What is the Wavelength of the Ejected Electron?

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The discussion revolves around calculating the wavelength of an ejected electron when a photon strikes a metal surface. Given the threshold frequency of 3.184 x 10^14 Hz and a photon wavelength of 368 nm, the energy of the incident photon is calculated using the equation E = hv. The kinetic energy of the ejected electron is determined by subtracting the threshold energy from the incident energy. Finally, the wavelength of the ejected electron is found using the de Broglie wavelength formula, λ = h/p, where p is the momentum derived from the kinetic energy. The solution emphasizes the importance of correctly applying these equations to arrive at the final answer.
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Homework Statement



The threshold frequency of a metal is 3.184 x 10^14 Hz. If a photon of wavelength 368 nm strikes the surface of the metal, what is the wavelength of the ejected electron?

Homework Equations


f (frequency) = c/(wavelength)
Eincident = Ethreshold + KE
E=fv(wavelength)

The Attempt at a Solution


Oh man... here it is. I bet I'm missing something huge but it's just been killing me.

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You're good up through the energy calculation. From there you can use:
$$KE = \frac{p^2}{2m}$$
to get a value for your momentum. Then just plug that number in to:
$$\lambda = \frac{h}{p}$$
and you get your answer.
 

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