Experimental setup & saturation current: photoelectric effect | Dual nature of light | Khan Academy

Let's explore how to measure the no. of photoelectrons emitted per second experimentally, in photoelectric effect. We will also draw a graph of the photocurrent vs anode potential.
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Created by Mahesh Shenoy

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Thank you for making the graph! It helped tremendously!

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[@kabirsethi2608]

Very well explained sir

[@halumgrr]

As we increase the voltage, the collector receives more electron per second and also they're being attracted with much greater force than before(because of more voltage), so the electrons should be received and circulating in the circuit quicker than before, shouldn't that result in higher circuit current regardless of the fact that we cannot increase the no. of electrons received per second anymore? ( therefore no thing as saturation current). What point am I missing?

[@jerven1128]

i think that, even though the electrons are circulating quicker than before in the circuit, the current should still depend on how fast they travel through the air gap between the emitter and the collector, for ex if 100 electrons go from the +ive terminal to the emitter and the incident photons just have enough energy to emit 10 photoelectrons from the emitter, it will take 10 seconds for them to pass that gap and get to the -ive terminal, thus restricting the current to 10e/s even though the potential was enough to move 100 electrons.
that sort of makes sense i guess????

[@rishavjain5087]

Path is broken dude ..the circuit is open...when u got the first batch of e- , they returned to the Zn plate....and can only pop out if they got energy from photons... Experiment's pretty full proof

[@ManojKumar-xk1wp]

we are increasing the potential the electrons speed also increases ,since i=nq/t u are saying n becomes constant but that does not mean q/t is constant then why current is getting saturation value

[@flip726]

Why does increasing the voltage increase the current here? Is it because you can treat it as a circuit and Ohm's law applies?

[@arhamdugar396]

i have a question
by increasing the voltage, we have decreased the work function and increased the kinetic energy of the emitted photoelectrons so why does photocurrent become constant and does not increase as velocity of electrons will increase with increase in potential difference?

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great explanation

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[@tendi.rutsate]

Thank you so much wow!

[@dravikumar4401]

Here, in saturation current topic we have seen photon ejected almost all the electrons
Then it must have efficiency like more than 90% right but why the efficiency of photoelectric effect is nearly 0.1%?

[@selin4393]

Thank you

[@sitcomshorts8749]

If electrons are collected using voltage that would result in more electron that is to say the pulling out of electron would not be impartial to light now.

[@divy_maurya]

Wow, amazing

[@okellokennedy6048]

By setting up a pd (the collector being positive with respect to emitter)

[@sainidhikarthikeyan2291]

Actually speaking, saturation current/e gives the number of photoelectrons/sec right?

[@VG__]

6:25 saturation current

[@xiaoruiwang4545]

excellent

[@naveenmydur9724]

Well explained

[@electroniciit9304]

what if the photoelectrons are just coming out with no kinetic energy . will the system still detect photoelectrons .

[@shahusardar2762]

Thank youu

[@muneerkhan1176]

Collector is which metal sir.

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