# Why does decreasing the wavelength of light while maintaining intensity decrease current in photo electric effect

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If we have more energetic photons, the photoelectrons have larger KE's. They should be arriving more frequently at the detector.

First, a photon's energy is not kinetic energy.

And the photon's energy doesn't affect the velocity it travels at. All photons travel at c in vacuum.

Finally, knowing the energy of the photons in a beam doesn't tell you anything about how many photons are reaching some point in the beam per unit time.

For that you have to measure the beam power. Then the rate of photon arrival is given by

$$r_p = \frac{P}{E_p} = \frac{P}{h\nu}$$

where $$r_p$$ is the arrival rate (in photons per second), $$P$$ is the beam power (in watts), and $$E_p$$ is the photon energy (in joules), which can also be written as $$h\nu$$.

As you can see, to keep the beam power constant while increasing the energy per photon, you must decrease the photon arrival rate.

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### Vishal Jain

Updated on August 01, 2022

• Vishal Jain over 1 year

I understand a photon with a smaller wavelength is more energetic so for a given intensity, less photons are incident on the electrons and so less photo electrons reach the detector per second. However, isnt current the rate of change of charge? If we have more energetic photons, the photoelectrons have larger KE's. They should be arriving more frequently at the detector. Does this not increase the current?

Why does the current necessarily decrease if the wavelength of incident light is decreased, with the sources intensity fixed. Does the decrease in photoelectrons produced win over the increased rate of arrival? Is there any mathematical model for this?

• Bill N over 4 years
You should edit your title. "increasing decreasing" doesn't make sense.
• Bill N over 4 years
Where have you gotten your information about the photoelectric effect? The second half of the first sentence is totally wrong. Your assumption in the second paragraph is wrong.
• Vishal Jain over 4 years
What part? If you change the wavelength of your light, your photons carry more energy, you are still supplying the same amount of energy per second, so you will have less photons being emitted from your light source per second. As the number of released photo electrons ( electrons emitted from the metal due to absorbing a photon ) is proportional to the number of photons reaching them per second, which is now lower, will also reduce. Hence less photo electrons will reach the detector per second.
• anna v over 4 years
This makes it clear hyperphysics.phy-astr.gsu.edu/hbase/mod2.html . It is the maximum photo electron energy that is plotted to show the work function. it is derived from the current as stated.
• John Rennie over 4 years
• Vishal Jain over 4 years
I was asking about the photo electrons, or the electrons released by the photons that are incident on them. My question was just asking that since the photons will be more energetic, they will release electrons with a larger KE, since the work function will not be affected. If electrons travel fasters, doesn't this increase the current detected since current is rate of charge flow wrt time.