## Brooke Yasuda's Wk 2 Workshop

$\lambda=\frac{h}{p}$

Nicole Huang 3F
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### Brooke Yasuda's Wk 2 Workshop

Hi, I'm confused on Q8 of Brooke's workshop from week 2; it says:
"You have a metal, Rubidium, which has a work function of 2.3 eV. If the electrons being ejected from the metal have a wavelength of 1.7x10^5 m, what is the wavelength of the incident light?

My process was to use De Broglie's equation to find the velocity of the electron, plug in the velocity into the kinetic energy equation to find the energy of the electron, and then add the energy of the electron with the work function (after converting it to Joules) to find the energy of the incident light. From there, I used E=h(c/wavelength) to find the wavelength. But, from the answer key, I found out that I was just supposed to plug in the energy of the work function into E=h(c/wavelength) and that wavelength I get is the wavelength of the incident light.

I was wondering why the energy of the incident light was equal to the energy of the work function if electrons were ejected, and why that process was used instead. Thanks!

Andrew Wang 1C
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Joined: Wed Sep 30, 2020 10:11 pm
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### Re: Brooke Yasuda's Wk 2 Workshop

I think the answer key had the energy of the incident light equal to the work function because the kinetic energy of the ejected electron is so small that it's basically negligible, so adding it to the work function wouldn't change anything. I'm pretty sure either way is fine and gives the same answer. Also, having the energy of incident light equal to the work function would still result in ejected electrons, they just would not be ejected with any kinetic energy.