IV curve on photoelectric effect

[KFC]

Well, according to Einstein's explanation of photoelectric effect, I know that above cutoff frequency, the high frequency the incident photon is, the more electrons will be strike out. Now we add the batter on the photoelectric apparatus. At some negative voltage (stop voltage) there is no photoelectric current. At that voltage we increase the voltage gradually, we will also see the current increase (nonlinearly), and if we keep increase the voltage and over some big value, the current almost doesn't change. I wonder why in the beginning the current change rapidly and nonlinearly and later it is also flat?

In addition, if we compare the I-V curve of two different incident light with different frequency but same intensity. According to experimental data, we see that the curves start from different stop voltage and the currents increases gradually. Simple analysis tells us the higher the frequency of the photon, the larger (absolute value) the stop voltage is. Since the intensity of these two lights are the same, and intensity is defined as

$$I = \frac{\textnormal{number of photon}}{\textnormal{sec}\cdot\textnormal{m}^2}h\nu$$

we see that if I unchanged, higher frequency will lead to less number of photon in unit time. So higher-frequency light will strike out less photoelectrons in unit time. If we consider the I-V curve at V=0, the current corresponding to high frequency should be lower that that corresponding to low frequency, but why the actual case is just the opposite?

 

[ZapperZ]

Well, according to Einstein's explanation of photoelectric effect, I know that above cutoff frequency, the high frequency the incident photon is, the more electrons will be strike out.

This is not completely true. The higher the frequency of the photon, the higher the energy of the photoelectrons. It isn't necessarily "more electrons", especially if the photon flux remains the same (and the quantum efficiency isn't a factor).

Now we add the batter on the photoelectric apparatus. At some negative voltage (stop voltage) there is no photoelectric current. At that voltage we increase the voltage gradually, we will also see the current increase (nonlinearly), and if we keep increase the voltage and over some big value, the current almost doesn't change. I wonder why in the beginning the current change rapidly and nonlinearly and later it is also flat?

Because in the beginning, you are losing electrons. Not all electrons emitted by the cathode make it to the anode. As you increase the bias, you are collecting more electrons. At some point, you will be collecting ALL of the electrons and increasing the bias doesn't give you more.

Zz.

 

[Entropia]

I would like to thank you for bringing up these interesting questions about the IV curve on photoelectric effect. The phenomenon of the IV curve may seem counterintuitive at first, but it can be explained by understanding the principles of the photoelectric effect and the behavior of electrons in a circuit.

Firstly, it is important to note that the IV curve in the photoelectric effect is a plot of the photocurrent (or the number of electrons emitted per unit time) against the applied voltage. This means that the y-axis, or the current, is directly related to the number of electrons emitted by the photocathode. As you correctly stated, the higher the frequency of the incident photons, the more electrons will be emitted. This is because higher frequency photons have more energy, and can therefore overcome the work function of the metal and release more electrons.

Now, let's consider the behavior of electrons in a circuit. When we apply a voltage to the circuit, we are essentially creating an electric field that can accelerate the electrons. At low voltages, the electrons may not have enough energy to overcome the opposing electric field created by the battery. This results in a lower photocurrent as the electrons are not able to reach the anode.

As we increase the voltage, the electric field becomes stronger and can overcome the opposing field, allowing more electrons to reach the anode. This is why we see an increase in current in the beginning of the curve. However, as we continue to increase the voltage, the electrons will eventually reach a maximum velocity and cannot be accelerated any further. This results in a plateau in the IV curve, as the current remains constant despite increasing the voltage.

As for the question about the comparison of two different frequencies with the same intensity, it is important to note that intensity is not the only factor that affects the photocurrent. The work function of the metal and the cutoff frequency also play a role. In the case of higher frequency light, although there may be fewer photons per unit time, the higher energy of the photons allows more electrons to be released from the metal. This can result in a higher photocurrent compared to lower frequency light with the same intensity.

In conclusion, the behavior of the IV curve in the photoelectric effect can be explained by understanding the principles of the photoelectric effect and the behavior of electrons in a circuit. It is important to consider all factors, such as frequency, intensity, and the work function of the metal, when comparing