# The bohr model - dusting off

I'm sitting with Physics Handbook and I'm trying to recall how the make out the orbitals in the energy levels. Lets take a look at the first levels.

n = 1 - sub-shell K
Number of electrons = $2n^2 \Rightarrow 2$ electrons.
Number of electrons = $l = 0,1... n-1 \Rightarrow l = 0$.

This is where I get stuck. I can't remember how to get the orbitals; s, p, d etc...

Edit, is n = 0 & l = 0 written as 1s?

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## Answers and Replies

Ah well, I think I've got it, using brute force on my F.H. One thing though.

I seem to remember using this model to decide between which orbitals a jump is possible. Is it, how would I do that if yes..?

Cheers!

Hmm... Thanks for the link!

I'm not sure of the terminology in English. Would this be called Electron Transition? The rules, how do I determine $\Delta m_l$ ? I added a picture of what I've got so far.

Rules

$\Delta l = \pm 1$
$\Delta m_l = 0, \pm 1$

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First write down the principle quatum no. n the azimuthal quatum number is l=n-1 and then magnetic quantum number is m=-l to l. n and l give energy states, and m gives orientation of degenerate states. n can never be zero here at in the which you are dealing with. n=1 l=0 is written as 1s.
for s, l=0
for p, l=1
for d, l=2
for f, l=3
for g, l=4 and so on..
Actually s, p, d, f, ... are initial alphabets of the spectral terms - sharp,principle,diffuse,fundamental.
general rule for writing quantum states is -(principle quntum no. n)(name of sub-shell s,p,d or f)and number of electron in that state in superscript.
One more thing s, p, d, f,... are subshells containing orbitals decided by m and each orbital contains at max. 2 electrons as per Pauli's exclusion principle.
s-> one orbital m=0,
p-> three degenerate orbitals m=-1,0,1
and so on

Thanks for the explanation. I tried to sketch possible transitions... Is any transitions but 1s --> 2p and 1s --> 3p likely to occur given that they have to stop at 2p?