Does intensity of photons affect threshold frequency?

AI Thread Summary
Increasing the intensity of photons while keeping energy constant does not lower the threshold frequency; rather, it increases the number of photons without changing their energy. The discussion clarifies that the stopping voltage remains consistent across different intensities at the same frequency, as illustrated in the referenced graph. The concept of "threshold frequency" is questioned, as it is not explicitly defined in the graph. The relationship between photon energy and frequency is emphasized, with the equation E = h nu highlighting that intensity relates to the number of photons rather than their energy. Overall, the graph's interpretation suggests a misunderstanding of how intensity and frequency interact in the context of the photoelectric effect.
Tommy1995
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If energy is kept constant and intensity of photons is increased, will the threshold frequency be lowered?

I ask because I'd like to understand this graph - http://www.flickr.com/photos/coachrobbo/3909285882/

"lower intensity same frequency" causes stopping voltage to be lowered, thus threshold frequency is lowered. Why is this so? It would be really helpful if someone could add in an equation showing that a change in photon intensity alters threshold frequency.

Regards - Tom
 
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Tommy1995 said:
"lower intensity same frequency" causes stopping voltage to be lowered, thus threshold frequency is lowered. Why is this so? It would be really helpful if someone could add in an equation showing that a change in photon intensity alters threshold frequency.

You've misunderstood the diagram. What is happening there is, the curve comes in at a different angle to the same intercept. So both the green and blue curves have zero current at -3 volts. The red curve crosses at -2 volts. So the stopping voltage is -3 volts for both the reference beam and the lower intensity, same frequency beam.

It's just that pixels are finite size, so you can't really see it correctly on the graph.

This corresponds to the idea that a photon is a bundle of energy. The size of that bundle is proportional to the frequency, E = h nu. A more intense beam of the same frequency is just more bundles of the same size.

It's not clear what you mean by "threshold frequency" since no such term is defined on the graph.
 
DEvens said:
You've misunderstood the diagram. What is happening there is, the curve comes in at a different angle to the same intercept. So both the green and blue curves have zero current at -3 volts. The red curve crosses at -2 volts. So the stopping voltage is -3 volts for both the reference beam and the lower intensity, same frequency beam.

It's just that pixels are finite size, so you can't really see it correctly on the graph.

This corresponds to the idea that a photon is a bundle of energy. The size of that bundle is proportional to the frequency, E = h nu. A more intense beam of the same frequency is just more bundles of the same size.

It's not clear what you mean by "threshold frequency" since no such term is defined on the graph.

Exactly yours (the OP) is not the same graph as is mostly shown to demonstrate the photoelectric effect. (As in this link.)
Also, all three lines on the graph relate to light which is above the threshold frequency.
 
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