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Photoelectric effect question (photoelectric effect over a period of time)

I know what the photoelectic effect is.

It's my understanding that the electron emission can occur due to photons (which have to have the right amount of energy or more), heat, and various other sources. If we talk about the photoelectric effect due to heat, I understand that over a period of time, the heat would build up and free the electron (whereas with photons, it couldn't use 2x photons which each had less energy than the work function - each photon has to have the correct energy).

However, I've just seen a question about the photoelectric effect involving UV photons, and it said that over a period of time, more electrons would be released. Why does time matter? Why would the photons keep coming from the UV source over a period of time, and not instantly? Is it because a UV source would constantly be emitting photons?
Reply 1
The shorter the wavelength of the more (kinetic) energy the electrons have, and so the faster they move. Hence UV light emits faster moving photons than, for example, yellow light would. However they both emit the same number of photons as long as they have the same light intensity.

Light intensity is what affects the number of electrons emitted, and a higher light intensity equates to more electrons being emitted (regardless of speed) in a given time.

Either way, electrons are constantly being emitted by the light source and hence the number of electrons must be measured in a given period of time, so it is all relative.

Edit: just rereading your question and I don’t think I properly answered it. I can’t be 100% sure on my terminology but I think the photons are referred to as “virtual” which essentially means they are both “there” and “not there” until you need them in which case they are there. Hence there is a constant source of “virtual” photons
(edited 6 years ago)
Reply 2
Original post by Pastelx
The shorter the wavelength of the more (kinetic) energy the electrons have, and so the faster they move. Hence UV light emits faster moving photons photoelectrons than, for example, yellow light would. However they both emit the same number of photons as long as they have the same light intensity.


Be careful, this isn't true, though it's a common error. Since the freq of UV is higher, each photon carries more energy than a photon of yellow light and therefore you need fewer UV photos to give the same energy and therefore the same intensity as the yellow light.

Also, of course, the photons all travel at the same speed, the speed of light, I suspect that was just a typo.




Hope this helps clarify.
(edited 6 years ago)
Reply 3
Original post by rsk
Be careful, this isn't true, though it's a common error. Since the freq of UV is higher, each photon carries more energy than a photon of yellow light and therefore you need fewer UV photos to give the same energy and therefore the same intensity as the yellow light.

Also, of course, the photons all travel at the same speed, the speed of light, I suspect that was just a typo.




Hope this helps clarify.


My AQA A Level course states that the intensity of light only affects the number of electrons emitted (regardless of the frequency of light), while frequency of light only affects the kinetic energy of the electrons?

Thanks for pointing out that typo! Clearly I wasn’t paying attention while I was typing haha
Reply 4
Original post by Pastelx
My AQA A Level course states that the intensity of light only affects the number of electrons emitted (regardless of the frequency of light), while frequency of light only affects the kinetic energy of the electrons?


If the frequency remains the same but the intensity changes, then you must have more photons per second so more electrons but with same av ke..

If the intensity stays the same but the frequency increases then you must have fewer photons (fewer photons needed to make same overall E) so you will have fewer electrons but each with greater ke.

Hope this helps
(edited 6 years ago)

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