Yet another chem problem

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Darwin420
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Yet another chem problem

Post by Darwin420 » Mon Nov 10, 2008 9:06 pm

Guys, I am struggling hard right now. I have no idea how to approach this question. Any insight?

The minimum energy for photoemission of electrons from a potassium surface is
3.69x10-19 J. What is the kinetic energy of electrons emitted from a potassium surface
when it is irradiated by UV light at 300 nm.?
(a) 7.60x10-20 J
(b) 3.69x10-19 J
(c) 2.93x10-19 J
(d) 6.62x10-19 J
(e) 1.03x10-18 J

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mith
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Post by mith » Mon Nov 10, 2008 11:23 pm

Calculate the amount of energy per photon. E = h*nu.

Convert using E=1/2 mv^2.

Energy of photon = Energy needed to remove an electron + Kinetic energy of the emitted electron

http://en.wikipedia.org/wiki/Photoelectric_effect
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Darwin420
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Re: Yet another chem problem

Post by Darwin420 » Tue Nov 11, 2008 5:33 pm

Thanks Mith, I was able to figure it out lastnight. Appreciated.

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