Chapter 3, Problem 3.85QA

Interpretation Introduction

To determine:

Which of the given combinations of quantum numbers is allowed

Answer to Problem 3.85QA

Solution:

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

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

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

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

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

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

2) Given:

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

3) Calculations:

a. $n=1,l=1,{m_l=0,m}_s=+\frac{1}{2}$

For $l=1$, the possible $m_l$ values range from $-1$ to $+1$. Therefore, this combination is allowed.

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

For $l=0$, the only possible $m_l$ value is $0$. Therefore, this combination is allowed.

c. $n=1,l=0,{m_l=1,m}_s=-\frac{1}{2}$

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

d. $n=2,l=1,{m_l=2,m}_s=+\frac{1}{2}$

For $l=1$, the possible $m_l$ values range from $-1$ to $+1$. Therefore, this combination is not allowed.

Conclusion:

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

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