Finding the wavelength from the work function in photoemission

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Homework Help Overview

The discussion revolves around the photoelectric effect, specifically relating to the calculation of maximum wavelength for photoemission given a metal's work function of 2.30 eV. Participants are exploring the relationship between energy, frequency, and wavelength in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula relating maximum electron energy to frequency and attempt to derive the maximum wavelength. Questions arise regarding the lack of given energy and how to calculate wavelength without it. Some participants consider the implications of maximum wavelength corresponding to minimum energy.

Discussion Status

There is an ongoing exploration of the relationship between wavelength and energy, with some participants suggesting that the maximum wavelength occurs when the energy is just sufficient to overcome the work function. Guidance is provided regarding the interpretation of maximum wavelength and its relation to minimum energy.

Contextual Notes

Participants note that the problem requires understanding the conditions under which photoemission occurs, particularly the significance of the work function and how it relates to the energy of incident light.

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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