Quantum Physics: Work Function and Electric Potential

In summary, a metal surface with a work function of 3.0eV and an electric potential 5V lower than infinity is illuminated with 200nm wavelength light. The maximum kinetic energy of the emitted photoelectrons can be determined using the equation K.Emax= hf - Φ, where f=c/λ. However, to accurately calculate the maximum kinetic energy, the impact of the electric potential must also be taken into account. This would cause the maximum kinetic energy of the photoelectrons to increase, as they would be attracted to the metal surface and require more energy to escape. To find the energy required to escape the surface, the work function must be converted from eV to joules.
  • #1
JohnGaltis
18
0

Homework Statement


Metal Surface is illuminated with 200nm wavelength light. Work Function of this metal is 3.0eV and its electric potential is 5V lower than a point of infinity.

Determine max K.E of photoelectrons, which are just emitted from the work surface.

Homework Equations


K.Emax= hf - Φ
f=c/λ
Not sure what other equations.

The Attempt at a Solution


I found f first by using f=c/λ. This way, I got a value of max K.E. But it's a really high mark question so it got to be more than just subbing in values.

I think I haven't yet include the impact the electric potential would have on the photoelectrons. It would cause the max K.E of photoelectrons to increase, right? Since it would be attracted to the metal surface and would need more energy to escape.

What equation would allow me to find the energy required to escape the surface?
 
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  • #2
Show your calculations. did you convert eV into joules?
 

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