How Is Maximum Kinetic Energy Calculated in the Photoelectric Effect?

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To calculate the maximum kinetic energy (Ek) of ejected electrons in the photoelectric effect, the equation Ek = (hc/lambda) - W is used, where W is the work function. A magnesium surface with a work function of 3.68 eV is struck by electromagnetic waves with a wavelength of 215 nm. A user initially miscalculated the energy, resulting in an incorrect value of -3.68*10^10 eV instead of the expected 2.10 eV. The error was identified as a mistake in converting the work function from eV to Joules. Correcting the calculation led to the accurate determination of the maximum kinetic energy.
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Homework Statement


A magnesium surface has a work.func of 3.68eV. El.mag. waves with wavelenght 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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