What Went Wrong with My Photoelectric Homework?

In summary, the problem is that the person doesn't know how to solve for the work function, and they are trying to use wavelengths to help.
  • #1
dontpunchme
4
0
So, I got the following question.

http://imageshack.us/photo/my-images/528/problemsl.jpg

This is figure 13.1.

http://imageshack.us/photo/my-images/100/131lg.jpg

C is the cathode material. S is the light. The voltage begins at A. The circuit carrying any current that makes it through the potential difference is everything else to the right.

So I think, okay, to identify the cathode material I need the work function value. Not too hard.

http://imageshack.us/photo/my-images/710/excelx.jpg

Note the formula I used for row 3.

http://imageshack.us/photo/my-images/4/attemptsolution.jpg

What did I do wrong? Why am I... what? These answers don't make any sense to me.

Thanks!

edit

Since the images aren't appearing, I just linked to them. Imageshack.
 
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  • #2
Would you please type in the problem and your attempt of solution?

ehild
 
  • #3
Alright.

First off, apologies to the boards. I didn't look for the "homework" subforum hard enough. (Being moved on my first post is embarrassing. :shy:)

So, here's my question, typed out.

I have an emitter electrode of unknown composition. There is a source of controllable light to the side, pointed right at the electrode. The cathode ray created by the electrode will have to travel through a potential difference between metal plates before reaching the anode where it is converted into a current.

My goal is to identify the emitter electrode composition. To determine this, I am given the following information.

I can vary the wavelength of the light created by the light source. I can also measure the minimum voltage (potential difference) required to stop the cathode ray from becoming a current completely.

If the following are the results of said experiment:
Code:
wavelength (nm)		250	300	350	400	450	500
minimum voltage (V)	2.70	1.90	1.30	0.80	0.50	0.10

This is my attempt at solving it.

(1/2)mv^2]max=eV'=(Planck's constant)(c/wavelength)-(work function)
Where V' is the minimum voltage.

So eV'+(work function)=(Planck's constant)(c/wavelength).

The only property here that is of the electrode is the work function, so...

(work function)=(Planck's constant)(c/wavelength)-(eV')

Now, since everything here has is measured in units of electron volts, I figure I might as well cancel all the e's from both sides.

(work function/e)=((Planck's constant)(c/wavelength)/e)-V'

Which makes things a bit simpler for the numbers. But it doesn't matter.

So I stuck this in Excel, hoping to come out with a single work function. But instead, my results were pretty much repeats of the minimum potential difference.

To which I said, "This can't be right."

(I'm only solving up to the work function in my answers, because finding the material composition from the work function is fairly trivial for this problem.)

Thanks in advance for the help!

(I need to learn LaTeX at some point.)
 
  • #4
Your method is all right, and I do not know what your problem can be without seeing the results you got for the work function. I got values scattering between 2.2 and 2.4 eV, 2.29 in average. Have you got a table of work function values of different solids? This is rather a low value, but there are metals with work function close to that.

ehild
 
Last edited:
  • #5
ehild said:
Your method is all right, and I do not know what your problem can be without seeing the results you got for the work function. I got values scattering between 2.2 and 2.4 eV, 2.29 in average. Have you got a table of work function values of different solids? This is rather a low value, but there are metals with work function close to that.

ehild

This is my input into Excel. (For the work function row.)

Code:
=((4.135667516*10^-15)*(299792458/A1))-A2

This is my result.

Code:
250		300		350		400		450		500
2.7		1.9		1.3		0.8		0.5		0.1
-2.699999995	-1.899999996	-1.299999996	-0.799999997	-0.499999997	-0.099999998
 
  • #6
Take care, the wavelengths are given in nanometers. So you need to have 250E-9,... and so on for A1.

ehild
 
  • #7
ehild said:
Take care, the wavelengths are given in nanometers. So you need to have 250E-9,... and so on for A1.

Oh. Okay.

Thanks.
 

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to light of a certain frequency. This effect was first observed by Heinrich Hertz in 1887 and was later explained by Albert Einstein in 1905.

2. How does the photoelectric effect work?

When light of a certain frequency, known as the "threshold frequency", falls on a material, it causes the electrons in the material to absorb the energy of the light and become excited. If the energy of the light is high enough, the electrons will be emitted from the material as photoelectrons.

3. What is the significance of the photoelectric effect?

The photoelectric effect is significant because it provided evidence for the particle nature of light and led to the development of the quantum theory. It also has practical applications in technologies such as solar cells and photocells.

4. What is the difference between the photoelectric effect and the photovoltaic effect?

The photoelectric effect refers to the emission of electrons from a material when it is exposed to light, while the photovoltaic effect refers to the conversion of light into electricity in a solar cell. Both processes involve the interaction between light and electrons, but they have different outcomes.

5. How can I calculate the energy of a photoelectron?

The energy of a photoelectron can be calculated using the equation E = hf - Φ, where E is the energy of the photoelectron, h is Planck's constant, f is the frequency of the incident light, and Φ is the work function of the material (the minimum energy required to remove an electron from the material).

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