Calculating Kinetic Energy of Photoelectrons in Ultraviolet Metal Interaction

In summary, the question is asking for the largest kinetic energy of photoelectrons generated when a beam of ultraviolet light is incident on a metal with a work function of 3.0eV, while the metal is applied with +1.0V with respect to the ground. The formula KEmax = hf - work function can be used to solve this, but the question remains whether the applied voltage affects KEmax. However, this voltage is not relevant to the calculation, as the stopping potential is determined by the energy difference between the incident light and the work function, not the applied voltage.
  • #1
JakeP
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< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

Hello guys, i need some help in clarifying a question.

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

Using the formula of KEmax = hf - work function, i can come up with an answer for it. However, from the question, it states the metal is applied with +1.0V with respect to the ground;Does this Voltage affect the KEmax? Because if i were to find stopping potential Vs using E=q|Vs|, Vs will result in approximately 6.211V, taking q to be 1.6 x 10^-19. So i believe that this voltage across the metal is redundant in the calculation.

Kindly advice.

Thank you guys!
 
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  • #2
JakeP said:
Using the formula of KEmax = hf - work function, i can come up with an answer for it. However, from the question, it states the metal is applied with +1.0V with respect to the ground;Does this Voltage affect the KEmax? Because if i were to find stopping potential Vs using E=q|Vs|, Vs will result in approximately 6.211V, taking q to be 1.6 x 10^-19.
So if the applied potential was 6.211 V instead of 1 V, what would you answer for KEmax?
 
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Likes xTyler
  • #3
Won't it be the same as KEmax?
 
  • #4
I'm quite confused, is it alright to assume that the applied +1V on the metal is also the stopping voltage? And does this applied voltage on the metal makes it harder for the electrons to break the bond from metal -> Vacuum -> Being ejected ?
 
  • #5
xTyler said:
I'm quite confused, is it alright to assume that the applied +1V on the metal is also the stopping voltage?
No. As said, the stopping potential is 6.211 V.

xTyler said:
And does this applied voltage on the metal makes it harder for the electrons to break the bond from metal -> Vacuum -> Being ejected ?
No. It changes what happens to the freed electron.
 
  • #6
JakeP said:
Won't it be the same as KEmax?
What is the definition of the stopping potential?
 
  • #7
I had this question figured out. Thanks!
 

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