Atomic sodium is studied by photoelectron spectroscopy. What is the maximum wavelength of the incident radiation that will ionize the most weakly held electrons in sodium and scatter them so that their de Broglie wavelength measures 3.091e-10 m?

Question

anonymous · Answer

Try using E=(mv^2)/2 to get the E, and then use E=HC/lambda :)

anonymous · Answer

I think that's what I did...

anonymous · Answer

Will you share your answer, please? Time is short :)

anonymous · Answer

Jajajaja sure, but try to do the calculations on your own so that you can understand it :) 1.12*10^-8

anonymous · Answer

I did. Head's exploding. And by the way: It's wrong! :)

anonymous · Answer

really? must have been a different one, cause I gave yu the answer I had for number 9...

anonymous · Answer

you*... Sorryy

anonymous · Answer

doesn't matter. I'm over the magic 60% already! :)

aaronq · Answer

deBroglie wavelength is described by: $$\lambda =\frac{ h }{ mv }$$