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A particular term in an atom in which LS coupling is a good approximation splits into three levels, each having the same L and same S but different J. If the relative spacings between the levels are in the proportion 5:3, find L and S.

Sagot :

ajeigbeibraheem

Answer:

Explanation:

From the information given;

Consider using Lande's Interval rule which can be expressed as:

[tex]\Delta E = E_{j+1} - E_jj \ = \alpha (j+1)[/tex]

here;

[tex]j+1[/tex]  = highest level of j

and

[tex]\dfrac{\Delta E_1}{\Delta E_2} = \dfrac{(j+2)}{(j+1)}[/tex]

[tex]\dfrac{5}{3} = \dfrac{(j+2)}{(j+1)}[/tex]

[tex]5(j+1) = 3(j+2)[/tex]

[tex]5j+5 = 3j+6[/tex]

[tex]2j = 1\\ \\ j = \dfrac{1}{2}[/tex]

recall that:

[tex]j = |S-L| \ \to \ |S+L |[/tex]

So;

[tex]S-L = \dfrac{1}{2} --- (1)[/tex]; &

[tex]S+L = \dfrac{5}{2} --- (1)[/tex]

Using the elimination method, we have:

[tex]2S = \dfrac{6}{2}[/tex]

[tex]S = \dfrac{3}{2}[/tex]

Since [tex]S = \dfrac{3}{2}[/tex]; then from (1)

[tex]\dfrac{3}{2} -L = \dfrac{1}{2}[/tex]

[tex]L = \dfrac{2}{2}[/tex]

[tex]L = 1[/tex]

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