Kinetic energy of the photoelectrons

AI Thread Summary
The discussion focuses on calculating the largest kinetic energy of photoelectrons generated by ultraviolet light with a wavelength of 200 nm incident on a metal with a work function of 3.0 eV, under a +1.0 V potential. The initial calculation used the formula Kmax = hf - phi, yielding a kinetic energy of 5.139 x 10^-19 J without considering the applied voltage. Participants emphasize the need to modify this kinetic energy calculation to account for the potential, which affects the energy of the released photoelectrons. The correct approach involves adjusting the kinetic energy based on the effect of the +1.0 V potential on the electrons. Understanding this adjustment is crucial for accurately determining the maximum kinetic energy of the photoelectrons.
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Homework Statement


A beam of ultraviolet light with wavelength of 200 nm is incident on a metal whose work function is 3.0 eV. Note that this metal is applied with +1.0 V with respect to the ground. Determine the largest kinetic energy of the photoelectrons generated in this process.

Homework Equations


Kmax = hf - phi

Kmax = q*|V|

The Attempt at a Solution



I am not entirely sure what the statement about the +1.0 V implies and hence I simply used the formula
Kmax = hf - phi . Substituting the appropriate values gives Kmax = 5.139*10^-19 J . Am I on the correct path ?
 
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You got the result in the absence of the applied potential. You must now modify the kinetic energy to take into account the potential.
 
DrClaude said:
You got the result in the absence of the applied potential. You must now modify the kinetic energy to take into account the potential.
how do you take into account the potential?
 
whatphysics said:
how do you take into account the potential?
By calculating its effect on the released photoelectron.
 

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