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(Follow the links to see the sources for the equations)

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- Thread starter otaKu
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- #1

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(Follow the links to see the sources for the equations)

- #2

DrSteve

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E

The factor of Z

In the first or second formula the

- #3

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Even if I were to replace h by h/2π the formulas remain unequal. The 1st one has a π in it as well. I'd want to believe that the second one is correct but unfortunately the book I am reading seems to use the third one over and over which is essentially different from the second one even after I replace the h in second one with h/2π.In the first or second formula thehshould beh bar(h/2Pi).

- #4

Vanadium 50

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The first uses different units than the second and third.

- #5

jtbell

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- #6

jtbell

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- #7

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The book's name is Optical Processes in Semiconductors by Pankove. He uses this equation to approximate the binding energy of an electron to it's parent impurity ion inside a solid by considering it as an hydrogen atom immersed in a dielectric medium. I'll post the image if it helps.

- #8

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what about the third one? It seems different than the rest of the two and I don't seem to figure out a way to link it to either of them.The first uses different units than the second and third.

- #9

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I substituted the corresponding values of the constants and the second one is definitely in the SI notation. The first one, as you say is in gaussian, though I need to validate it by putting in the values of the constants. I suppose the third one is wrong and we need to replace the 2 with 8 there to convert it to SI.

- #10

jtbell

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- #11

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I get it now. All three are correct. Actually if we observe the coulomb law in SI and its counterpart in CGS we find that CGS units are such that the value of 1/4πε in CGS is 1. That is, ε(in CGS) = 1/4π. Substituting it in the third equations reduces it to 1st. *sigh* This might not be the best approach to do it but I guess it is correct. This thing consumed a hell lot of time. :/_{i}in the third equation is a different energy from the E_{n}in the others. By letting E_{n}= - (13.6 eV)/n^{2}in the third equation, I get $$E_n = - \frac {mq^4} {2h^2 n^2}$$ which differs from the first equation by a factor of 1/(4π^{2}), and from the second equation by a factor of ##4 \varepsilon_0^2##. It might be yet another system of electromagnetic units, or it might be simply the result of a typographical error somewhere else.

Thank you everyone for their answers!

especially jtbell

- #12

jtbell

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