Finding the wavelength from the work function in photoemission

AI Thread Summary
The discussion focuses on calculating the maximum wavelength for photoemission based on the work function of a metal. The relevant formula for maximum electron energy is E = hf - σ, where E is the maximum energy, h is Planck's constant, and σ is the work function. To find the maximum wavelength, it is derived that λ = hc/(E + σ), with the understanding that the maximum wavelength corresponds to the minimum energy needed to overcome the work function. The calculation shows that the maximum wavelength for a work function of 2.30 eV is approximately 5.4 x 10^-7 m. This indicates that if the wavelength exceeds this value, photoemission will not occur.
EmmaLemming
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Homework Statement



a) Write down the formula that relates the maximum electron energy, Emax. to the
frequency of the incident light in the photoelectric effect.

b) Calculate the maximum wavelength of light for which photoemission occurs for light
incident on a metal whose workfunction is 2.30 eV.


2. The attempt at a solution

a) E = hf - σ

where E is the maximum energy, h is Planck's constant and σ is the work function.

b) E = (hc/λ) - σ

where λ is the wavelength and c is the speed of light in a vacuum

∴ λ = hc/(E+σ)

However I'm not given the energy so how am I to calculate λ?

Is there another way to calculate E?
I looked in my textbook but found nothing relevant :(
 
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EmmaLemming said:

Homework Statement



a) Write down the formula that relates the maximum electron energy, Emax. to the
frequency of the incident light in the photoelectric effect.

b) Calculate the maximum wavelength of light for which photoemission occurs for light
incident on a metal whose workfunction is 2.30 eV.2. The attempt at a solution

a) E = hf - σ

where E is the maximum energy, h is Planck's constant and σ is the work function.

b) E = (hc/λ) - σ

where λ is the wavelength and c is the speed of light in a vacuum

∴ λ = hc/(E+σ)

However I'm not given the energy so how am I to calculate λ?
The problem statement asked for "maximum wavelength". Don't forget, maximum wavelength corresponds to minimum frequency; thus minimum energy.

What happens if the wavelength is so large that hc/λ = σ ? :wink: What's the wavelength that would cause photoemission to occur, but without any leftover energy?
 
Okay, so I assume that minimum energy is 0 and then write,

hc/σ = λ

I get, λ = 5.4 x 10-7 m

That seems reasonable , right?

I don't know the answer to your last question.
 
EmmaLemming said:
Okay, so I assume that minimum energy is 0 and then write,

hc/σ = λ

I get, λ = 5.4 x 10-7 m

That seems reasonable , right?
'Looks right to me. :approve:
I don't know the answer to your last question.

What I mean is that if the light's wavelength is too big, there is not enough energy to overcome the work function and photo-emission doesn't take place at all. As you decrease the wavelength, the energy increases. Decrease it enough and photo-emission begins to occur. Decrease the wavelength even more and the leftover photon energy can end up becoming electron energy.

So the maximum wavelength is the wavelength where the maximum electron energy is zero.
 
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