# Energy Difference for Hydrogen Atom in 4f State in Magnetic Field ## Step 1: Find the Energy Difference Formula **Given:** - Hydrogen atom in the $4f$ state ($l = 3$). - Magnetic field $B = 1.08~\text{T}$. - Ignore electron spin (so, only orbital magnetic moment matters). The energy shift due to the Zeeman effect (ignoring spin) is: \[ \Delta E = \mu_B B m_l \] where $\mu_B$ = Bohr magneton = $9.274 \times 10^{-24}~\text{J/T}$, $m_l$ = magnetic quantum number ($-l, -l+1, ..., l$). **For $l = 3$ (4f state):** Possible $m_l$ values: $-3, -2, -1, , 1, 2, 3$. - **Greatest possible energy:** $m_l = +3$ - **Least possible energy:** $m_l = -3$ So, \[ \Delta E_{\text{max}} - \Delta E_{\text{min}} = \mu_B B (m_{l,\text{max}} - m_{l,\text{min}}) = \mu_B B (3 - (-3)) = 6 \mu_B B \] **Explanation:** The energy difference comes from the splitting of the magnetic sublevels in the external field. The largest difference is between $m_l = +3$ and $m_l = -3$. --- ## Step 2: Plug in Values and Calculate Plug in the values: - $\mu_B = 9.274 \times 10^{-24}~\text{J/T}$ - $B = 1.08~\text{T}$ \[ \Delta E = 6 \times (9.274 \times 10^{-24}~\text{J/T}) \times 1.08~\text{T} \] \[ = 6 \times 9.274 \times 1.08 \times 10^{-24}~\text{J} \] \[ = 6 \times 10.016 \times 10^{-24}~\text{J} \] \[ = 60.096 \times 10^{-24}~\text{J} \] \[ = 6.0096 \times 10^{-23}~\text{J} \] **Convert to eV:** $1~\text{eV} = 1.602 \times 10^{-19}~\text{J}$ \[ \Delta E~(\text{eV}) = \frac{6.0096 \times 10^{-23}~\text{J}}{1.602 \times 10^{-19}~\text{J/eV}} \] \[ = 3.75 \times 10^{-4}~\text{eV} \] **Explanation:** We substituted the known constants and calculated the energy difference both in joules and eV. --- ## Final Answers - **Energy difference (J):** $\boxed{6.01 \times 10^{-23}~\text{J}}$ - **Energy difference (eV):** $\boxed{3.75 \times 10^{-4}~\text{eV}}$ --- ## Level Diagram  *Zeeman splitting of the $4f$ state of hydrogen in a magnetic field. The largest energy difference is between $m_l = -3$ and $m_l = +3$.*