Question

The wave number of the first line of the Balmer series of hydrogen is $15200c{m^{ - 1}}$. The wave number of first line of Balmer series for $L{i^{2 + }}$ ion is:
A.$15200c{m^{ - 1}}$
B.$60800c{m^{ - 1}}$
C.$76000c{m^{ - 1}}$
D.$136800c{m^{ - 1}}$

Answer 1

Hint:Wave number is defined as the number of waves in a unit distance. It is reciprocal of wavelength (which is length of single cycle in a wave). Wavelength can be measured as velocity of the wave divided by its frequency. Hence wave number can be measured as frequency of the wave divided by the velocity of the wave.

Complete step by step answer:
First of let us talk about the wave number, wavelength and five series of hydrogen:
Wave number is defined as the number of waves in a unit distance. It is reciprocal of wavelength (which is length of single cycle in a wave). Wavelength can be measured as velocity of the wave divided by its frequency. Hence wave number can be measured as frequency of the wave divided by the velocity of the wave.
The five series are: Balmer, Lyman, Paschen, Brackett and Pfund.
When the spectrum (i.e. the arrangement of different wavelengths in either increasing order or decreasing order) of hydrogen was studied then these five series were found in the spectra.
Rydberg formula: The relation between energy difference between different energy levels and wavelength of absorbed or emitted photons is given by Scientist Rydberg that’s why the relation is known as Rydberg formula which is as: $\dfrac{1}{\lambda } = R{Z^2}[\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_h}^2}}]$ where
$\lambda $ is wavelength of the wave
$R$ is Rydberg constant i.e. $1.097{m^{ - 1}}$
$Z$ is atomic number of the element
${n_i}$ is lower energy level and ${n_h}$ is higher energy level.
Lyman series: By the Rydberg formula if ${n_i}$ (i.e. the lower energy level) is $1$ and ${n_h}$ (i.e. the higher energy levels) are $2,3,4,..$, then the series of wavelength will be in Lyman series. If ${n_h}$ is two then it will represent the first line of the Lyman series and if ${n_h}$ is three then it will represent the second line of Lyman series and so on.
Balmer series: By the Rydberg formula if ${n_i}$ (i.e. the lower energy level) is $2$ and ${n_h}$ (i.e. the higher energy levels) are $3,4,5,..$, then the series of wavelength will be in Balmer series. If ${n_h}$ is three then it will represent the first line of the Balmer series and if ${n_h}$ is four then it will represent the second line of the Balmer series and so on.
Paschen series: By the Rydberg formula if ${n_i}$ (i.e. the lower energy level) is $3$ and ${n_h}$ (i.e. the higher energy levels) are $4,5,6,..$, then the series of wavelength will be in Paschen series. If ${n_h}$ is four then it will represent the first line of the Paschen series and if ${n_h}$ is five then it will represent the second line of the Paschen series and so on.
Brackett series: By the Rydberg formula if ${n_i}$ (i.e. the lower energy level) is $4$ and ${n_h}$ (i.e. the higher energy levels) are $5,6,7,..$, then the series of wavelengths will be in the Brackett series. If ${n_h}$ is five then it will represent the first line of the Brackett series and if ${n_h}$ is six then it will represent the second line of the Brackett series and so on.
Pfund series: By the Rydberg formula if ${n_i}$ (i.e. the lower energy level) is $5$ and ${n_h}$ (i.e. the higher energy levels) are $6,7,8,..$, then the series of wavelength will be in Pfund series. If ${n_h}$ is six then it will represent the first line of the Pfund series and if ${n_h}$ is seven then it will represent the second line of Pfund series and so on.
So here we are given with the first line of Balmer series of hydrogen as $15200c{m^{ - 1}}$ and we know that atomic number is hydrogen is one i.e. ${v_H} = \dfrac{1}{{{\lambda _H}}} = R \times {1^2}[\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}] = 15200c{m^{ - 1}}$
Now, we have to find the first line of the Balmer series of lithium. We know that the atomic number of lithium is three. So, $\dfrac{1}{{{\lambda _{Li}}}} = R \times {3^2}[\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}] = 9{v_H} = 9 \times 15200$
${v_{L{i^{2 + }}}} = 138600c{m^{ - 1}}$.

So option D is correct.

Note:
The elements or species i.e. ions which have only one electron follows these spectrum i.e. Balmer, Lyman, Paschen, Brackett and Pfund. For example: $H{e^ + },L{i^{2 + }}$ and so many which have only electrons in their shells.