Light-wavelength calculation

  • Thread starter Kahsi
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  • #1
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Hi.

How do I calculate this:

How long is the shortest light-wavelenght that can breake loose a elektron from a natrium surface?
 

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  • #3
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Thank you for the link, clive.
This is what I've got now:

[tex]eV_{0} = \frac{hc}{\lambda}-\phi[/tex]
[tex]h = 6,6261 * 10^{-34}[/tex]
[tex]c = 2,99792458*10^8[/tex]
[tex]\phi =2,28eV[/tex]

I need [tex]eV_0[/tex] to calculate [tex]\lambda[/tex]. How do I get it?

And ofcourse I want the longest wavelength and not the shortest as I said before. :smile:
 
  • #4
Doc Al
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Kahsi said:
I need [tex]eV_0[/tex] to calculate [tex]\lambda[/tex]. How do I get it?
[tex]eV_0[/tex] is the maximum KE of the ejected electron; to find the minimum energy photon (longest wavelength) needed to eject an electron, set [tex]eV_0[/tex] to 0.
 
  • #5
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Okay so we got

[tex]\frac{hc}{\lambda} = \phi[/tex]
[tex]h = 6,6261*10^{-34}[/tex]
[tex]c\approx 3*10^8[/tex]
[tex]\phi = 2,28eV[/tex]

This gives us

[tex]\lambda\approx 8,7*10^{-26}[/tex]

Is this correct? Maybe i should turn eV to joule first?

Btw: Is the answer in meter? (I'm new to frequency)
 
Last edited:
  • #6
Doc Al
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use standard units

Kahsi said:
Maybe i should turn eV to joule first?
You'd better use consistent units throughout. Since h = 6.625 x 10-34 J-s, you'd better convert eV to Joules!

Btw: Is the answer in meter? (I'm new to frequency)
The standard unit for length is the meter, so that equation for [itex]\lambda[/itex] will give you meters.
 
  • #7
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Another question. If I'm lighting up a natriumsurface with a quicksilver-lamp electrons will "release". But which formulas should I use to calculate which energy the fastest electrons will get?

Thank you in advance.
 
  • #8
Doc Al
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Kahsi said:
But which formulas should I use to calculate which energy the fastest electrons will get?
See post #4 in this thread. You have the formula.
(Photon energy) - (energy needed to eject an electron) = maximum KE of ejected electron

This may help you: http://en.wikipedia.org/wiki/Photoelectric_effect#Explanation
 

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