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Advanced Physics Homework Help
Clarification on J and F (total angular momentum quantum numbers)
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[QUOTE="Higgy, post: 4545250, member: 153394"] Right, I think this is resolved. Here is my solution. I would appreciate it if someone looked it over and confirmed that it is correct. [B]Example:[/B] The ground state of [itex]^{39} K[/itex] is the [itex]4s[/itex] state. (i.) To calculate [itex]J[/itex], observe that an [itex]s[/itex] state has [itex]L = 0 [/itex], and that the spin of the electron is [itex]S = 1/2[/itex]: [tex]|L-S| ≤ J ≤ L+S \\→ |0-\frac{1}{2}| ≤ J ≤ 0 + \frac{1}{2} \\→ \frac{1}{2} ≤ J ≤ \frac{1}{2} \\→ J = \frac{1}{2}[/tex]So [itex]J[/itex] only takes one value in this case, although it can take more values, just not negative values. (ii.) To calculate the values of [itex]F[/itex], observe that the nucleus of the [itex]^{39} K[/itex] atom has spin of [itex]3/2[/itex], so [itex]I = 3/2[/itex]: [tex]|J-I| ≤ F ≤ J+I \\→ |\frac{1}{2}-\frac{3}{2}| ≤ F ≤ \frac{1}{2} + \frac{3}{2} \\→ 1 ≤ F ≤ 2 \\→ F = 1, 2[/tex] So the ground state has two hyperfine states labelled by [itex]F=1[/itex] and [itex]F=2[/itex]. (iii.) We calculate the [itex]``J"[/itex] Lande g-factor for our only value of [itex]J[/itex]: [tex]g_{J} ≈ \frac{3}{2} + \frac{ \frac{3}{4} - L(L+1) }{ 2J(J+1) } \\→ g_{J} ≈ \frac{3}{2} + \frac{ \frac{3}{4} - 0(0+1) }{ 2(\frac{1}{2})(\frac{1}{2}+1) } \\→ g_{J} = 2[/tex] (iv.) We have two [itex]F[/itex] states, so we calculate [itex]g_{F}[/itex] separately for each. [tex]g_{F} = g_{J} \frac{ F(F+1) + J(J+1) - I(I+1) }{ 2F(F+1) } \\→ g_{F=2} = (2) \frac{ 2(2+1) + \frac{1}{2}(\frac{1}{2}+1) - \frac{3}{2}(\frac{3}{2}+1) }{ 2(2)(2+1) } = \frac{1}{2} \\→ g_{F=1} = (2) \frac{ 1(1+1) + \frac{1}{2}(\frac{1}{2}+1) - \frac{3}{2}(\frac{3}{2}+1) }{ 2(1)(1+1) } = -\frac{1}{2}[/tex] (v.) Finally, we find the slopes of the Zeeman states for each hyperfine state by multiplying by [itex]m_{F}[/itex]. For [itex]F=2[/itex]:[tex]\Delta E_{F=2,m_{F}=2} = \mu _{B} (\frac{1}{2})(2) B _{z} = \mu _{B} B _{z}[/tex][tex]\Delta E_{F=2,m_{F}=1} = \mu _{B} (\frac{1}{2})(1) B _{z} = \frac{1}{2} \mu _{B} B _{z}[/tex][tex]\Delta E_{F=2,m_{F}=0} = \mu _{B} (\frac{1}{2})(0) B _{z} = 0[/tex][tex]\Delta E_{F=2,m_{F}=-1} = \mu _{B} (\frac{1}{2})(-1) B _{z} = -\frac{1}{2} \mu _{B} B _{z}[/tex][tex]\Delta E_{F=2,m_{F}=-2} = \mu _{B} (\frac{1}{2})(-2) B _{z} = -\mu _{B} B _{z}[/tex]For [itex]F=1[/itex]:[tex]\Delta E_{F=1,m_{F}=1} = \mu _{B} (-\frac{1}{2})(1) B _{z} = -\frac{1}{2} \mu _{B} B _{z}[/tex][tex]\Delta E_{F=1,m_{F}=0} = \mu _{B} (-\frac{1}{2})(0) B _{z} = 0[/tex][tex]\Delta E_{F=1,m_{F}=-1} = \mu _{B} (-\frac{1}{2})(-1) B _{z} = \frac{1}{2} \mu _{B} B _{z}[/tex] I've attached a plot of these states. [/QUOTE]
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Clarification on J and F (total angular momentum quantum numbers)
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