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)

1 Answer
Oct 23, 2016

Phosphorus has atomic number #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.