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Photoelectric effect/finding kinetic energy.

  • Thread starter xxabr
  • Start date
  • #1
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Homework Statement


A photoelectric effect experiment uses a light of frequency 8x1014Hz and a metal with a work function of 1.2eV. A 1.5V stop potential is also applied. What is the maximum kinetic energy of the emmited electrons?

Homework Equations


I was thinking about using Ek= hf-W

The Attempt at a Solution


h= 6.63x10-14
W= 1.2eV
f= 8x1014

If I were to use that equation I would get:
Ek=hf-W
= (6.63x10-14)(8x1014) - (1.2)
= 53.04 - 1.2
= 51.84 J

* I have never used eV before so I'm not sure if I had to do anything with it...
Also it gives me the stop potential and I'm not sure if or how I'm supposed to use it.
Please help, thanks. (:
 

Answers and Replies

  • #2
Redbelly98
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Homework Statement


A photoelectric effect experiment uses a light of frequency 8x1014Hz and a metal with a work function of 1.2eV. A 1.5V stop potential is also applied. What is the maximum kinetic energy of the emmited electrons?

Homework Equations


I was thinking about using Ek= hf-W

The Attempt at a Solution


h= 6.63x10-14
Look up h again. What number should go in the exponent, 10???, and what are the units that go with h?
W= 1.2eV
f= 8x1014

If I were to use that equation I would get:
Ek=hf-W
= (6.63x10-14)(8x1014) - (1.2)
= 53.04 - 1.2
Watch the units. What units go with "53.04"? What units go with "1.2"?
= 51.84 J

* I have never used eV before so I'm not sure if I had to do anything with it...
It has a lot to do with it. You need to decide what energy units you are going to use -- either Joules or eV -- and convert as necessary.

Also it gives me the stop potential and I'm not sure if or how I'm supposed to use it.
You can think of the energy associated with the stop potential as adding to the work function, reducing the kinetic energy of the electron.
 
  • #3
14
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Oops yeah, I made a mistake. h is 6.63x10-34Js.

And I got it.

Thank yoou
 
Last edited:

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