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Q7P

Expert-verifiedFound in: Page 1247

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An electron in a multi-electron atom has ${{\mathbf{m}}}_{{\mathbf{l}}}{\mathbf{=}}{\mathbf{+}}{\mathbf{4}}$ For this electron, what are (a) the value of I, (b) the smallest possible value of n, and (c) the number of possible values of ${{\mathbf{m}}}_{{\mathbf{s}}}$?**

- The value of is 4 .
- The smallest possible value of n is 5.
- The number of possible values of ${\mathrm{m}}_{\mathrm{s}}$ is 2.

An electron in a multi-electron atom has ${\mathrm{m}}_{1}=+4$.

**For every value of the orbital quantum number, there exist (21+1) values of magnetic quantum number ranging from -I to +I. Thus, the maximum value of this magnetic quantum number is equal to the value of the orbital quantum number. Now, for every value of the principal quantum number, there are n values of I ranging from 0 to (n-1) . Thus, the smallest value of n is given by (I+1). Now, every electron has two spin orientations, thus the value of the magnetic quantum number is determined by these spins.**

Formula:

The smallest possible value of n is

n =I +1 ….. (1)

The value of $\mathcal{l}$ considering the table is given by,

$\mathrm{I}={\left({\mathrm{m}}_{\mathrm{I}}\right)}_{\mathrm{max}}\phantom{\rule{0ex}{0ex}}=4$

Hence, the value is 4.

Using the above value of I=4 in equation (1), we can get the smallest possible value of n as follows:

n = 4 + 1

=5

Hence, the smallest possible value of is 5.

Using the concept, we know that there are two possible spin orientations of electron,

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

Hence, there are two possible values.

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