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May 11, 2014 at 14:35 vote accept Jesse
S May 11, 2014 at 12:13 history suggested CommunityBot
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May 11, 2014 at 12:11 review Suggested edits
S May 11, 2014 at 12:13
May 11, 2014 at 3:22 comment added Jesse yeah i fixed it.
May 11, 2014 at 3:22 comment added LDC3 Oh, I found it, it should be: $$\frac{r}{\lambda} = \alpha \pi Z$$ since $\alpha = \frac{ke^2}{\hbar c}$
May 11, 2014 at 3:22 history edited Jesse CC BY-SA 3.0
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May 11, 2014 at 3:21 comment added Jesse The $m_e$ goes away, $n^2$ goes, $Z^2$ becomes Z. That leaves $\frac{\hbar}{ke^2}\alpha^2\pi Z c$ which, since $\alpha=\frac{ke^2}{\hbar c}$ means it's $\alpha \pi Z$. I think that should work, but I may have messed it. :-)
May 11, 2014 at 3:08 comment added LDC3 You have: $$\frac{r}{\lambda} = \frac{n^2 \hbar^2} {Zm_e k e^2}\alpha^2 \frac{\pi Z^2 m_ec}{ n^2 \hbar}$$ $$=\alpha \pi Z c$$ I must have missed a step since I don't see how you cancelled out so many terms. Where did $\frac {\hbar}{ke^2}$ go to?
May 11, 2014 at 2:50 answer added DavePhD timeline score: 3
May 11, 2014 at 1:33 history edited Jesse CC BY-SA 3.0
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May 11, 2014 at 1:19 history asked Jesse CC BY-SA 3.0