# Homework Help: Photoelectric effect

1. Oct 21, 2014

### hunc

1. The problem statement, all variables and given/known data
I am doing the photoelectric effect experiment. And we were trying to verify the relationship between photoelectric current and distance is
$I \times d^2 = k$, where k is a constant.(At least I believe this to be true.)

But from the data, I kind of get a $I= k d^{-2} + b$. And I got a really big b...
I was using light of $\lambda = 436nm,$ and diaphragm of $\phi=2mm$. Here is the data.

i = [310, 198, 131] (10^-10A)
d^-2 = [1./30**2, 1./35**2, 1./40**2] (cm^-2)
And I got a relation like$$I = 369221.118818 d^{-2} - 101.137960591$$. I can't understand why is the b like this, big and negative.

2. Relevant equations

3. The attempt at a solution

• I am not alone. My mates have the same problem. And I checked with many of them.(I realize that three points is too small for data...)

• If distance grows really big, then current should curve. But that should put b higher than origin, while I got a minus sign. Also the line is really really linear.

• The "apparatus" was OK. Not great, but OK. I measured Planck constant right (to one part in a thousand), and I verified relation between I and \phi (of diaphragm), which has a interception of 0(to one part in a thousand again).

• My pals and myself can only come up with reasons for the interception to be positive.

Last edited: Oct 21, 2014
2. Oct 21, 2014

### ehild

What distance do you speak about? Distance of what from what???

ehild

3. Oct 21, 2014

### hunc

The distance between the photon emitter (the bulb) and the electron emitter (the metal).

But strictly speaking, both of them have protective coverings. So I was really measuring the distance between them.
(There are arrows supposed to imply the exact position of the real emitters.)

4. Oct 21, 2014

### ehild

So it can be imagined that your measured distance (dm) values are less then the real ones. d=dm+x.

Then instead of the theoretical intensity I=I0/d2 you measure I(dm). I(dm) = I0/(dm+x)2.
You can assume that x<<dm.
You fit a function to the I(dm) data. $$I=\frac{I_0}{dm^2(1+\frac{x}{dm})^2}$$ If x/dm<<1 the function can be approximated as $$I=\frac{I_0}{dm^2}(1-2 \frac{x}{dm})$$,
that is $$I=I_0/dm^2 - A$$ where A is some average of 2xI0/dm3.

ehild

5. Oct 21, 2014

### hunc

Thanks so much!

This idea came up earlier today. But I didn't find it plausible. For if x << dm, then we get to keep the linearity, but A then must be suffiently small. Note that it takes quite some distance to reach 100e-10 A. One the other hand, if x is only somewhat smaller than dm, then it's hard to keep the linearity.(At least that's what I thought!)

But I just take the leap and draw the curve. The second guess turns out to be a correct line for the eye. So I find my answer.

6. Oct 21, 2014

### ehild

I just looked at your data. They are very few and the distances very close to get a really good fit. You should have measured at least at 5 distances, and in a broader range, starting from about 15-20 cm.

ehild

7. Oct 21, 2014

### hunc

I'll try to do that today.