# Energy levels of electrons excited by photons

1. Jul 2, 2008

### xArcherx

I've been researching the change of energy levels of electrons when excited by photons. The equations I have found are...

E_n = (-2pi^2 * m * e^4 * Z^2)/(n^2 * h^2)

Which gives us the energy of the electron at a particular energy level.

m = mass of electron
e = electric charge of electron
Z = number of proton (atomic number)
h = Planck's Constant
n = Energy level (1, 2, 3, ....)

Then there is...

v = (E_n' - E_n)/h

Which gives the frequency of the photon emitted due to a specified drop.

So if we have an excited electron at level 6 and the electron drops all the way to level 1 then it is...

v = (E_6 - E_1)/h

If you combine the two then you get...

v = ((-2pi^2 * m * e^4 * Z^2)/(6^2 * h^2) - (-2pi^2 * m * e^4 * Z^2)/(1^2 * h^2))/h

Are these equations correct? Will it give me a photon frequency such that if such a photon were to get absorbed by an electron in the first energy level that it will jump up to the sixth energy level?

Where v = c/λ then λ = c/v and this gives me the wavelength of the photon and thus the corresponding color along the spectrum.

2. Jul 2, 2008

### Mindscrape

Re: Confirmation

Looks good to me based on a first inspection. The ideas and original equations are right at least. Well, keep in mind that this only applies to atom's based on the Bohr model, which is ultimately wrong.

If you want some more grounding in this, you should check out Hydrogen spectroscopy.

3. Jul 3, 2008

### xArcherx

Re: Confirmation

Thanks, I'll check it out...

4. Jul 3, 2008

5. Jul 5, 2008

### xArcherx

Re: Confirmation

Awesome, thanks a lot.

Now I used the calculator on the sight and got a wavelength of 93.02514591068145 nm when Z = 1, n_2 = 7 and n_1 = 1.

So an ultraviolet photon of ~93 nm is emitted when there is a drop from level 7 to level 1.

So a question I have is where did I go wrong? Seriously wrong...

E_7 = (-2pi^2 * m * e^4 * Z^2)/(n^2 * h^2)
E_7 = (-19.739208757068045 * 9.10938188e-31 * 6.5893319466295382533529270269186e-76 * 1)/(49 * 4.39047898626754816e-67)
E_7 = (-1.1848408938972153629875267059781e-104)/(2.1513347032710985984e-65)
E_7 = -5.507468884761018301895046351223e-40

E_1 = (-2pi^2 * m * e^4 * Z^2)/(n^2 * h^2)
E_1 = (-19.739208757068045 * 9.10938188e-31 * 6.5893319466295382533529270269186e-76)/(4.39047898626754816e-67)
E_1 = (-1.1848408938972153629875267059781e-104)/(4.39047898626754816e-67)
E_1 = -2.6986597535328989679285727120993e-38

v = (E_7 - E_1)/h
v = ((-5.507468884761018301895046351223e-40) - (-2.6986597535328989679285727120993e-38))/(6.62606896e-34)
v = (2.6435850646852887849096222485868e-38)/(6.62606896e-34)
v = 3.9896733351916228546308733988576e-5

λ = c/v
λ = 299792458/3.9896733351916228546308733988576e-5
λ = 7514210633628.2555594458185616074

I've copy and pasted to window's calculator so it cut off the last digit in some of the numbers. However, with the scale of the error, I very much doubt that the cutting off of the last digit would be the cause.

6. Jul 5, 2008

### alphysicist

Re: Confirmation

Hi xArcherx,

This equation is not correct when you are using SI units. A more general formula for would be:

$$E_n = \frac{-2 \pi^2 k^2 m e^4 Z^2}{n^2 h^2}$$

where $k$ is Coulomb's constant. (In the cgs system, the numeric value of Coulomb's constant is 1.)

So if you want to use your formula, you need to use the mass, charge, and Planck's constant in the cgs units (they are easy to find), and when you get the answer, it will be in the energy units of ergs.

Or you can use the above formula with the numeric value of $k=8.99\times 10^9\, \mbox{N m}^2/\mbox{C}^2$.

Whichever you use, a really helpful fact to remember is that the ground state of hydrogen $E_1$ is -13.6 eV.

7. Jul 7, 2008

### xArcherx

Re: Confirmation

Thanks, so it was the use of SI units that messed me up and I was missing the k^2 altogether. I try it again and see if my answer matches that of the site.