# Millikan’s photoelectric experimentWatch

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#1
This is in reference to Millikan’s photoelectric experiment.

I couldn't attach the photo, so, well it's line any other Millikan photoelectric experiment except for that there is a potential divider in the curcuit.

The book says:
The potential divider is now adjusted until none of the electronsreaches the collector. We can now use the law of conservationof energy to find the KE of the fastest electrons.

Loss of KE = gain in electrical PE
0.5m v2=V e
So maximum KE, KEmax = Ve

But then won't V be zero because none of the electrons reach the collector?
0
5 years ago
#2
This is in reference to Millikan’s photoelectric experiment.

I couldn't attach the photo, so, well it's line any other Millikan photoelectric experiment except for that there is a potential divider in the curcuit.

The book says:
The potential divider is now adjusted until none of the electronsreaches the collector. We can now use the law of conservationof energy to find the KE of the fastest electrons.

Loss of KE = gain in electrical PE
0.5m v2=V e
So maximum KE, KEmax = Ve

But then won't V be zero because none of the electrons reach the collector?

They don't reach the collector because they have lost all their kinetic energy in moving through that pd.
An electron can be accelerated or decelerated by a field.
In both cases the kinetic energy gained (when accelerated) or lost (when decelerated) is equal to Ve, as you say.
In Millikan's experiment, you equate the energy they had at the start (max ke) to the energy lost moving through the field.
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#3
(Original post by Stonebridge)
They don't reach the collector because they have lost all their kinetic energy in moving through that pd.
An electron can be accelerated or decelerated by a field.
In both cases the kinetic energy gained (when accelerated) or lost (when decelerated) is equal to Ve, as you say.
In Millikan's experiment, you equate the energy they had at the start (max ke) to the energy lost moving through the field.
Thanks!
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