How to calculate the work function of a metal?

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Homework Help Overview

The discussion revolves around calculating the work function of a metal when illuminated by light of a specific wavelength. The original poster presents an equation involving kinetic energy and attempts to derive the work function from given values.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between energy, kinetic energy, and work function, questioning unit conversions and the accuracy of the wavelength used in calculations. There are discussions about the necessity of converting between joules and electron volts.

Discussion Status

The discussion is ongoing, with participants providing guidance on unit conversions and checking the calculations. There is an emphasis on ensuring the correct application of formulas and the importance of consistent units.

Contextual Notes

There is uncertainty regarding the required units for the work function, as the original poster notes that the book does not specify this information.

madayag1
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HI I am doing this equation : 14. When a certain metal is illuminated at 3.50x10^2nm the maximum kinetic energy of the ejected electrons is 1.20 eV. Calculate the work function of the metal.
I did this:
E=Ek+W
W=E+Ek
W=hc/Lambda -Ek
=(6.63E-34J.s)(3.00E8m/s)/3.50E2m)-1.20eV
=-1.2

That answer is wrong the correct one is 2.35 eV. what did i do wrong, PLS help me ty and show me what I should of done so i Dont repeat this mistake again.
 
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madayag1 said:
W=hc/Lambda -Ek
=(6.63E-34J.s)(3.00E8m/s)/3.50E2m)-1.20eV

Hello madayag1
Look again at the figure you've used for the wavelength - you forgot the nm.
Also, the first part of the calculation is in J, but you've subtracted a value in eV from it.
 
Did you convert the units, Joule to eV?
 
The result of 'W=hc/Lambda " is in joules is it not? You cannot subtract ev from joules wiothout a unit conversion.

Also, you have ente3red your wavelength incorrectly.
 
srry it should be W=( (6.63E-34J.s)(3.00E8m/s) / (3.50E2nm) ) - 1.9E-19J = -1.9E-19
still the wrong answer tho.
 
madayag1 said:
srry it should be W=( (6.63E-34J.s)(3.00E8m/s) / (3.50E2nm) ) - 1.9E-19J = -1.9E-19
still the wrong answer tho.

You've written the nm part for the wavelength, but did you include it in the calculation?
 
Check your math. The answer is not negative. Also, what units are reque3sted?
 
book doesn't say what units is requested
 
I think work function is normally in ev. Check your math however.
 

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