Quantum Theory - The Photo-Electric Effect

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Homework Help Overview

The discussion revolves around a problem related to the photoelectric effect, specifically involving the interaction of white light with a potassium surface. Participants are tasked with calculating the maximum kinetic energy of ejected electrons and identifying the frequency range for which no electrons are emitted, given the work function of potassium.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply Planck's equation to find the maximum kinetic energy of electrons but encounters difficulties with unit conversions and calculations. Some participants suggest clarifying the relationship between photon energy, work function, and kinetic energy. Others raise questions about the assumptions regarding frequency ranges and the conditions for electron ejection.

Discussion Status

Participants are actively engaging with the problem, offering clarifications on the concepts involved, particularly the work function and energy calculations. There is a recognition of the need for careful unit management and the implications of frequency on electron ejection, but no consensus has been reached on the specific calculations or interpretations.

Contextual Notes

Participants note the importance of using consistent units when calculating energy and emphasize the need to consider both the minimum and maximum frequencies in relation to the work function. The original poster's approach to the problem is acknowledged, but there are indications that further exploration of the concepts is necessary.

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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.
 

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