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Potential Difference

  • Thread starter a.a
  • Start date
  • #1
a.a
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Homework Statement



How would we find the potential difference given the frequency of the incident light and the kenetic energy of an the emitted electrons?

Homework Equations



E= hf-W
V= w/q
e= 1.602*10^-19

The Attempt at a Solution



Given: f= 7.4*10^15 Hz
KE= 1.9eV
Required: potential difference

Solution: KE= 1.9eV * 1.602*10^ -19 J/eV = 3.0438*10^-19

V= 3.0438*10^-19/ 1.602*10^ -19 = 1.9

The answer is supossedd to be 2.7, I dont think I'm doing this the correct way.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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The stopping potential would be 1.9V, not sure what p.d. they are asking for though.
 
  • #3
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
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Use the 1st equation you wrote,

E = hf - W

To get W, you need to calculate E and hf.
 
  • #4
a.a
127
0
Even when you do that, you wouldnt get 2.7..
 

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