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)

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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)

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Answer 1 · 2016-10-23T04:27:51

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

The following electronic configurations could be excited states

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

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

$n = 3$ $n = 3$ , $l = 1$ $l = 1$ , $m_{l} = - 1, 0, 1 and m_{s} = - \frac{1}{2}, + \frac{1}{2}$ $m_l = -1,0,1 and m_s = -1/2, +1/2$ .

And the $4 s$ $4s$ orbital has

$n = 4$ $n = 4$ , $l = 0$ $l = 0$ , $m_{l} = 0 and m_{s} = - \frac{1}{2}, + \frac{1}{2}$ $m_l = 0 and m_s = -1/2, +1/2$

And the $3 d$ $3d$ orbital has

$n = 3$ $n= 3$ , $l = 2$ $l = 2$ , $m_{l} = - 2, - 1, 0, 1, 2 and m_{s} = - \frac{1}{2}, + \frac{1}{2}$ $m_l=-2,-1,0,1,2 and m_s =-1/2, +1/2$

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

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

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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)

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