Question
The electrons identified by quantum numbers n and l:
(a) n = 4, l = 1 (b) n = 4, l = 0 (c) n = 3, l = 2 (d) n = 3 , l = 1
Can be placed in order of increasing energy as
-
(c) < (d) < (b) < (a)
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(d) < (b) < (c) < (a)
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(b) < (d) < (a) < (c)
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(a) < (c) < (b) < (d)
Solution
B.
(d) < (b) < (c) < (a)
a.) (n + l) = 4 + 1 = 5
b.) (n + l) = 4 + 0 = 4
c.) (n + 1) = 3 + 2 = 5
d.) (n + 1) = 3 + 1 = 4
Higher the value of (n+1), higher the energy. if (n+1) are same, sub orbit with a lower value of n has lower energy. Thus,
3p<4s<3d<4p
