Calculating Lowest Frequency of EMR to Initiate Photocurrent

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The discussion focuses on calculating the lowest frequency of electromagnetic radiation (EMR) required to initiate a photocurrent when UV radiation at 250 nm liberates electrons with a maximum kinetic energy of 1.00*10^-19 J. The initial calculations presented are correct, leading to a final frequency of 1.05*10^15 Hz. However, there is a critique regarding the grouping of the work function (Ø) and kinetic energy (K) as total energy, emphasizing that Ø represents the negative change in energy for the electron. The conversation also touches on the conservation of energy principles related to this calculation. Overall, the calculations are validated, but the conceptual understanding of energy changes is questioned.
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Can some please review this question and tell me if the working looks correct? Please advise of any mistakes i may have made. Thanks

Q: UV radiation (wavelength = 250 nm) falls on a metal target, and electrons are liberated. If the maximum kinetic energy of these electrons is 1.00*10^-19 J, what is the lowest frequency EMR that will initiate a photocurrent on this target?

A: E = Ø + K = h*(c/λ)

Therfore Ø = ((h*c)/250*10^-9) - 1*10^-19
Which = 6.96*10^-19 J
= h*f

Therefore f = 1.05*10^15 Hz

Is this what i am supposed to do? Thanks for any help! :smile:
 
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Your final answer is correct.

I find it peculiar that you group " Ø + K " together
- it's NOT the total Energy " E " of any configuration!
Rather, it's the NEGATIVE Energy change of the electron
(- E_initial + E_final) ... so it does equal E_photon,initial.

Most of us learn conservation laws as
"final = initial + change" or "change = final - initial" ,
once in a while as "in - out = change".
Is there a word for this "negative change"?
 
This is what my demonstrator told me to use?
 
Do you see how it comes from E conservation?
Ø is just the negative of the electron's initial PE, etc.
 

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