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A photocell is lit by a violet lamp. The lamp is dimmed until it sends one quarter of its light…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A photocell is lit by a violet lamp. The lamp is dimmed until it sends one quarter of its light, with the same colour. Say what happens to (a) the saturation current, (b) the stopping potential, (c) the largest kinetic energy of the photoelectrons and (d) the delay before the first electron leaves.
Need a reminder? The key ideaShow
In a photocell the saturation current is in proportion to the intensity. The largest kinetic energy of the electrons is , and it depends on the frequency, not on the intensity. There is no emission below the threshold frequency, and no delay above it.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. The colour is the same, so the frequency and the energy of each photon are the same. Only the number of photons each second changes.
Step 1. The saturation current. It is in proportion to the intensity. One quarter of the light gives one quarter of the current: of the old value.
Step 2. The stopping potential. has no intensity in it. The stopping potential stays the same, so the fraction is .
Step 3. The largest kinetic energy. It follows the stopping potential, so it does not change.
Step 4. The delay. There is still none. Each photon gives its energy at once, however few photons arrive.
Answer: (a) one quarter, (b) unchanged, (c) unchanged, (d) still no delay.
Watch out. A fainter light does not make slower electrons. It makes fewer of them.
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