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# Chem homework help!

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This is probably much easier than I think, but I need to calculate wavelength using velocity. What is the formula for that? Thanks!

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ok never mind, just figured it out, thanks anyway!

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The question I'm currently having trouble with is:

The work function for lithium is 279.7 kj/mol. What is the maximum wavelength of light that can remove one electron from one atom of lithium metal?

and...

The ionization energy of gold is 890.1 kj/mol. Is light with a wavelength of 225nm capable of ionizing a gold atom?

I'm yet to wrap my head around this material, clearly. Any help would be appreciated!

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This is probably much easier than I think, but I need to calculate wavelength using velocity. What is the formula for that? Thanks!

Wavelength using valocity of what? An electron?

If so, the kinetic energy will probably have to be equated with photon energy. Maybe you'll have to add the work function to the KE of the electron.

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The question I'm currently having trouble with is:

The work function for lithium is 279.7 kj/mol. What is the maximum wavelength of light that can remove one electron from one atom of lithium metal?

First convert your work function to electron volts. You can do this easy enough knowing one eV equals 96.485 kJ/mol.

Then you can use this equation:

$\lambda = \frac{hc}{\phi}$

• where [imath]\lambda[/imath] is the wavelength (what you're looking for)
• c is the speed of light (use 2.9979 x 10^17 nanometers/second to get your answer in nanometers)
• h is plank's constant (4.13566 X 10^-15 eV/second)
• and [imath]\phi[/imath] is the work function in electron Volts - what we are given

That will give you the largest wavelength that can eject an electron from a lithium plate. You can both check your results and get a good description of the photoelectric effect here.

By check your results, I mean you can input eV toward the bottom of the page and get a wavelength.

and...

The ionization energy of gold is 890.1 kj/mol. Is light with a wavelength of 225nm capable of ionizing a gold atom?

You can use the same formula as above. Instead of "work function" being the energy in eV or [imath]\phi[/imath], it is "ionization energy" - but the equation is the same. See if the resulting wavelength is larger or smaller than 225 nm.

-modest

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