QuarkDecay
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The Atom of Helium is doubly excited in 2p2 1D
Can someone explain to me how these energy symbols work? I have a problem with what the 1D means specifically. I know 2p2 means two electrons in the 2p state. The 1 in 1D could be referring to electron being in a singleton, but I don't understand what the D means exactly, and what does it have to do with the 2p2.
The full question is, if the Helium is in this state above and it's decay time is in pico seconds, which one of the below will we get;
(a) H*( 2p2 1D ) → H+ (2s) + eA
(b) H*( 2p2 1D ) → H* (1s5p) + γ
(c) H*( 2p2 1D ) → H+ (1s) + eA + γ
(d) H*( 2p2 1D ) → H(1s2 1S) + γ1 + γ 2
Can someone explain to me how these energy symbols work? I have a problem with what the 1D means specifically. I know 2p2 means two electrons in the 2p state. The 1 in 1D could be referring to electron being in a singleton, but I don't understand what the D means exactly, and what does it have to do with the 2p2.
The full question is, if the Helium is in this state above and it's decay time is in pico seconds, which one of the below will we get;
(a) H*( 2p2 1D ) → H+ (2s) + eA
(b) H*( 2p2 1D ) → H* (1s5p) + γ
(c) H*( 2p2 1D ) → H+ (1s) + eA + γ
(d) H*( 2p2 1D ) → H(1s2 1S) + γ1 + γ 2