Work Function and Wavelength Relationship in Photoelectric Effect

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SUMMARY

The discussion centers on the relationship between work function and wavelength in the context of the photoelectric effect. The maximum kinetic energy of photoelectrons is initially 2.8 eV, which decreases to 1.1 eV when the wavelength is increased by 50%. The equations used include K.E.max = hc/(lambda) - work function, leading to two equations that can be solved simultaneously to find both the work function and the initial wavelength. The user successfully identifies the need for two equations to solve for the two unknowns.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with the equation K.E.max = hc/(lambda) - work function
  • Basic knowledge of energy units (eV and Joules)
  • Ability to solve simultaneous equations
NEXT STEPS
  • Research the concept of work function in photoelectric materials
  • Learn about the Planck constant (h) and its role in quantum mechanics
  • Study the implications of wavelength changes on kinetic energy in photoelectric experiments
  • Explore advanced topics in quantum physics related to electron emission
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Students and professionals in physics, particularly those studying quantum mechanics and the photoelectric effect, as well as educators looking for practical examples of energy and wavelength relationships.

kingwinner
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1) The maximum kinetic energy of photoelectrons is 2.8eV. When the wavelength of the light is increased by 50%, the maximum energy decreases to 1.1 eV. What are the "work function" of the cathode and the "initial wavelength"?
K.E.initial=K.E.max=2.8eV=4.48x10^-19J
lambda.final=1.5*lambda.initial
Final energy = 1.1 eV (<---is this purely kinetic energy? why?)

K.E.max = hc/(lambda.initial) - work function
But now I have 2 unknowns: work function and lambda.initial, what can I do to solve for both?

I am stuck here...

Does anyone have any idea or insight? Any help is greatly appreciated.
 
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Can someone please help me? I am sure that there are a lot of genius here...
 
kingwinner said:
K.E.max = hc/(lambda.initial) - work function

Make that:

K.E.max.initial = hc/(lambda.initial) - work function

Then how about

K.E.max.final = hc/(lambda.final) - work function

as a second equation?
 
Thanks for your help! I got it!
 

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