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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)

Md Ariful Islam

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Md Ariful Islam

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

The following electronic configurations could be excited states

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

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

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

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

$n = 3$, $l = 1$, $m_l = -1,0,1 and m_s = -1/2, +1/2$.

And the $4s$ orbital has

$n = 4$, $l = 0$, $m_l = 0 and m_s = -1/2, +1/2$

And the $3d$ orbital has

$n= 3$, $l = 2$, $m_l=-2,-1,0,1,2 and m_s =-1/2, +1/2$

For a transition from $3p$ to $4s$, electron need to loose 1 quantum of angular momentum. This means absorbing a photon with spin $-1$.
For a transition from $3p$ to $3d$, 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 $4s$ while there are three in the $3p$ state. This make the transition more probable - due to greater number of states available to end up in.

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