0
$\begingroup$

In photoelectric effect if we look at two different cross sections inside the tube at saturation current. Then the average drift velocity will be different as the electrons are accelerated by external battery.

So does this mean average current is different at different cross sections. According to me it (current) should remain the same at both cross section. But average drift speed is different. So by $$I=neAv_d$$ Either n should be different at different cross sections or i should be different. But if n is different is there any explanation why does electron density change.

Note:- I am assuming cross sectional area constant through the tube

$\endgroup$
1
$\begingroup$

Either n should be different at different cross sections or i should be different. But if n is different is there any explanation why does electron density change.

The current should be the same everywhere- otherwise there would be some charge accumulation, which would not be sustainable.

The number density, n, is decreasing because, as the electrons accelerate, they spread out, like cars on a highway after a toll booth.

$\endgroup$
  • $\begingroup$ but the no of electrons entering one part equals the no entering other part same with no of electrons leaving. I am actually getting little idea that acceleration will change n. But my mind is still fully not convinced. $\endgroup$ – ATHARVA Sep 19 '18 at 18:45
  • 1
    $\begingroup$ @ATHARVA Imagine a bunch of cars moving at $1$m/s, at $10$ m intervals. First car crosses the start line and accelerates. Second car crosses the start line $10$ sec later and accelerates, etc. The cars will be crossing the finish every $10$ seconds, but the distance between them will be greater, since their speeds are greater. So the cars move faster, with lower density, but we still have one car crossing the start line and finish line every $10$ seconds. $\endgroup$ – V.F. Sep 19 '18 at 19:23
  • $\begingroup$ @No problem, glad to help. $\endgroup$ – V.F. Sep 20 '18 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.