0
$\begingroup$

This question already has an answer here:

I understand a photon with a smaller wavelength is more energetic so for a given intensity, less photons are incident on the electrons and so less photo electrons reach the detector per second. However, isnt current the rate of change of charge? If we have more energetic photons, the photoelectrons have larger KE's. They should be arriving more frequently at the detector. Does this not increase the current?

Why does the current necessarily decrease if the wavelength of incident light is decreased, with the sources intensity fixed. Does the decrease in photoelectrons produced win over the increased rate of arrival? Is there any mathematical model for this?

$\endgroup$

marked as duplicate by Bill N, John Rennie, Jon Custer, GiorgioP, Kyle Kanos May 27 at 2:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You should edit your title. "increasing decreasing" doesn't make sense. $\endgroup$ – Bill N May 23 at 20:59
  • $\begingroup$ Where have you gotten your information about the photoelectric effect? The second half of the first sentence is totally wrong. Your assumption in the second paragraph is wrong. $\endgroup$ – Bill N May 23 at 22:56
  • $\begingroup$ What part? If you change the wavelength of your light, your photons carry more energy, you are still supplying the same amount of energy per second, so you will have less photons being emitted from your light source per second. As the number of released photo electrons ( electrons emitted from the metal due to absorbing a photon ) is proportional to the number of photons reaching them per second, which is now lower, will also reduce. Hence less photo electrons will reach the detector per second. $\endgroup$ – Vishal Jain May 24 at 6:06
  • $\begingroup$ This makes it clear hyperphysics.phy-astr.gsu.edu/hbase/mod2.html . It is the maximum photo electron energy that is plotted to show the work function. it is derived from the current as stated. $\endgroup$ – anna v May 24 at 6:57
  • $\begingroup$ Also Photocurrent's dependence on frequency $\endgroup$ – John Rennie May 24 at 9:11
0
$\begingroup$

If we have more energetic photons, the photoelectrons have larger KE's. They should be arriving more frequently at the detector.

First, a photon's energy is not kinetic energy.

And the photon's energy doesn't affect the velocity it travels at. All photons travel at c in vacuum.

Finally, knowing the energy of the photons in a beam doesn't tell you anything about how many photons are reaching some point in the beam per unit time.

For that you have to measure the beam power. Then the rate of photon arrival is given by

$$r_p = \frac{P}{E_p} = \frac{P}{h\nu}$$

where $r_p$ is the arrival rate (in photons per second), $P$ is the beam power (in watts), and $E_p$ is the photon energy (in joules), which can also be written as $h\nu$.

As you can see, to keep the beam power constant while increasing the energy per photon, you must decrease the photon arrival rate.

$\endgroup$
  • $\begingroup$ I was asking about the photo electrons, or the electrons released by the photons that are incident on them. My question was just asking that since the photons will be more energetic, they will release electrons with a larger KE, since the work function will not be affected. If electrons travel fasters, doesn't this increase the current detected since current is rate of charge flow wrt time. $\endgroup$ – Vishal Jain May 24 at 5:52
0
$\begingroup$

Let say we are projecting "x" number of golf balls per min. at some standing beer cans. Each ball has the minimum energy necessary to knock over a can. Any ball with extra energy will knock the can it hits farther. Now lets replace the golf balls with an equal number of marshmallows. None of the marshmallows has enough energy to knock over a can. What would happen if we started replacing golf balls with marshmallows?

$\endgroup$

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