Question

All other things being equal, shouldthe spin-orbit interaction be a larger or smaller effect in hydrogen as increases? Justify your answer

Short answer

Therefore, the interaction energy is proportional to$n^{-6}$ , the spin-orbit interaction gets smaller as n increases in hydrogen

Step-by-step solution

Given data

Effect in hydrogen over n increment

Concept of spin orbit interaction energy

The interaction energy U between the spin dipole vector $♢s$ and magnetic field due to the proton that the electron sees ${\text{B}}_L$is, $U=-{\mu }_s.B_L$.

Here${\mu }_s$ represents spin dipole moment of electron and$B_L$ represents magnetic field due to the proton that the electron

Step 3:Find the equation for the interaction energy, magnetic field, and allowed radii

In order to determine whether the spin-orbit interaction gets bigger or smaller in hydrogen as n increases, equation for the interaction energy, magnetic field, and allowed radii are needed.

The interaction energy U between the spin dipole vectorand magnetic field due to the proton that the electron sees${\text{B}}_L$is,. $U=-{\mu }_s.B_L$ ……. (1)

The magnetic field B that the electron sees from the proton is:

$B=\frac{{\mu }_{0}e}{4\pi m_e{r}^3}L$ .……. (2)

Here ${\mu }_0$ is the permeability of free space, e is the charge on the electron, m is the mass of the electron, r is the distance of the electron from the proton, and L is the angular momentum vector of the proton.

The Bohr model can be used to get an approximation for the allowable radiiof the electron from the proton is,$r_n=a_e{n}^2$. ……. (3)

Here $a_0$ is being the Bohr radius, and n being the principal quantum number.

Determine the effect on spin-orbit interaction in hydrogen as  n  increases 

Substitute the value offrom equation (2) to equation (1).

$\begin{array}{rcl}U& =& -{\mu }_s·B_L\\ & =& {\mu }_s·\left(\frac{{\mu }_{0}e}{4\pi m_e{r}^3}L\right)\\ & =& \frac{{\mu }_{0}e}{4\pi m_e{r}^3}{\mu }_{s·}L\end{array}$

Substitute the value of r from equation (3) to above equation.

$\begin{array}{rcl}U& =& -\frac{{\mu }_0{e}^e}{4\pi m_e{r}^3}{\mu }_s.L\\ & =& -\frac{{\mu }_0{e}^2}{4\pi m_e{\left(a_0{n}^2\right)}^3}{\mu }_s.L\\ & =& -\frac{{\mu }_0{e}^2}{4\pi m_e{a}_0^3{n}^6}{\mu }_s.L\\ & & \end{array}$

Therefore, the interaction energy is proportional to $n^{-6}$, the spin-orbit interaction gets smaller as n increases in hydrogen.