# Photoelectric effect experiment help

1. Jul 4, 2006

### al_201314

Hi guys,

I've just returned from my exams.. I thought it was pretty alright but I think it could've been better. I was stuck with this question.

A photoelectric experiment was set up. Monochromatic light is incident on a metal plate, and the photo-electrons are collected at a electrode made of the same metal in a vacuumed glass. The difference in potential between the electrode and the metal plate is V. The power of the light is V, the current flowing through the tube is I and frequency of light is f.

Sketch the graph of current I against P when f constant.
Sketch graph of current I against V when power P and frequency f constant.

How should the graphs look like and why? I have no idea I drew a straight line graph with a +ve gradient for both. Couldn't see the picture.

Anyone who could enlighten? Many thanks. If I had prepared more for this I would have donw better.

Thanks guys!

2. Jul 4, 2006

### Staff: Mentor

See these -

http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html

http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html

Below a certain frequency, there will be no photoelectrons - so the current would be zero.

The current will be determined by the rate of electrons being ejected from the surface, so for a given frequency, above a certain threshold, the current will be proportional to P - the rate of electrons ejected = rate of photon incident upon the electron.

As for I vs V, what will happen as V increases? P is constant so the number of photons striking the photocathode is constant.

3. Jul 4, 2006

### al_201314

Thanks Astro.

I know as intensity increases, the current increases for the same reason. But how does the intensity of light relate to power of the light? Power is proportional to intensity?

As for I vs V, how does one increase V? After the photo-electrons has moved to the photocathode to create a net difference in charges? So is it correct to say that it would be a horizontal straight line?

4. Jul 4, 2006

### Hootenanny

Staff Emeritus
Intensity is equal to the power per unit area and has SI units of watts per square meter. The magnitude of the intensity at a distance r from an ideal point source of power P is given by;

$$|I| = \frac{P}{4\pi r^2}$$

Note that intensity is a vector and is a measure of the average energy flux through a surface.
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5. Jul 4, 2006

### al_201314

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How does it make intensity to be a vector since power and area are both scalars? Also it brings me to another question I just thought of. Why is current not considered a vector since it has direction and magnitude?

6. Jul 4, 2006

### Hootenanny

Staff Emeritus
Technically, the power is obtained by finding the product between energy density and the velocity at which the energy is moving (through a surface), thus resulting in a vector function.
Current, does not in fact have a direction, only a magnitude. It is defined as the rate of flow of charge; it matter not a jot the direction in which the electrons are travelling (as they will always be travelling in the same direction).

7. Jul 4, 2006

### Staff: Mentor

Simply, Power = photons/time * energy/photon = energy/time

V (an electrostatic potential) can be induced by a voltage source (e.g. battery or rectified AC source). The voltage source would collect the current and move them back to the photocathode.

With respect to increasing V, what about the reduction in the energy necessary to remove photoelectrons?

8. Jul 4, 2006

### al_201314

I'm still a bit confused about the potential difference issue. Before the light was shown on the metal, the 2 metal plates should have a potential difference of 0V since there is no net charges between them, am I right? The potential difference increases only when photo-electrons flow from the metal plate to the cathode causing there to be a difference in charges between the 2 plates?

Assuming I'm right, wouldn't the graph be a horizontal straight line for a constant current in I vs V? My reasoning is that as the current is constant, charges flow continually and the difference in charges between the photocathode and the metal plate increases? And current is constant because frequency and Power is constant? How then does the work function come into play here?

Thanks for the continued help :)

9. Jul 4, 2006

### Staff: Mentor

Think of a capacitor - apply a potential difference and what happens to the charges (electrons)? There are conduction electrons, which may become photoelectrons if they interact with a photon if the energy is sufficient.

10. Jul 6, 2006

### al_201314

Thanks guys plus a bit of reading up I think I got it