Question

What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series?

Short answer

The required ratio is 4.

Step-by-step solution

Describe the energy of the photon emitted by a hydrogen atom jumping from one state to another:

The energy in a hydrogen atom depends on the energy of the electron. When an electron changes levels, it lowers energy and the atom emits photons. A photon is emitted when an electron moves from a higher energy level to a lower energy level.

The energy is given by,

$$\begin{gathered} \mathbf{E}\mathbf{=}\mathbf{A}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ \mathbf{Where}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{level}\mathbf{}{\mathbf{n}}_{\mathbf{2}}\mathbf{>}{\mathbf{n}}_{\mathbf{1}}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{A}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{eV}\mathbf{.} \end{gathered}$$ .....(1)

Find the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series:

Multiply both sides of the equation (1) with $\frac{1}{hc}$

$\frac{\mathrm{E}}{\mathrm{hc}}=\frac{\mathrm{A}}{\mathrm{hc}}\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

It is known that

$\frac{1}{\mathrm{\lambda }}=\frac{\mathrm{E}}{\mathrm{hc}}$

Substitute $\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$for E in the above eqution.

$\frac{1}{\mathrm{\lambda }}=\frac{\mathrm{A}}{\mathrm{hc}}\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

According to this equation, it can be seen that the shortest wavelength is when the left-hand side of this equation is the greatest, the Lyman series is associated with transitions to or from the n = 1 level of the hydrogen atom, so the shortest wavelength occur between ${\mathrm{n}}_1=1\mathrm{and}{\mathrm{n}}_2=\infty$.

The wavelength is,

$$\begin{gathered} \frac{1}{{\mathrm{\lambda }}_{\mathrm{L}}}=\frac{\mathrm{A}}{\mathrm{hc}}\left(1-\frac{1}{\infty }\right) \\ \frac{1}{{\mathrm{\lambda }}_{\mathrm{L}}}=\frac{\mathrm{A}}{\mathrm{hc}} \\ {\mathrm{\lambda }}_{\mathrm{L}}=\frac{\mathrm{hc}}{\mathrm{A}} \end{gathered}$$

The Balmer series is associated with transitions to or from the n = 2 level of the hydrogen atom, so the shortest wavelength occur between ${\mathrm{n}}_1=2\mathrm{and}{\mathrm{n}}_2=\infty$.

The wavelength is,

$$\begin{gathered} \frac{1}{{\mathrm{\lambda }}_{\mathrm{B}}}=\frac{\mathrm{A}}{\mathrm{hc}}\left(\frac{1}{4}-\frac{1}{\infty }\right) \\ \frac{1}{{\mathrm{\lambda }}_{\mathrm{B}}}=\frac{\mathrm{A}}{4\mathrm{hc}} \\ {\mathrm{\lambda }}_{\mathrm{B}}=\frac{4\mathrm{hc}}{\mathrm{A}} \end{gathered}$$

Find the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series.

$$\begin{gathered} \frac{{\mathrm{\lambda }}_{\mathrm{B}}}{{\mathrm{\lambda }}_{\mathrm{L}}}=\frac{{\displaystyle \frac{4\mathrm{hc}}{\mathrm{A}}}}{{\displaystyle \frac{\mathrm{hc}}{\mathrm{A}}}} \\ =\frac{4\mathrm{hc}}{\mathrm{A}}\times \frac{\mathrm{A}}{\mathrm{hc}} \\ =4 \end{gathered}$$

Therefore, the required ratio is 4.