In a photo cell 4 unit photo electric current is flowing,the distance between source and cathod is 4 unit.now distance between source and cathod becomes 1 unit.what will be photo electric current now?
Answers
Answer:
Let's first do the theoretical part. Consider two conducting plates separated by some distance (d), each having equal and opposite charge (q). Then we know that the electric potential between the plates is directly proportional to the distance between the plates. Let this potential be 'V'. This would be the potential that an electron has to overcome when moving across the plates. If now without changing any other parameters of this system, I increase the distance between the plates than the electric potential would also increase accordingly. So, now if an electron had to move across the plates it has to overcome a greater potential. Now let's move to the problem.
The photoelectrons now have to overcome this increased potential. And the photons incident on the plate have to compensate this extra energy need. This implies, we are going to observe photocurrent at a larger frequency of the photons i.e. more energetic photons. Once current is stabilized i.e. saturated, the rest is same as any photoelectric experiment.
Mathematically, kinetic energy of the photoelectron = energy of incident photon - (work function + the potential across the plates).
Answer:
Intensity of photon (means no. Of photo electron emitted )
Intensity inversely proportional to square of distance. Which means on decreasing the distance increase the intensity.
Hence it will become 64.