Question

The ratio of the minimum wavelength of the Lyman and Balmer series will be:

• A. $10$
• B. $0.5$
• C. $0.25$
• D. $1.25$

Answer 1

The correct option is C

$0.25$

The wavelength, Blamer Series, and Lyman series.

1. Wavelength- The distance between the crests and troughs of a wave motion is specified as the wavelength.
2. Balmer Series- It is the part of the hydrogen emission spectrum that shows electron transitions from $n>2$ to $n=2$ energy levels.
3. Lyman Series- It is the part of the hydrogen emission spectrum that shows electron transitions from $n>1$ to $n=1$ energy levels.

Derivation:

Expression of wavelength = $\frac{1}{\mathrm{\lambda }}=R\left[\frac{1}{{{\mathrm{n}}_1}^2}-\frac{1}{{{\mathrm{n}}_2}^2}\right]$

where R is Rydberg constant,

The last line has the shortest wavelength.

For the shortest wavelength, take, $n_2=\infty$,

$\frac{1}{\mathrm{\lambda }}=R\left[\frac{1}{{{\mathrm{n}}_1}^2}\right]$

${\mathrm{\lambda }}_{\mathrm{Series}\mathrm{limit}}=\frac{{{\mathrm{n}}_1}^2}{R}$

For the Lyman series $n_1=1$ , and for the Balmer series $n_1=2$.

Therefore ratio of the shortest wavelengths = $\frac{{\lambda }_{lyman}}{{\lambda }_{balmer}}$

$$\begin{gathered} \frac{{\left(n_{1Lyman}\right)}^2}{{\left(n_{1Balmer}\right)}^2}=\frac{1^2}{2^2} \\ =\frac{1}{4} \\ =0.25 \end{gathered}$$

Therefore, the correct option is (C) $0.25$.