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The first excited state corresponds to n = 3.
true
false

2007-02-04 14:20:06 · 2 answers · asked by Moe 2 in Education & Reference Homework Help

2 answers

that depends on the atom; if we are talking about hydrogen the answer is false, since the principle quantum #, n, is 1 and corresponds to the k shell or 1s orbital, the next allowable quantum state would be n=2 in a the empty p orbital, or l shell.
The answer would be true if we are talking about Ne, for example, which has a full octet and electron config. of 1s2 2s2 2p6, since the highest energy electrons are in the 2p orbital with the quantum #s being n=2, l=1, and m=+-1, 0 promotion can only be to the n=3 shell or the empty d orbital.

2007-02-07 18:30:26 · answer #1 · answered by Anonymous · 0 0

Think its false. N should be 2.

The formula is 2n^2. For the first excited state,n=1. so 2*1^2 which is 2

Not very certain though. Its been years since I studied chem.

2007-02-05 00:15:42 · answer #2 · answered by Josh 3 · 0 0

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