Relating Work Function and Wavelength

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Homework Help Overview

The problem involves determining the longest wavelength of light that can cause electron emission from barium, given its work function of 2.48 eV. The subject area pertains to quantum mechanics and the photoelectric effect.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the work function to wavelength using relevant equations but expresses confusion at a certain point. Some participants question the relationship between frequency and wavelength, prompting further exploration of the equations involved.

Discussion Status

Some participants have provided guidance by confirming the relationship between frequency and wavelength, and one participant claims to have solved the problem, suggesting a productive direction. However, there is no explicit consensus on the correctness of the final answer.

Contextual Notes

The discussion includes references to specific equations and constants, indicating a reliance on established physics principles. There is an implicit assumption that participants are familiar with the concepts of work function and the photoelectric effect.

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Homework Statement


Barium has a work function of 2.48 eV. What is the longest wavelength of light that will cause electrons to be emitted from barium?

Homework Equations


W = hf○
hf = W + Ek
E = hf = hc/λ

The Attempt at a Solution


1 eV/1.6e-19 C = 2.48/X
X = 3.97e-19

W=hf○
3.97e-19/6.63e-34f
f=5.99e14 Hz

Now I'm lost...Anybody care to help? :)
 
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So what's the relationship between frequency and wavelength? You've already used it (look at equation 3).
 
hage567 said:
So what's the relationship between frequency and wavelength? You've already used it (look at equation 3).

I think I've figured out the problem using v=fλ, 3.0e8 = 5.98e14λ, λ=5.02e-7m.

Also the answer to your question is that as frequency increases, wavelength decreases.

Would be awesome if you could tell me I'm on the right track, and of course if I'm not, as well.
 
Yes, you are right.
 
Thanks much.
 

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