Answer :
Answer:
These energy shifts split the [tex]\( d \)-state into two levels due to the spin-orbit interaction, with the \( j = \frac{5}{2} \) state being higher in energy than the \( j = \frac{3}{2} \) state. This splitting is crucial in explaining fine structure in atomic spectra.[/tex]
Explanation:
To solve the problem, we need to understand the interaction of the magnetic dipole moment associated with the electron's spin and the magnetic field generated by the electron's orbital motion. This is known as the spin-orbit interaction. Let's go through the details step-by-step.
Hydrogen Atom in a D-state
- The quantum number l = 2 corresponds to the d-state.
- The magnetic quantum number [tex]\( m_l \)[/tex] can take values -2, -1, 0, 1, 2.
Spin-Orbit Interaction
The Hamiltonian gives the spin-orbit interaction:
[tex]\[ H_{SO} = \xi(r) \mathbf{L} \times \mathbf{S} \][/tex]
where:
[tex]- \( \xi(r) \) is a function of the radial distance \( r \).\\- \( \mathbf{L} \) is the orbital angular momentum.\\- \( \mathbf{S} \) is the spin angular momentum.[/tex]
Energy Levels and Quantum Numbers
The total angular momentum \( \mathbf{J} \) is given by:
[tex]\[ \mathbf{J} = \mathbf{L} + \mathbf{S} \][/tex]
For l = 2 and [tex]\( s = \frac{1}{2} \)[/tex]:
[tex]- \( j \) can be \( \frac{3}{2} \) or \( \frac{5}{2} \).[/tex]
Energy Shift Due to Spin-Orbit Coupling
The energy shift due to spin-orbit coupling is:
[tex]\[ \Delta E_{SO} = \frac{\xi(r)}{2} \left[ j(j+1) - l(l+1) - s(s+1) \right] \][/tex]
For a hydrogen atom, [tex]\( \xi(r) \)[/tex] can be approximated, but let's use the general formula for now.
Calculating the Energy Shifts
For [tex]\( j = \frac{5}{2} \):[/tex]
[tex]\[ \Delta E_{SO}(j = \frac{5}{2}) = \frac{\xi(r)}{2} \left[ \frac{5}{2} \left( \frac{5}{2} + 1 \right) - 2(2+1) - \frac{1}{2} \left( \frac{1}{2} + 1 \right) \right] \][/tex]
[tex]\[ = \frac{\xi(r)}{2} \left[ \frac{5}{2} \times \frac{7}{2} - 6 - \frac{1}{2} \times \frac{3}{2} \right] \][/tex]
[tex]\[ = \frac{\xi(r)}{2} \left[ \frac{35}{4} - 6 - \frac{3}{4} \right] \]\[ = \frac{\xi(r)}{2} \left[ \frac{35 - 24 - 3}{4} \right] \]\[ = \frac{\xi(r)}{2} \left[ \frac{8}{4} \right] \]\[ = \frac{\xi(r)}{2} \times 2 \]\[ = \xi(r) \][/tex]
For [tex]\( j = \frac{3}{2} \):[/tex]
[tex]\[ \Delta E_{SO}(j = \frac{3}{2}) = \frac{\xi(r)}{2} \left[ \frac{3}{2} \left( \frac{3}{2} + 1 \right) - 2(2+1) - \frac{1}{2} \left( \frac{1}{2} + 1 \right) \right] \][/tex]
[tex]\[ = \frac{\xi(r)}{2} \left[ \frac{3}{2} \times \frac{5}{2} - 6 - \frac{1}{2} \times \frac{3}{2} \right] \]\[ = \frac{\xi(r)}{2} \left[ \frac{15}{4} - 6 - \frac{3}{4} \right] \]\[ = \frac{\xi(r)}{2} \left[ \frac{15 - 24 - 3}{4} \right] \]\[ = \frac{\xi(r)}{2} \left[ \frac{-12}{4} \right] \]\[ = \frac{\xi(r)}{2} \times (-3) \]\[ = -\frac{3}{2} \xi(r) \][/tex]
Summary of Energy Shifts
[tex]- For \( j = \frac{5}{2} \): Energy shift \( \Delta E_{SO} = \xi(r) \)\\- For \( j = \frac{3}{2} \): Energy shift \( \Delta E_{SO} = -\frac{3}{2} \xi(r) \)[/tex]
These energy shifts split the [tex]\( d \)-state into two levels due to the spin-orbit interaction, with the \( j = \frac{5}{2} \) state being higher in energy than the \( j = \frac{3}{2} \) state. This splitting is crucial in explaining fine structure in atomic spectra.[/tex]
