Question

The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series in the spectrum of hydrogen is _______.

Answer 1

Wavelength:

The distance between two successive crests or troughs of a wave is known as its wavelength.

Lyman series:

It is the series of transitions and resulting emission lines of the hydrogen atom, as an electron goes from n > 1 to n = 1.

Balmer series:

It is the series of transitions and resulting emission lines of the hydrogen atom, as an electron goes from n > 2 to n = 2.

The spectrum of hydrogen:

1. When electrons in an atom or molecule absorb energy and become excited, they leap from a lower to a higher energy level, and when they return to their original states, they emit radiation.
2. This phenomenon also accounts for the hydrogen emission spectrum, sometimes known as the hydrogen emission spectrum.

Calculation:

Step 1: Given data

$n_1=1$

$n_2=2$

$n_3=3$

Step 2: Formula used

The wavelength for Lyman series,

${\lambda }_L=E_{n_2}-E_{n_1}\alpha (-\frac{1}{{n_1}^2}+\frac{1}{{n_2}^2})$

The wavelength for Balmer series,

${\lambda }_B=E_{n_3}-E_{n_2}\alpha (-\frac{1}{{n_1}^2}+\frac{1}{{n_2}^2})$

Step 3: Calculating the ratio

When the electron's energy transitions from n=2 to n=1, the Lyman Series wavelength will be the longest.

${\lambda }_L=E_2-E_1\alpha (-\frac{1}{1^2}+\frac{1}{2^2})=\frac{3}{4}$

Here,

$E_2$ and $E_1$ are the energy of shell 2 and 1 respectively.

When the electron's energy transitions from n=3 to n=2, the Balmer Series wavelength will be the longest.

${\lambda }_B=E_3-E_2\alpha (-\frac{1}{3^2}+\frac{1}{2^2})=\frac{5}{36}$

Here,

$E_3$ and $E_2$ are the energy of shells 3 and 2 respectively.

The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series

$$\begin{gathered} \frac{{\lambda }_L}{{\lambda }_B}=\frac{E_2-E_1}{E_3-E_2} \\ =\frac{\frac{3}{4}}{\frac{5}{36}} \\ =\frac{27}{5} \end{gathered}$$

Hence, $\frac{{\lambda }_L}{{\lambda }_B}=\frac{5}{27}$ as $E\propto \frac{1}{\lambda }$

Therefore, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series in the spectrum of hydrogen is $\frac{5}{27}$.