## De Broehli [ENDORSED]

$c=\lambda v$

Sabrina Fardeheb 2B
Posts: 51
Joined: Sat Jul 22, 2017 3:00 am
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### De Broehli

I have a question that relates to De Broeli. In chapter 1, for #33, why can't I combine and manipulate the equations for KE, E, and c in order to derive the wavelength? What's the difference between solving for the wavelength of light that ejects electrons and solving for the wavelength of the electron itself?

Kevin Liu 3G
Posts: 28
Joined: Fri Sep 29, 2017 7:07 am

### Re: De Broehli  [ENDORSED]

For question 33, you set the KE and the energy needed to emit an electron equal to each other. 1/2mv^2 = hv - (symbol) then you solve for frequency.

Then to calculate wavelength you would use the equation lambda = hc/KE or hc/(1/2)mv^2. You should get 33.7 nm.

AtreyiMitra2L
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### Re: De Broehli

To derive the equation, you basically have to do with the person before me told you to do. Now for the second part of your question, they are the same. We use light in our attempt to understand atoms better. This is one of those many instances. Also, the values for what is emitted and absorbed are basically the same thing. Therefore, you would get the same thing.