Photoelectric effect and mercury

AI Thread Summary
The photoelectric effect does not occur when mercury is illuminated with UV light at a wavelength of 300 nm, as this wavelength exceeds the cutoff of 250 nm, indicating lower energy and frequency. The relationship between wavelength and energy shows that longer wavelengths correspond to lower energy, which is insufficient to eject electrons. For calculating the cutoff frequency of an x-ray tube operating at 44 kV, the correct conversion from kV to eV is essential, as 1 kV equals 1,000 eV. The formula used for frequency, which involves the work function and Planck's constant, may yield incorrect results if the energy conversion is mishandled. Understanding these principles is crucial for accurately determining the conditions under which the photoelectric effect can occur.
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Homework Statement



a. Does the photoelectric effect take place if mercury is illuminated with UV light with a wavelength λ = 300 nm? The cutoff wavelength for mercury is 250 nm.


Homework Equations



ft= WF/h(planks constant)
E=h(f)

The Attempt at a Solution



im not sure how to start this because the work function is not given and neither is the frequency i know that i can get the work function if i have the frequency by multiplying it with planks constant, but I am not sure how to get it with just the wavelength.
 
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You are given the cut off wavelength, and wavelength and frequency are related.
But you don't need to calculate anything - just a bit of reasoning.

Your light is longer wavelength than the cut off, is this higher or lower energy?
Which would it have to be to eject an electron?
 
o ok, so this means that the energy is lower because the frequency would be lower, so photoelectric effect would not take place. is that right?
 
Correct, longer wavelength = lower frequency = lower energy
So no emmission
 
alright thanks
 
How would you calculate the cutoff frequency of an x-ray tube operating at 44kV? I am using frequency = work function/h. I have (44000eV x 1.602E-19J)/1.626E-34 but my answer is wrong. I'm thinking that I converted kV to eV incorrectly, but I'm not sure.
Thanks.
 
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