## Solution Given: - Temperature, \( T = 500 \) K - Find **relative population** of hydrogen atoms in \( n = 1, 2, 3, 4 \) (energy levels) - **Multiplicity** must be included. ### 1. **Boltzmann Distribution** The relative population of atoms in energy level \( i \) is given by: \[ \frac{N_i}{N_1} = \frac{g_i}{g_1} \exp\left(-\frac{E_i - E_1}{kT}\right) \] Where: - \( N_i \): Number of atoms in level \( i \) - \( g_i \): Degeneracy (multiplicity) of level \( i \) - \( E_i \): Energy of level \( i \) - \( k \): Boltzmann constant (\( 8.617 \times 10^{-5} \) eV/K) - \( T \): Temperature in Kelvin --- ### 2. **For Hydrogen Atom:** - **Energy levels:** \( E_n = -13.6\,\text{eV}/n^2 \) - **Multiplicity:** \( g_n = 2n^2 \) (factor of 2 for spin) --- #### Calculate \( E_n \): \[ \begin{align*} E_1 &= -13.6\, \text{eV} \\ E_2 &= -13.6/4 = -3.4\, \text{eV} \\ E_3 &= -13.6/9 = -1.511\, \text{eV} \\ E_4 &= -13.6/16 = -.85\, \text{eV} \\ \end{align*} \] --- #### Calculate \( g_n \): \[ \begin{align*} g_1 &= 2 \times 1^2 = 2 \\ g_2 &= 2 \times 2^2 = 8 \\ g_3 &= 2 \times 3^2 = 18 \\ g_4 &= 2 \times 4^2 = 32 \\ \end{align*} \] --- #### Calculate \( \Delta E_{n1} = E_n - E_1 \): \[ \begin{align*} \Delta E_{21} &= -3.4 - (-13.6) = 10.2\, \text{eV} \\ \Delta E_{31} &= -1.511 - (-13.6) = 12.089\, \text{eV} \\ \Delta E_{41} &= -.85 - (-13.6) = 12.75\, \text{eV} \\ \end{align*} \] --- ### 3. **Plug into Boltzmann Equation** Boltzmann constant: \[ k = 8.617 \times 10^{-5} \ \text{eV/K} \] \[ kT = 8.617 \times 10^{-5} \times 500 = .431 \ \text{eV} \] Now compute ratios: \[ \frac{N_n}{N_1} = \frac{g_n}{g_1} \exp\left(-\frac{\Delta E_{n1}}{kT}\right) \] --- #### **For \( n=2 \):** \[ \frac{N_2}{N_1} = \frac{8}{2} \exp\left(-\frac{10.2}{.431}\right) = 4 \times \exp(-23.68) \] \[ \exp(-23.68) \approx 5.17 \times 10^{-11} \] \[ \frac{N_2}{N_1} \approx 2.07 \times 10^{-10} \] --- #### **For \( n=3 \):** \[ \frac{N_3}{N_1} = \frac{18}{2} \exp\left(-\frac{12.089}{.431}\right) = 9 \times \exp(-28.05) \] \[ \exp(-28.05) \approx 6.62 \times 10^{-13} \] \[ \frac{N_3}{N_1} \approx 5.96 \times 10^{-12} \] --- #### **For \( n=4 \):** \[ \frac{N_4}{N_1} = \frac{32}{2} \exp\left(-\frac{12.75}{.431}\right) = 16 \times \exp(-29.58) \] \[ \exp(-29.58) \approx 1.36 \times 10^{-13} \] \[ \frac{N_4}{N_1} \approx 2.18 \times 10^{-12} \] --- ## **Summary Table** | Level | \( g_n \) | \( E_n \) (eV) | \( \Delta E_{n1} \) (eV) | \( N_n/N_1 \) | |-------|--------|----------------|-------------------------|--------------------| | 1 | 2 | -13.6 | | 1 | | 2 | 8 | -3.4 | 10.2 | \( 2.07 \times 10^{-10} \) | | 3 | 18 | -1.511 | 12.089 | \( 5.96 \times 10^{-12} \) | | 4 | 32 | -.85 | 12.75 | \( 2.18 \times 10^{-12} \) | --- ## **Conclusion** - **Almost all hydrogen atoms are in the ground state (\( n = 1 \)) at 500 K.** - **Very few are in \( n=2, 3, \) or \( 4 \).** Let me know if you want the absolute numbers or further explanation!