As stated in class, for one$-$electron atoms and ions, the spin$-$orbit parameter, ${}_{nl},$ is

given by:

${}_{nl}=\frac{Z^4(g_e-1)}{2n^{3}l(l+\frac{1}{2})(l+1)}\frac{e^2{}^2}{m_e^2{c}^2{a}_0^3}.$

However, this expression is given in cgs units, where the unit of charge is the

electrostatic unit. The proper expression in SI units must be divided by $4{}_0,$ giving:

${}_{nl}=\frac{Z^4(g_e-1)}{2n^{3}l(l+\frac{1}{2})(l+1)}\frac{e^2{}^2}{4{}_0{m}_e^2{c}^2{a}_0^3}.$

Using $e=1\mathrm{.}60217657{10}^{-19}C,m_e=9\mathrm{.}1093897{10}^{-31}kg,c=2\mathrm{.}99792458{10}^{8}m{s}^{-1},$

$g_e=2\mathrm{.}0023193,a_0=5\mathrm{.}29177249{10}^{-11}m,=1\mathrm{.}05457266{10}^{-34}J*s,$ and

${}_0=8\mathrm{.}85419{10}^{-12}{C}^2{J}^{-1}{m}^{-1},$ this expression will provide ${}_{nl}$ in units of Joules $($J$).$

This may be converted into wavenumber units $(c{m}^{-1})$ using the conversion

factor $1J=5\mathrm{.}03445{10}^{22}c{m}^{-1}.$

$($a$)$ Evaluate the physical constants to provide an expression for ${}_{nl}$ in $c{m}^{-1}$ units, in terms of

$Z,n,$ and $l.$

$($b$)$ For the hydrogen atom $(Z=1),$ evaluate the spin$-$orbit constant, ${}_{nl},$ for the $2p,3p,4p$;$3d,$

$4d$; and $4f$ orbitals. Also evaluate the expected energy splitting between the associated

${?}^2{P}_{\frac{3}{2}u}^2{P}_{\frac{1}{2}u,}^2{D}_{\frac{5}{2}g}^2$ File Preview ${?}_{\frac{7}{2}u}^{-2}{F}_{\frac{5}{2}u}$ pairs of states.

How do your results compare to the measured splittings:

$2p$ states: $E{(}^2{P}_{\frac{3}{2}u})-E{(}^2{P}_{\frac{1}{2}u})=0\mathrm{.}365c{m}^{-1}$

$3p$ states: $E{(}^2{P}_{\frac{3}{2}u})-E{(}^2{P}_{\frac{1}{2}u})=0\mathrm{.}108c{m}^{-1}$

$4p$ states: $E{(}^2{P}_{\frac{3}{2}u})-E{(}^2{P}_{\frac{1}{2}u})=0\mathrm{.}046c{m}^{-1}$

$3d$ states: $E{(}^2{D}_{\frac{5}{2}g})-E{(}^2{D}_{\frac{3}{2}g})=0\mathrm{.}0361c{m}^{-1}$

$4d$ states: $E{(}^2{D}_{\frac{5}{2}g})-E{(}^2{D}_{\frac{3}{2}g})=0\mathrm{.}0152c{m}^{-1}$

$4f$ states: $E{(}^2{F}_{\frac{7}{2}u})-E{(}^2{F}_{\frac{5}{2}u})=0\mathrm{.}0076c{m}^{-1}?$

$($c$)$ Use the expression above to evaluate ${}_{2p}$ and the ${?}^2{P}_{\frac{3}{2}u}-{?}^2{P}_{\frac{1}{2}u}$ separation of the $2p^1$

configuration in lithium dication, $L{i}^{2+},$ which has only one electron. How does this

compare to the experimentally measured separation of $E{(}^2{P}_{\frac{3}{2}u})-E{(}^2{P}_{\frac{1}{2}u})=29\mathrm{.}6c{m}^{-1}$ in

the $L{i}^{2+}$ ion?

$($d$)$ In the neutral lithium atom the $1s^{2}2p^1,{?}^2{P}_u$ levels lie at energies of

$E(1s^{2}2p^1,{?}^2{P}_{\frac{3}{2}}u)=14904\mathrm{.}00c{m}^{-1}$ and $E(1s^{2}2p^1,{?}^2{P}_{\frac{1}{2}u})=14903\mathrm{.}66c{m}^{-1}.$

From these data, calculate ${}_{2p}$ for the neutral lithium atom. Why is this value so greatly

reduced from that of the $L{i}^{2+}$ ion?