# Quantum Theory - The Photo-Electric Effect

Hi, I got this question and I have spent a while trying to crack it but I'm still stuck.

White light, with frequencies ranging from 4.00 x10^14 Hz to 7.85x10^14 Hz, is incident on a potassium surface. Given that the work function of potassium is 2.24 eV, find the following values.
(a) the maximum kinetic energy of electrons ejected from this surface
eV
(b) the range of frequencies for which no electrons are ejected
minimum -
maximum -

For part a) I used the formula e=hv, as described by Planck. For Planck's Constant h I used a value of 6.62x10^-34 and for my frequency v I used the highest value in the range i.e. 7.85x10^14 in order to yield the highest K.E. but I came up with an answer of 5.205x10^-19 which is wrong...can someone point me in the right direction? Thanks.

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The work function is the energy needed to eject an electron. So the photons incident on the surface will use their energy first to knock out an electron (i.e. the work function) and the rest of the energy will go into kinetic energy of the electron.

So for (a), work out the energy of the photon, as you have done, then subtract the work function and what is left is the kinetic energy of the electron.

Beware to use the right units - if your frequency is in Hertz and you Planck constant in Joules per second (ie. the number you quoted) then your energy of the photon will be in Joules. You need to convert to eV using 1 eV = 1.602 176 53 (14)×10^−19 J.

Ah that clears things up, that's great thanks. OK I managed to do part b)i) it was deceptively simple, the answer was just the lowest value in the frequency range given (i.e. 4.00 x10^14 Hz). But I don't think that the answer to b)ii) will just be the highest value in the frequency range...

You need to think what the KE of an would electron be, if it has not been ejected from the potassium. Then, what would be the f which gave this KE, using the relationship between E of photon, KE of electron and work function.