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Q8P

Expert-verifiedFound in: Page 1247

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?**

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

The quantum number of the subshell is I=3.

**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)

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.

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

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

Hence, there are two states available with that value.

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

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

Hence, there are 14 electron states in the subshell.

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