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Chapter 8: Q4CQ (page 338)

Compare and contrast the angular momentum and magnetic moment related to orbital motion with those that are intrinsic.

Short Answer

Expert verified

The magnetic moment and orbital angular momentum are related by

$P_{L} - \frac{e}{2 m_{0}} L$

The magnetic moment and intrinsic angular momentum are related by

$P_{5} = g \frac{q}{2 m} S$

Step by step solution

01

Magnetic moment

The orbital and intrinsic angular momentums are the first kind of angular momenta, having the same physical quantity, capable of adding as vectors, and are quantized.

The permitted values of orbital angular momentum depend on the spatial state of rotating electrons, governed by quantum number $I, m_{1}$ . The values of the $m_{1}$ can go from $- 1$ to $+ I .$

02

Angular momentum.

The permitted values of intrinsic angular momentum depend on the kind of the particle, governed by the quantum numbers $s$ , $m_{s}$ . The values of $m_{s}$ can go from $- s$ to $+ s$ .

The magnetic moment and orbital angular momentum are related by.

$P_{L} - \frac{e}{2 m_{0}} L$

The magnetic moment and intrinsic angular momentum are related by

$P_{5} = g \frac{q}{2 m} S$

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Most popular questions from this chapter

Summarize the connection between angular momentum quantization and the stem-Gerlach experiment.

The radius of cesium is roughly $0.26 nm$ .

(a) From this estimate the effective charge its valence electron orbits

(b) Given the nature of the electron's orbit. is this effective nuclearcharge reasonable?

(c) Compare this effective Zwith that obtained for sodium in Example 8.3. Are the values at odds with the evidence given in Figure $8.16$ that it takes less energy to remove an electron from cesium than from sodium? Explain.

Whether adding spins to get total spin, spin and orbit to get total angular momentum, or total angular momenta to get a "grand total" angular momentum, addition rules are always the same: Given $J_{1} = 鈩� \sqrt{j_{1} (j_{1} + 1)}$ and $J_{2} = 鈩� \sqrt{j_{2} (j_{2} + 1)}$ . Where is an angular momentum (orbital. spin. or total) and a quantum number. the total is $J_{T} = 鈩� \sqrt{j_{T} (j_{T} + 1)}$ , where $j_{T}$ may take on any value between $| j_{1} - j_{2} |$ and $j_{1} + j_{2}$ in integral steps: and for each value of $J_{螕} J_{T z} = {m_{蟺}}^{f}$ . where $m_{蟺}$ may take on any of $2 j_{r} +$ I possible values in integral steps from $- j_{T}$ for $+ j_{T}$ Since separately there would be $2 j_{1} + 1$ possible values for $m_{11}$ and $2 j_{2} +$ I for $m_{蚁^{2}}$ . the total number of stales should be $(2 j_{1} + 1) (2 j_{2} + 1)$ . Prove it: that is, show that the sum of the $2 j_{T} + 1$ values for $m_{i t}$ over all the allowed values for $j_{7}$ is $(2 j_{1} + 1) (2 j_{2} + 1)$ . (Note: Here we prove in general what we verified in Example $8.5$ for the specialcase $j_{1} = 3, j_{2} = \frac{1}{2}$ .)

To investigate the claim that lowerimplies lower f energy. consider a simple case: lithium. which has two $n = 1$ electrons and alone $n = 2$ valence electron.

(a)First find the approximate orbit radius, in terms of $a_{0}$ . of an $n = 1$ electron orbiting three protons. (Refer to Section 7.8.)

(b) Assuming the $n = 1$ electrons shield/cancel out two of the protons in lithium's nucleus, the orbit radius of an $n = 2$ electron orbiting a net charge of just $+ e$ .

(c) Argue that lithium's valence electron should certainly have lower energy in a 25 state than in a $2 p$ stale. (Refer Figure 7.15.)

The angles between S and $渭_{S}$ and between L and $渭_{L}$ are 180^o. What is the angle between J and $渭_{J}$ in a $2 p_{\frac{3}{2}}$ state of hydrogen?

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