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Topic: de Broglie wave length calculation for an ejected electron  (Read 4978 times)

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Offline LawlKake

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de Broglie wave length calculation for an ejected electron
« on: November 04, 2010, 10:34:17 PM »
"Suppose that a particular sample of H atoms has all of the atoms in the n=5 state. Photons emitted from the sample are used as a light source to dislodge electrons from the surface of Mg(s). What is the minimum de Broglie wave length of an ejected electron? Work function for Mg(s) is 5.86e-19 J. Mass electron = 9.109e-31 kg."

My attempt:

Code: [Select]
KE = 1/2mv^2
5.86e-19 = 1/2 9.109e-31 v^2
v = 1134301.365
wavelength = h/mv
                = h / 9.109e-31 x 1134301.365
                = 6.41e-10 m

Apparently that's not the right answer (Online Quiz).

Offline LawlKake

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Re: de Broglie wave length calculation for an ejected electron
« Reply #1 on: November 04, 2010, 11:24:58 PM »
Yes, I've tried simply calculating the wavelength using W = hf = hc/lambda, wavelength = 3.39e-7 m and submitting it as an answer, but it was not correct.

Offline Borek

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Re: de Broglie wave length calculation for an ejected electron
« Reply #2 on: November 05, 2010, 03:58:31 AM »
You need to use two equations - Rydberg equation for energy of photon emitted by hydrogen, and equation for photoelectric effect.
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