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How do I calculate this:

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

- Thread starter Kahsi
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- #1

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How do I calculate this:

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

- #2

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- #3

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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

- #4

Doc Al

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[tex]eV_0[/tex] is the maximum KE of the ejected electron; to find theKahsi said:I need [tex]eV_0[/tex] to calculate [tex]\lambda[/tex]. How do I get it?

- #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)

[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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You'd better use consistent units throughout. Since h = 6.625 x 10-34Kahsi said:Maybe i should turn eV to joule first?

The standard unit for length is the meter, so that equation for [itex]\lambda[/itex] will give you meters.Btw: Is the answer in meter? (I'm new to frequency)

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Thank you in advance.

- #8

Doc Al

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See post #4 in this thread. You have the formula.Kahsi said:But which formulas should I use to calculate which energy the fastest electrons will get?

(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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