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Quantum Numbers

The electrons...

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) $n = 3, l = 1$
Arrange the following in the order of increasing energy.

(A) < (C) < (B) < (D)

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(D) < (B) < (C) < (A)

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(B) < (D) < (A) < (C)

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Solution

The correct option is D (D) < (B) < (C) < (A)
Concept: Greater the $n + l$ value greater is the energy associated. In case where the $n + l$ values are same the energy will be higher for the higher n value.
Hence the order would be - (D) < (B) < (C) < (A)
Option C.

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n

l

n = 4, l = 1

n = 4, l = 0

n = 3, l = 2

n = 3, l = 1

n

l

n = 4, l = 1

n = 4, l = 0

n = 3, l = 2

n = 3, l = 1

Using s, p, d notations, describe the orbital with the following quantum numbers.

(a) n = 1, l = 0; (b) n = 3; l =1 (c) n = 4; l = 2; (d) n = 4; l =3.

n = 4, l = 0, m = 0, s = + \frac{1}{2}

n = 3, l = 1, m = 1, s = - \frac{1}{2}

n = 3, l = 2, m = 0, s = + \frac{1}{2}

n = 3, l = 0, m = 0, s = - \frac{1}{2}

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