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Found in: Page 1247

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

In the subshell I =3, (a) what is the greatest (most positive) ${{\mathbf{m}}}_{{\mathbf{1}}}$ value, (b) how many states are available with the greatest ${{\mathbf{m}}}_{{\mathbf{1}}}$ value, and (c) what is the total number of states available in the subshell?

1. The greatest (most positive) ${\mathrm{m}}_{1}$ value is 3 .
2. The number of states available with the greatest ${\mathrm{m}}_{1}$ value is 2 .
3. The total number of states available in the subshell is 14 .
See the step by step solution

Step 1: The given data

The quantum number of the subshell is I=3.

Step 2: Understanding the concept of magnetic angular quantum number

Using the concept of the possible value of magnetic quantum number for every value of the orbital quantum number. The maximum value of m is + I . Thus, every electron has two spin orientations and thus there are two states.

Now, for every value of I, there are ( 2I +1) magnetic angular quantum number values. Thus, considering the case of two spins for every electron, have 2(2I+1) electron states.

Formula:

The total possible electron states for every value is,

${\mathrm{N}}_{\mathrm{I}}=2\left(2\mathrm{I}+1\right)$ ….. (1)

Step 3: a) Calculation of the greatest ml value

Using the concept, the greatest ${\mathrm{m}}_{1}$ value for I = 3 is as follow.

${\mathrm{m}}_{1}=+\mathrm{I}\phantom{\rule{0ex}{0ex}}=3$

Hence, the greatest value is 3.

Step 4: b) Calculation of the states available with the greatest ml value

Using the concept of spins orientation, the states available with the greatest ${\mathrm{m}}_{\mathcal{l}}\left(=3\right)$ value is given by:

${\mathrm{m}}_{\mathrm{s}}=±\frac{1}{2}$

Hence, there are two states available with that value.

Step 5: c) Calculation of the total number of states available in the subshell

Using the given value I = 3 in equation (1), the total number of states available in the subshell is as given by,

${\mathrm{N}}_{\mathrm{I}}=2\left(2×3+1\right)\phantom{\rule{0ex}{0ex}}=14$

Hence, there are 14 electron states in the subshell.