Frequency of light to eject electrons from gold

AI Thread Summary
To eject electrons from a gold surface with a maximum kinetic energy of 6.48×10(−19) J, the frequency of light must be calculated based on the work function of gold, which is 4.58 eV. The energy of a photon is related to its frequency, and the total energy required is the sum of the work function and the kinetic energy. Converting the work function from eV to joules is necessary for accurate calculations. The correct approach involves using the formula for maximum kinetic energy to solve for frequency. The discussion highlights the importance of understanding energy conversions in photon interactions with metals.
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What frequency of light must be used to eject electrons from a gold surface with a maximum kinetic energy of 6.48×10(−19)J ?

The work function of gold is 4.58 eV.

f=
 
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You have to at least attempt an answer to a homework question.
Whats the energy of a photon, in terms of frequency?
 


I used the formula for K(max) and solved for f and got 6.91 x 10(33)...and that not the right answer
 


So you need to supply enough energy to overcomethe work function and then supply the KE.
So convert the work function into joules and add to the ke.
Then you just need the energy of aphoton in terms of frequency.

oops - you didn't post in the HW forum, so didn't get the list of things to do.
 


Nevermind...I figured it out
 
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