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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?

Vishal Jain
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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.

The Photon
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Let say we are projecting "x" number of golf balls per min. at some standing beer cans. Each ball has the minimum energy necessary to knock over a can. Any ball with extra energy will knock the can it hits farther. Now lets replace the golf balls with an equal number of marshmallows. None of the marshmallows has enough energy to knock over a can. What would happen if we started replacing golf balls with marshmallows?

Metaman
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