Interpreting I-V graphs in Photoelectric effect

[Curious12345]

Homework Statement

In photoelectric effect I-V graphs, for the same intensity but with different light frequencies (f2>f1), the I-V graph has the same max photocurrent but |Vs2| > |Vs1|. If we concentrate between the cutoff and the saturation regions of the I-V graph, we can see that at any given applied voltage, the photocurrent from the case with light freq f2 is higher than that with f1. Seems the saying increase freq only increases max KE but not the photocurrent cannot explain this observation.

Homework Equations

The Attempt at a Solution

I know that there are lost of discussions in this and other forums about frequency only increase KE of liberated electrons but not the photocurrent. However, it seems to me that they are talking about the max photocurrent, i.e. the saturation region of the I-V graph but not the "linear" region of the I-V graph. Could the photocurrent for f2 higher than f1 due to the fact that space-charge still exist in this region and therefore the probability of an electron arriving at anode increases with increased KE (specially when the applied voltage is negative)?

 

[Andrew Mason]

If you think of I in terms of dq/dt what does that tell you about the the relationship between current and the speed of a charge?

AM

 

[ehild]

I know that there are lost of discussions in this and other forums about frequency only increase KE of liberated electrons but not the photocurrent. However, it seems to me that they are talking about the max photocurrent, i.e. the saturation region of the I-V graph but not the "linear" region of the I-V graph. Could the photocurrent for f2 higher than f1 due to the fact that space-charge still exist in this region and therefore the probability of an electron arriving at anode increases with increased KE (specially when the applied voltage is negative)?

Yes, the maximum KE of the liberated electrons increase with the frequency of the light, but they have to reach the anode to contribute to the photocurrent. If the anode is at high enough negative potential, the electrons lose the KE when they travel away from the cathode and fall back into it. Even at zero or low positive applied voltage , there is a space charge region, where the electrons are attracted back to the cathode, and only the fastest reach the anode. At high enough positive potential of the anode, all the liberated electrons can leave the neighborhood of the cathode and we get the saturation current. With increasing frequency of light, the average KE of the electrons becomes higher, and more of them can get over the space-charge region, as you said.

 

[Curious12345]

If you think of I in terms of dq/dt what does that tell you about the the relationship between current and the speed of a charge?

AM

Hi Andrew, Thanks for replying. In terms of rate of change of charge, it doesn't explained why higher frequency gives higher photocurrent when the applied voltage is between the stopping voltage and saturation voltage.