How Is Maximum Kinetic Energy Calculated in the Photoelectric Effect?

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Homework Help Overview

The discussion revolves around calculating the maximum kinetic energy of electrons ejected from a magnesium surface when exposed to electromagnetic waves, specifically in the context of the photoelectric effect. The original poster provides a work function value and a wavelength, aiming to find the maximum kinetic energy in electron volts.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to convert units and apply the photoelectric equation but encounters a negative value for kinetic energy. Other participants request to see the full calculation to identify errors and suggest checking the power of ten for the work function in Joules.

Discussion Status

Participants are actively engaged in identifying the source of the original poster's error in calculations. Guidance has been offered to check specific aspects of the calculations, indicating a productive direction in the discussion.

Contextual Notes

The original poster's calculations involve unit conversions from eV to J and nm to m, which may be contributing to the confusion. The work function is expressed in electron volts, and there is a need to ensure proper conversion to Joules.

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Homework Statement


A magnesium surface has a work.func of 3.68eV. El.mag. waves with wavelength of 215nm strike the surface and eject electrons. Find the max Ek of the ejected electrons in eV.

Homework Equations


E=Ek-W <=> Ek=(hc/lambda) - W

The Attempt at a Solution


I convert eV to J and nm to m but i get -3.68*10^10eV, but the answer is suppoused to be 2.10eV.
I even re did it but got the same answer, where em I going wrong?

Thanks
 
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Hello.
In order to see where you are going wrong, we need to see your entire calculation.
 
Ek=(hc/lambda)-W

((6.63*10^-34)*(3*10^8))/(2.15*10^-7) - (5.888*10^-9)

(9.25*10^-19) - ( (5.888*10^-9) = -5.887*10^-9 J / (1.6*10^-19) = -3.68*10^10eV
 
mss90 said:
Ek=(hc/lambda)-W

((6.63*10^-34)*(3*10^8))/(2.15*10^-7) - (5.888*10^-9)

(9.25*10^-19) - ( (5.888*10^-9) = -5.887*10^-9 J / (1.6*10^-19) = -3.68*10^10eV

Check the power of 10 for the work function expressed in Joules.
 
ah ye, there we go. thanks :)
 

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