Hey can someone explain stopping voltage to me and its relevant formula (max KE = qv) or something
Thanks!
Hey I just learnt this

Stopping voltage (V
s) - voltage required to stop current.
Remember when you were investigating the photoelectric effect at school (or if not, you can still read the following explanation and hopefully it'll make sense) you had a set up like this?

So the cathode here is the stopping electrode. We are able to make the cathode more and more negative, by increasing the voltage supplied by the external supply. This then begins to repel the photoelectrons (because like charges repel) and eventually, the flow of charge will be stopped. Hence the name stopping electrode. When it is high enough, the voltage will be such that electrons will be able to just reach the cathode but not flow into the external circuit. When this happens, the voltage being applied is known as the stopping voltage i.e. V = V
s.
Now onto making sense of: E
k(max) = qV
sRecall that

.
E
k(max) refers to the kinetic energy of the fastest moving photoelectron. The electrons which are furthest away from the nucleus are held the weakest and so require the least amount of energy to be released when a photon strikes the metal surface. When an incoming photon of light (carrying E = hf) has sufficiently high frequency, and so energy, these outermost electrons will be emitted with the greatest kinetic energy.
Spoiler
For further clarity and so, to continue: This is because the photon's energy is larger than the energy required to overcome the electrostatic attraction acting between the electron and nucleus. Because not all of the photon's energy is absorbed in the process of liberating the electron (it only needs enough to overcome the attractive force) it still has some energy left over which is then converted into kinetic energy. The outermost electrons will have the greatest amount of energy left after interacting with the incoming photon of light because the attractive force is smallest and so the energy required to overcome this force is the smallest.
At the stopping voltage, V
s, only the fastest moving electrons are reaching the (stopping) anode (note, they're just reaching, not flowing into external circuit so current reading is 0), because all the slower moving ones don't have enough E
k to overcome the large potential difference (i.e. they are repelled). These emitted electrons (photoelectrons) have the highest kinetic energy as:
} \propto v)
} = qV{s})
Really hoped that helped convey a more conceptual explanation - it really puts the formulas into context and saves you the time of memorising the formula (plus it'll have more meaning to you too)
Let me know if you have any more questions
If you need me to, i'm happy to also explain the formula:
} = hf - \phi )