Calculate the wavelength of the radiation emitted when an electron in a hydrogen atom transitions from $n = 4$ to $n = 2$ (Balmer series). (Rydberg constant $R_H = 1.097 \times 10^7$ m⁻¹)

$$\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = 1.097\times10^7\left(\frac{1}{4} - \frac{1}{16}\right)$$

$$= 1.097\times10^7 \times \frac{3}{16} = 2.057\times10^6 m^{-1}$$

$$\lambda = \frac{1}{2.057\times10^6} \approx 486 nm$$

(This is the blue-green H$_\beta$ line.)