Question

Draw an energy level diagram for hydrogen atom. Calculate the wavelengths of (i) H$_\alpha$, (ii) H$_\beta$, and (iii) Series limit of Balmer series.

Hint:

Balmer series lines always end at $n_1=2$; H$_\alpha$ (red), H$_\beta$ (blue-green).

Answer 1

Step 1: Formula (Rydberg equation).
\[ \frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] For Balmer series: $n_1 = 2$, $n_2 = 3,4,5,...$ where $R_H = 1.097 \times 10^7 \, m^{-1}$.
Step 2: For H$_\alpha$ line ($n_2 = 3 \to 2$).
\[ \frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{9}\right) \] \[ = 1.097 \times 10^7 \left(\frac{5}{36}\right) \approx 1.523 \times 10^6 \, m^{-1} \] \[ \lambda_{H_\alpha} \approx 6563 \, \text{\AA} \]
Step 3: For H$_\beta$ line ($n_2 = 4 \to 2$).
\[ \frac{1}{\lambda} = R_H \left(\frac{1}{4} - \frac{1}{16}\right) = 1.097 \times 10^7 \times \frac{3}{16} \] \[ = 2.06 \times 10^6 \, m^{-1} \quad \Rightarrow \quad \lambda_{H_\beta} \approx 4861 \, \text{\AA} \]
Step 4: Series limit (when $n_2 \to \infty$).
\[ \frac{1}{\lambda} = R_H \left(\frac{1}{4} - 0\right) = \frac{R_H}{4} \] \[ \lambda = \frac{4}{R_H} \approx \frac{4}{1.097 \times 10^7} \approx 3.646 \times 10^{-7} \, m \] \[ \lambda \approx 3646 \, \text{\AA} \]
Step 5: Conclusion.
- H$_\alpha$: $6563 \, \text{\AA}$
- H$_\beta$: $4861 \, \text{\AA}$
- Balmer limit: $3646 \, \text{\AA}$