# Frequency, Velocity, Voltage, and Intensity

Illuminitwit

## Homework Statement

I have to calculate the frequency of incident light shot at a metal plate.

## Homework Equations

I know the voltage, intensity, and wavelength.

wavelength = velocity/frequency
frequency = velocity/wavelength
voltage = current • resistance

## The Attempt at a Solution

There isn't any specific math involved in this problem, really. I mean, there is because I have the data in a chart that I made after having to experiment with incident light and different kinds of metal plates observing the photoelectric effect. All I need to know in order to finish the table and data by myself is how to get the frequency when I have the intensity, wavelength, and voltage. I don't have a velocity, unless I want to use the velocity of light, which I'm considering. Is it even possible to use the voltage in such calculations to end up at the frequency? I don't think the voltage has a whole lot to do with frequency of incident light. Input is greatly appreciated!

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Illuminitwit
Waiiit... Can't I just use the velocity of light in a vacuum? I mean, I'd have to assume the photoelectric effect in a vacuum, but I guess that'll work... Sorry to waste a post! Unless anyone has any extra advice. Thanks! :)

Mentor

## Homework Statement

I have to calculate the frequency of incident light shot at a metal plate.

## Homework Equations

I know the voltage, intensity, and wavelength.

wavelength = velocity/frequency
frequency = velocity/wavelength
voltage = current • resistance

## The Attempt at a Solution

There isn't any specific math involved in this problem, really. I mean, there is because I have the data in a chart that I made after having to experiment with incident light and different kinds of metal plates observing the photoelectric effect. All I need to know in order to finish the table and data by myself is how to get the frequency when I have the intensity, wavelength, and voltage. I don't have a velocity, unless I want to use the velocity of light, which I'm considering. Is it even possible to use the voltage in such calculations to end up at the frequency? I don't think the voltage has a whole lot to do with frequency of incident light. Input is greatly appreciated!

Current isn't really relevant here, and neither is voltage. You need to figure out the frequency of the incident light required to exceed the work function of the particular metal...

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

.

Illuminitwit
Thanks! I didn't think voltage and current stuff was necessary... It isn't really about electrical circuits or anything, sooo...

The three metals I have are Potassium, Calcium, and Uranium... Should I just look up the work function online?

Mentor
Thanks! I didn't think voltage and current stuff was necessary... It isn't really about electrical circuits or anything, sooo...

The three metals I have are Potassium, Calcium, and Uranium... Should I just look up the work function online?

Yes, you should get the work functions for your calculations for the lab report. Also, how were you measuring the photoemissions? Maybe that's where the V and I results come in? Can you describe your lab setup and measuring instruments?

Illuminitwit
I found a website that has the listed work functions in eV for various metals regarding photoelectric effect. I just have to convert that using 1.6 x 10-19 J•s. So I'm set.

Actually, it's pretty pathetic because it's not a "real" lab. It's considered a virtual simulation online and I have to setup a bunch of different values for wavelength, voltage, et cetera. It's for my Physics class. Sometimes we have real labs and sometimes we're told to go to a certain URL and do an experiment. Basically, I just have controls to change the type of metal, voltage, wavelength, and intensity of the incident light and the photoemissions travel along a little route through this tank and light up a little bulb at the other end (if they reach that far; sometimes the photoemissions just shoot out for a little bit and then go back towards the metal and never reach the bulb). The photoemissions aren't measured, really. I have to calculate them based on what the little simulation window says. I think I have it figured out now, though. Thanks! :)

Mentor
Yeah, be sure to read and understand the wikipedia link about Photoemission, or your textbook. You should understand what to expect with the online experiments. Like, you should be able to calculate what voltage you need to apply to make the emitted electrons turn around and go back to the emitting plate.

Quiz Question -- how would you calculate that voltage, based on the metal and the incident light's wavelength?

Illuminitwit
I have no idea... That isn't covered anywhere in my textbook or lessons. I'll see if I can figure it out, though...

The voltage in the simulation is retarding voltage, I think...

According to http://web.mit.edu/zchen/www/PHOTOELECTRICPAPERDRAFT.pdf,
V(retarding) = (hf - phi)/e
KE = hf - phi
phi = work function of the metal
h = Planck's constant
f = frequency
E = hf
e = charge of an electron

"From this equation, we can see that if V is plotted against f, the slope of the line is h/e. Planck's constant h is simply the slope of the Voltage vs. frequency line multiplied by e (charge of an electron)."

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Mentor
I have no idea... That isn't covered anywhere in my textbook or lessons. I'll see if I can figure it out, though...

The voltage in the simulation is retarding voltage, I think...

According to http://web.mit.edu/zchen/www/PHOTOELECTRICPAPERDRAFT.pdf,
V(retarding) = (hf - phi)/e
KE = hf - phi
phi = work function of the metal
h = Planck's constant
f = frequency
E = hf
e = charge of an electron

"From this equation, we can see that if V is plotted against f, the slope of the line is h/e. Planck's constant h is simply the slope of the Voltage vs. frequency line multiplied by e (charge of an electron)."

That's all correct. Hopefully you can figure out why, before you do the lab. It's best to understand the material and have predictions for the results, before actually doing the lab.

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