Finding the wavelength from the work function in photoemission

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SUMMARY

The discussion focuses on calculating the maximum wavelength for photoemission from a metal with a work function of 2.30 eV. The relevant formula derived is λ = hc/σ, where λ is the wavelength, h is Planck's constant, c is the speed of light, and σ is the work function. The calculated maximum wavelength is 5.4 x 10-7 m, corresponding to the condition where the photon energy equals the work function, allowing for photoemission to occur without any excess energy. This establishes a clear relationship between wavelength, work function, and the energy of incident light.

PREREQUISITES
  • Understanding of the photoelectric effect and its equations
  • Familiarity with Planck's constant (h) and the speed of light (c)
  • Knowledge of electron energy concepts in quantum mechanics
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the photoelectric effect equations
  • Learn about the implications of work function in different metals
  • Explore the relationship between frequency and wavelength in electromagnetic radiation
  • Investigate experimental setups for measuring photoemission
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Students studying quantum mechanics, physics educators, and researchers interested in the photoelectric effect and its applications in material science.

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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