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Physics question the photo electric effect

5mW is directed at the photocathode of a vacuum photocell, which is connected to a microammeter which reads 0.4micro Amperes. calculate the number of photons per second incident on a photocathode.

if i do the number of photoelectrons per seconds is i/e where e is the charge of the electron.
Doing it this way gives : 2.5 x 10^12

The other way is if using the equation(FOR number of photons):
Power of the beam = nhf
h=6.63 x 10^-34
f=3 x10^8/590 x 10^-9 using (c=f x lambda)
so
0.5 x 10^-3/(6.63 x 10^-34 x f) = n
(this way i get n=1.5 x 10^15)

the second answer is the correct answer, but why is the first one wrong :frown:

thank you in advance :smile:

Reply 1

lordaxil

The first calculation gives the number of ELECTRONS flowing through the ammeter per second, which has no direct relation to the number of photons incident on the photocathode (without knowing its efficiency).

Reply 2

username3477548

Original post by lordaxil

Thank you

But isn't the no. Of electrons= the number of photons incident? Each photon affects one electron?

Reply 3

lordaxil

Original post by Qxi.xli

But isn't the no. Of electrons= the number of photons incident? Each photon affects one electron?

Not at all - the ratio is known as the quantum efficiency, and can be as low as

10^{-6}

10^{-5}

depending on the metal.

https://physics.stackexchange.com/questions/193622/quantum-efficiency-of-photoelectric-effect

Reply 4

username3477548

Original post by lordaxil

Not at all - the ratio is known as the quantum efficiency, and can be as low as

10^{-6}

10^{-5}

depending on the metal.

https://physics.stackexchange.com/questions/193622/quantum-efficiency-of-photoelectric-effect

ooh ok thanks :smile:

But in my AQA physics textbook, it says one photon affects one electron? and my teacher also said that same? xx

Reply 5

lordaxil

It requires at least one photon to eject an electron (and it must be of a certain minimum frequency) but it is not the case that every photon results in ejection of an electron. Can you send a photo of the relevant page in textbook?

(edited 3 years ago)

Reply 6

username3477548

Original post by lordaxil

ohh ok thanks :smile:

please find attached, its the first bullet point

Untitled document (73).pdf276.2KB

Reply 7

lordaxil

Thanks. I'm quoting the relevant text here for ease of reference:

When light is incident on a metal surface, an electron at the surface absorbs a single photon from the incident light and therefore gains energy equal to hf, where hf is the energy of a light photon

The emphasis here is that the energy for ejecting a photoelectron must come from a single photon with energy hf (e.g. two photons with energy hf/2 would not be sufficient). However, this does not imply that every single photon ejects a photoelectron.

Although I wouldn't take it too literally, the diagram on right of the text actually shows multiple photons incident on the surface, and only one photoelectron being emitted. In fact, the yields are much lower than that in practice, due to the fact that photoelectron is often recaptured before it can escape from the metal (for that reason, using a thin film of metal can increase the quantum yield)

Reply 8

username3477548

Original post by lordaxil

Thanks. I'm quoting the relevant text here for ease of reference:

The emphasis here is that the energy for ejecting a photoelectron must come from a single photon with energy hf (e.g. two photons with energy hf/2 would not be sufficient). However, this does not imply that every single photon ejects a photoelectron.

Although I wouldn't take it too literally, the diagram on right of the text actually shows multiple photons incident on the surface, and only one photoelectron being emitted. In fact, the yields are much lower than that in practice, due to the fact that photoelectron is often recaptured before it can escape from the metal (for that reason, using a thin film of metal can increase the quantum yield)