# B Binding energy of the electron in Hydrogen atom.

#### otaKu

This website here says that the expression for binding energy for an electron is:

This pdf by MIT calculates it quantum mechanically to give:
The book I was reading optoelectronics from says that the energy binding the electron to the impurity(ionic nucleus) is
I am totally confused as to why there is a discrepancy between these results. Don't they mean the same(not talking about the inclusion of atomic number in first and effective mass in third)? Since all of these denote the energy of the electron bound to the core(nucleus) why do they differ?! Am I overlooking something and all three of these are correct? I would highly value any advice or explaination on this discrepancy. I've referred multiple sources and feel completely lost even though it is something very fundamental and basic. Thank you!

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#### DrSteve

Gold Member
Essentially they are all saying the same thing, that the energy of the hydrogen atom with the electron in the nth state is given by

En= -13.6 eV/n2

The factor of Z2 refers to hydrogen-like atoms.

In the first or second formula the h should be h bar (h/2Pi). The last formula you presented is slightly more complicated because the lattice gives the electron an effective mass different from its isolated mass. I wouldn't worry about this until you take solid state physics.

#### otaKu

In the first or second formula the h should be h bar (h/2Pi).
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.

Staff Emeritus
The first uses different units than the second and third.

#### jtbell

Mentor
The first equation uses Gaussian units. See this page of the lectures from which the first equation was taken, and notice how he writes Coulomb's law. Compare it to how the MIT notes write Coulomb's law.

#### jtbell

Mentor
Google doesn't let me view the page from the third book, that you linked to, and I'm not familiar with that book, or with that field, so I'd better not try to guess what's happening here.

#### otaKu

Google doesn't let me view the page from the third book, that you linked to, and I'm not familiar with that book, or with that field, so I'd better not try to guess what's happening here.
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.

#### otaKu

The first uses different units than the second and third.
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.

#### otaKu

Google doesn't let me view the page from the third book, that you linked to, and I'm not familiar with that book, or with that field, so I'd better not try to guess what's happening here.
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.

#### jtbell

Mentor
The Ei in the third equation is a different energy from the En in the others. By letting En = - (13.6 eV)/n2 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/(4π2), 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.

#### otaKu

The Ei in the third equation is a different energy from the En in the others. By letting En = - (13.6 eV)/n2 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/(4π2), 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.
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. :/
Thank you everyone for their answers!
especially jtbell

#### jtbell

Mentor
Those different systems of electromagnetic units can be a real pain in the <insert least favorite part of the anatomy here>.

"Binding energy of the electron in Hydrogen atom."

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