D lines of sodium (Z = 11)

[boyu]

Consider the D1 and D2 transitions (3p --> 3s) of the sodium atom (Z = 11).

How to calculate the magnetic field experienced by the valence electron arising from its orbital motion, given that the wavelength difference between the D lines?

The formula that I have from lecture notes for spin-orbit interaction energy is

[tex]<V_{SL}>=\frac{Z^{4}\alpha^{2}}{n^{3}}E_{0}\frac{j(j+1)-l(l+1)-\frac{3}{4}}{l(l+1)(2l+1)}[/tex]

But I can't see any method to calculate the magnetic field from it.

 

[read]

Spin orbit splitting is simply -\mu*B, where B is magnetic field acting on 3p-electron, \mu is the electron magnetic moment.

 

[boyu]

Thanks for your reply, but I guess we still need to consider the angle between [tex]\mu[/tex] and [tex]B[/tex], right?

And this angle is also given by the angle between [tex]L[/tex] and [tex]S[/tex], since [tex]\mu[/tex] is parallel to [tex]S[/tex] and [tex]B[/tex] is parallel to [tex]L[/tex].

Thus, [tex]cos\theta=\frac{j(j+1)-l(l+1)-s(s+1)}{2\sqrt{l(l+1)s(s+1)}}[/tex],

since [tex]J^{2}=L^{2}+S^{2}+2L\cdot S=l(l+1)+s(s+1)+2\sqrt{l(l+1)s(s+1)}cos\theta[/tex]

But I am still not sure about this answer...

 

[read]

Thanks for your reply, but I guess we still need to consider the angle between [tex]\mu[/tex] and [tex]B[/tex], right?

And this angle is also given by the angle between [tex]L[/tex] and [tex]S[/tex], since [tex]\mu[/tex] is parallel to [tex]S[/tex] and [tex]B[/tex] is parallel to [tex]L[/tex].

Thus, [tex]cos\theta=\frac{j(j+1)-l(l+1)-s(s+1)}{2\sqrt{l(l+1)s(s+1)}}[/tex],

since [tex]J^{2}=L^{2}+S^{2}+2L\cdot S=l(l+1)+s(s+1)+2\sqrt{l(l+1)s(s+1)}cos\theta[/tex]

But I am still not sure about this answer...

Well, angle is not well defined value. Splitting is \delta=2\mu_B*C*(L*S), The mag field B=C*L. The quantum numbers here are S^2, L^2, J^2 and J_z. Thus B^2=C^2*L^2. C is some constat proportional to el. fiel gradient.