Photoelectric Effect: Solving the Mystery of Big Negative Interception

[hunc]

Homework Statement

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.

Homework Equations

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.

4. Thanks in advance!

 

[ehild]

Homework Statement

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.)

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

ehild

 

[hunc]

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

ehild

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.)

 

[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

 

[hunc]

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

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.

 

[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

 

[hunc]

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

I'll try to do that today.