# What is the electron configuration for an excited atom of phosphorous, ready to emit a photon of energy and return to ground state? (configuration has to follow all rules except be in lowest energy)

Oct 23, 2016

Phosphorus has atomic number $15$.
As such its electronic configuration in ground state is
$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{3}$

The following electronic configurations could be excited states

1. $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1} 3 {p}^{4}$
Here one of $3 s$ electrons has been promoted to $3 p$ sub level

2. $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{2} 3 {d}^{1}$
Here one of $3 p$ electrons has been promoted to $3 d$ sub level

3. $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{2} 4 {s}^{1}$
Another possibility is that one electron from the $3 p$ promoted to the $4 s$ sub-level
However, as explained below probability of transition listed at 2. is more than transition listed at 3.

The $3 p$ orbital has the following states for electron filling

$n = 3$, $l = 1$, ${m}_{l} = - 1 , 0 , 1 \mathmr{and} {m}_{s} = - \frac{1}{2} , + \frac{1}{2}$.

And the $4 s$ orbital has

$n = 4$, $l = 0$, ${m}_{l} = 0 \mathmr{and} {m}_{s} = - \frac{1}{2} , + \frac{1}{2}$

And the $3 d$ orbital has

$n = 3$, $l = 2$, ${m}_{l} = - 2 , - 1 , 0 , 1 , 2 \mathmr{and} {m}_{s} = - \frac{1}{2} , + \frac{1}{2}$

For a transition from $3 p$ to $4 s$, electron need to loose 1 quantum of angular momentum. This means absorbing a photon with spin $- 1$.
For a transition from $3 p$ to $3 d$, a photon of spin $+ 1$ needs to be absorbed and there are all three ${m}_{l}$ states.

There is only one ${m}_{l}$ level for the $4 s$ while there are three in the $3 p$ state. This make the transition more probable - due to greater number of states available to end up in.