Question

The first member of the Balmer series of the Hydrogen atom has a wavelength of $6561\mathrm{\AA }$. The wavelength of the second member of the Balmer series (in nm) is

Answer 1

Balmer series:

1. One of a group of six named series that describe the spectral line emissions of the hydrogen atom is the Balmer series, often known as the Balmer lines in atomic physics.
2. The Balmer formula, an empirical equation developed by Johann Balmer in 1885, is used to compute the Balmer series.
3. The wavelength is given by the general formula $\frac{1}{\mathrm{\lambda }}=R_H(\frac{1}{4}-\frac{1}{n^2})$, where n is the level of the orbital.

Using the formula,

$$\begin{gathered} \Delta E=\frac{hC}{\lambda } \\ =\Delta E_o\left[\frac{1}{4}-\frac{1}{9}\right]\mathit{.}\mathit{.}\mathit{.}..\left(1\right) \\ \frac{hC}{{\lambda }^{\text{'}}}=\Delta E_o\left[\frac{1}{4}-\frac{1}{16}\right].....\left(2\right) \end{gathered}$$

From equations (1) and (2),

$$\begin{gathered} \Rightarrow \frac{{\lambda }^{\text{'}}}{\lambda }=\frac{5\times 16}{9\times 4\times 3} \\ \Rightarrow {\lambda }^{\text{'}}=\frac{5\times 4\times 656.1}{9\times 3} \\ =486\mathrm{nm} \end{gathered}$$

Hence, the wavelength of the second member of the Balmer series is 486 nm.