What is the work function of the photoelectric material?

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Homework Help Overview

The problem involves determining the work function of a photoelectric material based on the momentum of incident photons and the measured cutoff voltage. The context is within the realm of photoelectric effect and energy calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between photon momentum and energy, and how to correctly apply the equations related to the photoelectric effect. There are attempts to clarify the conversion of units for consistency in calculations.

Discussion Status

Some participants have provided guidance on unit conversions, and there is acknowledgment of an error in the initial calculations. The discussion reflects a collaborative effort to refine the approach to finding the work function, although no consensus on the final value has been reached.

Contextual Notes

There is an emphasis on ensuring that units are consistent throughout the calculations, particularly regarding the speed of light and momentum. The participants are navigating through the implications of their calculations on the work function.

ikihi
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Homework Statement



Photons with momentum p= 7.88 x 10-18 strike a photoelectric material which is in the configuration shown in the figure. The cutoff voltage is measured to be Vstop= 1.75 V. What is the work function of the photoelectric material?

Homework Equations



p = Ephoton / c
E0 = (c ⋅ p) - (Vstop ⋅ e)

The Attempt at a Solution



E0 = (3.00 x 108 m/s)(7.88 x 10-18 (ev ⋅ s)/nm) - (1.75 V ⋅ 1.602 x 10-19 J) = 2.36 eV
 

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Last edited:
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you should convert the speed of light to nm/s to match the momentum unit (ev.s/nm)
 
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Likes   Reactions: ikihi
patric44 said:
you should convert the speed of light to nm/s to match the momentum unit (ev.s/nm)

Thanks for catching that error. The new answer I got is 2.36 eV
 
ikihi said:
Thanks for catching that error. The new answer I got is 2.36 eV
you are welcome.

thats still the energy of the photon
subtract that from the ke
 

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