Photoelectric Effect and ozone layer

AI Thread Summary
The discussion centers on whether direct sunlight can produce photoelectrons from a penny, specifically focusing on the copper coating with a work function of 4.7 eV. The ozone layer blocks most solar radiation below 320 nm, but some 200 nm light reaches the Earth's surface. To determine if photoelectrons can be emitted, the energy of a 200 nm photon needs to be calculated using the formula E = hc/wavelength. The calculated energy per photon at 200 nm is crucial for comparing it to the work function of copper. Understanding this relationship will clarify if photoelectrons can be generated from the sunlight reaching the penny.
fireandwater
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Hi. I hope someone can help. Here is the problem:

[Pennies are made of zinc coated with copper. Copper has a work function of 4.7eV. The ozone layer blocks nearly all solar radiation with wavelength shorter than 320nm, but a very small amount of 200nm light still manages to reach the Earth's surface. Can direct sunlight produce photoelectrons from a penny?]

I converted the 4.7eV into joules: 4.7eV * 1.602e-19J = 7.529e-19J
I found the minimum kinetic energy needed to for the electrons to escape from the copper:
KE = e * V_o => 1.602e-19 * 7.529e-19 = 1.206e-37J

I'm thinking that I need to use the 200nm to find out whether the kinetic energy I calculated will be produced by the sunlight, but I don't see how to do that.
 
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Your reasoning is incorrect. KE = E - W (E is energy per photon, W is work function, KE is max kinetic energy of photoelectron).

Stay in eV, it's easier. Use 4.14 x 10^-15 eVs as h.

Energy per photon is hf =hc/wavelength.

How much energy in a 200 nm photon?
 
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Thank you!
 
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