Photoelectric Effect, Wave Length To Area Relationship

Whatever are you talking about?

Hi John, yes. Ive been to some whatever these days. Hehe hope your well.

Photoelectric?

Wave length?

Area?

Maybe describe what you are trying to do.

So far light does not get around corners. So the area of "projection" is the same area that you project the light at.

This should be your starting point until you can come up with something else.

It's a simple equation:

Wavelength = K • ( Projected Area ), where K is in nanometers per steradian.

Except, there is no such constant as K.

Because there is no actual relationship between wavelength and projected area.

I mean like if you want to design an antenna for FM radio frequencies how would you size it? Is it small or big one? How should it be size? Compare to am radio? High frequency, shorter wave length needs long or big antennas? Or low frequencies long wave length needs just small antenna? What you think?

You might want to look into optics a bit deeper.

Your initial question is not what you ask here.

So what exactly is the application?

Is this for photometric analysis?

The energy of a photon is inversely proportional to the wavelength. It takes a certain amount of energy, the work function, to remove an electron from a metal surface so a photon has to have a short enough wavelength to exceed this energy. (The work function depends on the metal.) The intensity of the light (number of photons per unit time) with wavelength shorter than the critical wavelength determines the number of electrons (current) that will be removed from the metal.

So if you have a certain intensity light shorter than the critical wavelength (photons per area per unit time) then the amount of photocurrent would be proportional to the area.

https://en.wikipedia.org/wiki/Photoelectric_effect

Or to put it another way, "if the wavelength's too long, the area is irrelevant".

Like spectrum 400 to 900 nm wavelengths, how would i approach the smallest receptor/absorber for it? Id like to do the math, but i ain got the right foundation yet. I really hope someone could help, if he she be kind. Thanks in advance.

First, you have to have the right kind of metal. Look on the left side of the periodic table. The alkalai metals don't hold on to their electrons to tightly.

900 nm = 1.37760 ev

400 nm = 3.09961 ev

http://www.calctool.org/CALC/other/converters/e_of_photon

Here is a table of work functions for common metals:

http://www.knowledgedoor.com/2/elements_handbook/photoelectric_work_function.html

The energy of the electrons coming off is the photon energy minus the work function for the chosen metal.

The shorter the wavelength the greater the voltage (electron energy in ev)

The brighter the light, the more the current (number of electrons).

The amount of light energy per unit time would be the product of the area and the light intensity: Energy = area x (Energy/area) and would be roughly equal to (electron energy + work function) x current

Or you just need a bigger antenna!

Thought we were talking about the photoelectric effect. But who reads thread titles, anyway?

A o yes, smart ideas should be keep close in discretion, thanks John.

Yeah....

Are we?

I read along in the thread. From OP at #6

I mean like if you want to design an antenna for FM radio frequencies how would you size it? Is it small or big one?

Thanks Rix, youre such a someone to rely on.

As Rixter and another pointed out the photoelectric effect is a different phenomena from radio frequency reception. The area of the sensor, photo cell, etc. will not be a critical factor with sub-micron wave lengths of light.

What you seem to interested in is called solid state theory. The link is to the Rössler introduction textbook at Springer. Don't let the "introduction" title fool you. This textbook is for a post graduate class. Digesting this one textbook would likely take the complete semester.

Springer has been releasing many of their older textbooks as completely free to download. I'm not sure if that link will work for everyone.

Hi redfred. Yeah, this is good. Thank you.

Get yourself a copy of the Amateur Radio Relay League (ARRL) Handbook, it has all the information/formulas you want/need and it's written so that non-math/non-engineering type people can understand it, if they make the effort.