the metal does not matter in general, as they move in any random direction away from the metal.

The number of ejected electrons is determined by the number of photons striking the metal and therefore by the intensity of the light, not by its frequency.

I hope this helps you.


When you say the stopping voltage determines the kinetic energy of the electrons you have that the wrong way round. For a given kinetic energy of photoelectron you must apply a potential to bring it to rest.

So it is true that increasing the intensity of the incident light will increase the photocurrent as you have more electrons ejected from the metal, provided the incident light is of a high enough frequency to overcome the work function of the metal.

If the energy of the photons (and therefore frequency) isn't large enough to overcome the work function there will be no photocurrent regardless of the incident intensity of the light.

Increasing the energy of the incident light increases the number of electrons in the metal which can be excited out into the vacuum. For higher photon energy states that are further away from the Fermi level can be excited into the photocurrent. Increasing the frequency of the incident light also increases the kinetic energy that the photoelectrons have when they reach the vacuum.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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