Photoelectric Effect: Calculating Potential Difference

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SUMMARY

The discussion focuses on calculating the potential difference required to stop the fastest photoelectrons emitted from a nickel surface when exposed to ultraviolet light with a wavelength of 220 nm. The work function of nickel is given as 5.00 eV. The relevant equation used is h[c]/[\lambda] = Φ + eV, where h is Planck's constant (6.6 x 10^-34 Js) and c is the speed of light (3.0 x 10^8 m/s). The calculation reveals that the work function must be converted from eV to Joules, and the final answer must be expressed in volts, not meters per second.

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Homework Statement


What potential difference must be applied to stop the fastest photoelectrons by a nickel surface under the action of ultraviolet light of wavelength 220 nm? work function of nickel is 5.00eV.

Homework Equations


h[c]/[\lambda] = \Phi + eV

The Attempt at a Solution


h=6.6x10^-34 Js
c=3.0x10^8 m/s
\lambda=200x10^-9 m
\Phi=5.00eV
e=1.6x10^-19
V=?

[6.6x10^-34(3.0x10^8)]/[/200x10^-9] = 5.00 + 1.6x10^-19V
9.9x10^-37 = 5.00 + 1.6x10^-19V
9.9x10^37/5.00 = + 1.6x10^-19V
1.98x10^37 / - 1.6x10^-19 =V
v= 1.23x10^.18m/s

not sure if this is right
 
Last edited:
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You need to convert the work function from eV to Joules before you plug it in.
 
Also: really watch your units. As you have the equation solved... you're giving an answer in m/s... which doesn't make any sense. Carry your units along with the math... the problem is looking for a potential difference (and the solution should therefore be in volts).
 

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