Problem 1: First Balmer Line Wavelength

Given: Electron transitions from n=3 to n=2 in hydrogen Find: Wavelength

Step 1: Identify the formula: $\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

Step 2: Substitute values: $\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{9}\right)$

Step 3: Compute bracket: $\frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$

Step 4: Multiply: $\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{5}{36} = 1.524 \times 10^6 \text{ m}^{-1}$

Step 5: Invert: $\lambda = \frac{1}{1.524 \times 10^6} = 6.56 \times 10^{-7} \text{ m} = \boxed{656 \text{ nm (red)}}$