Chapter 3, Problem 3.86QA

Interpretation Introduction

To determine:

which of the given combinations of quantum numbers are allowed

Answer to Problem 3.86QA

Solution: The combinations of quantum numbers that are allowed are as follows:

No.	Quantum Numbers	Allowed
a.	$n=3,l=2,{m_l=0,m}_s=-\frac{1}{2}$	Yes
b.	$n=5,l=4,{m_l=4,m}_s=+\frac{1}{2}$	Yes
c.	$n=3,l=0,{m_l=1,m}_s=+\frac{1}{2}$	No
d.	$n=4,l=4,{m_l=1,m}_s=-\frac{1}{2}$	No

Explanation of Solution

1) Concept: Each orbital can contain a maximum of 2 electrons. The number of orbitals for a given set of numbers can be calculated using the following rules:

i. For a given value of $m_s$, there can only be one electron for fixed values of $n$,$l$ and $m_l$.

ii. A given value of $m_l$ corresponds to a single orbital.

iii. For a given value of $l$, $(2l+1)$ values of $m$ are possible, ranging from $+l$ to $-l$.

iv. For a given value of $n$, $(n-1)$ values of $l$ are possible.

2) Given Information:

No.	Quantum Numbers
a.	$n=3,l=2,{m_l=0,m}_s=-\frac{1}{2}$
b.	$n=5,l=4,{m_l=4,m}_s=+\frac{1}{2}$
c.	$n=3,l=0,{m_l=1,m}_s=+\frac{1}{2}$
d.	$n=4,l=4,{m_l=1,m}_s=-\frac{1}{2}$

3) Calculations:

a) $n=3,l=2,{m_l=0,m}_s=-\frac{1}{2}$

For $l=2$, the possible $m_l$ values range from $-2$ to $+2$. Therefore, this combination is allowed.

b) $n=5,l=4,{m_l=4,m}_s=+\frac{1}{2}$

For $l=4$, the possible $m_l$ values range from $-4$ to $+4$. Therefore, this combination is allowed.

c) $n=3,l=0,{m_l=1,m}_s=+\frac{1}{2}$

For $l=0$, the only possible $m_l$ value is $0$. Therefore, this combination is not allowed.

d) $n=4,l=4,{m_l=1,m}_s=-\frac{1}{2}$

For $n=4$, the possible $l$ values are $0$ to $3$. Therefore, this combination is not allowed.

Conclusion: The combinations of quantum numbers that are allowed are as follows:

No.	Quantum Numbers	Allowed
a.	$n=3,l=2,{m_l=0,m}_s=-\frac{1}{2}$	Yes
b.	$n=5,l=4,{m_l=4,m}_s=+\frac{1}{2}$	Yes
c.	$n=3,l=0,{m_l=1,m}_s=+\frac{1}{2}$	No
d.	$n=4,l=4,{m_l=1,m}_s=-\frac{1}{2}$	No