Chapter 9, Problem 15P

To determine

The magnitude of spin-orbit energy of an electron present in $2p$ state.

Answer to Problem 15P

The magnitude of spin-orbit energy of an electron present in $2p$ state is $3.2\times {10}^{-5}\text{eV}$.

Explanation of Solution

Consider an electron present in the $2p$ state revolves in a circular orbit or loop of radius $r$. The motion of the electron around the nucleus gives rise to a current which produces magnetic field and magnetic dipole moment.

Write the expression for the magnetic moment of the electron.

  $\mu =iA$

Substitute $\frac{e}{T}$ for $i$, $\frac{2\pi r}{v}$ for $T$ and $\pi r^2$ for $A$ in above expression and simplify.

  $\mu =\frac{evr}{2}$

Substitute $\frac{L}{m_e}$ for $vr$ and $\sqrt{l\left(l+1\right)}\hslash$ for $L$ in above equation and simplify.

  $\mu =\left(\frac{e\hslash }{2m_e}\right)\sqrt{l\left(l+1\right)}$

Substitute ${\mu }_B$ for $\frac{e\hslash }{2m_e}$ in above expression.

  $\mu ={\mu }_B\sqrt{l\left(l+1\right)}$        (I)

Here, $\mu$ is the magnetic moment of electron, ${\mu }_B$ is Bohr’s magnetron and $l$ is azimuthal; quantum number.

Write the expression for the energy.

  $U=-{\mu }_s\cdot B$

Substitute $g\left(\frac{e}{2m}\right)S_z$ for ${\mu }_s$ in above equation.

  $U=-g\left(\frac{e}{2m}\right)S_{z}B$

Substitute $m_s\hslash$ for $S_z$ in above equation and simplify.

  $U=-g{\mu }_B{m}_{s}B$

Here, $U$ is the energy, $g$ is the gyromagnetic ratio of electron, $m_s$ is azimuthal spin quantum number and $B$ is the magnetic field.

The electron is present in the $2p$ state and the distance of the $2p$ o0rbit from the centre is $4a_0$.

Write the expression for magnetic field at the centre of current carrying loop.

  $B=\frac{2k_m\mu }{r^3}$

Substitute $4a_0$ for $r$ in above equation.

  $B=\frac{2k_m\mu }{{\left(4a_0\right)}^3}$        (II)

Here, $B$ is the magnetic field, $k_m$ is a magnetic constant and $a_0$ is Bohr radius.

For a material, which is a simple one electron system, there exist only two possible energy states corresponding to $m_S=\pm \frac{1}{2}$. When this material is placed in the presence of external magnetic field, the spin of the electron gets excited to higher energy state.

It means that the lower energy level which was initial aligned with the direction of magnetic field, angle between magnetic moment and magnetic field is $0°$, will get excited and get aligned opposite to the direction of magnetic field, angle between magnetic moment and magnetic field will become $180°$.

Therefore, the change in the energy is the sum of energy corresponding to each state.

Write the expression for the change in energy.

  $\Delta U=g{\mu }_{B}B$        (III)

Here, $\Delta U$ is change in energy.

Conclusion:

Substitute 1 for $l$ and $9.273\times {10}^{-24}\text{J}/\text{T}$ for ${\mu }_B$ in equation (I).

  $\begin{array}{c}\mu =\left(9.273\times {10}^{-24}\text{J}/\text{T}\right)\sqrt{1\left(1+1\right)}\\ =1.31\times {10}^{-23}\text{J}/\text{T}\end{array}$

Substitute $0.053\text{nm}$ for $a_0$, ${10}^{-7}\text{N}/{\text{A}}^{\text{2}}$ for $k_m$ and $1.31\times {10}^{-23}\text{J}/\text{T}$ for $\mu$ in equation (II).

  $\begin{array}{c}B=\frac{2\left({10}^{-7}\text{N}/{\text{A}}^{\text{2}}\right)\left(1.31\times {10}^{-23}\text{J}/\text{T}\right)}{{\left(4\left(0.053\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)\right)}^3}\\ =0.276\text{T}\end{array}$

Substitute 2 for $g$, $0.276\text{T}$ for $B$ and $9.273\times {10}^{-24}\text{J}/\text{T}$ for ${\mu }_B$ in equation (III).

  $\begin{array}{c}\Delta U=2\left(9.273\times {10}^{-24}\text{J}/\text{T}\right)\left(0.276\text{T}\right)\\ =5.12\times {10}^{-24}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =3.2\times {10}^{-5}\text{eV}\end{array}$

Thus, the magnitude of spin-orbit energy of an electron present in $2p$ state is $3.2\times {10}^{-5}\text{eV}$.