Question

The Wave Number Of The First Line Of Balmer Series Of Hydrogen Is 15200 cm⁻¹ . The Wave Number Of The First Line Of Balmer Series For Li ²⁺ Ion Is

Answer 1

Wave number:

• The number of waves present in a unit distance or the reciprocal of wavelength is termed a wave number.
• It can be measured as the frequency of a wave divided by its velocity.
• The five series of Hydrogen or Hydrogen-like elements are: Lyman, Balmer, Paschen, Brackett, and Pfund.

Rydberg formula:

• A scientist named Rydberg established the relation between the energy difference between different energy levels and the wavelength of absorbed/ emitted photons.
• The following is known as the Rydberg formula.

$\overline{\nu }=\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]$ , where

$\overline{\nu }$= Wave number of the wave.

$\lambda$=Wavelength of the wave.

$R_H$=Rydberg Constant i.e. 1.097 m^-1

$Z$= Atomic number of the element

$n_1$= Lower energy level

$n_2$= Higher energy level

Step 1: Specifying the given credentials and assumptions:

• The wave number of the first line of the Balmer series of Hydrogen($\mathrm{H}$) = ${\overline{\nu }}_H$ = 15200 cm^-1
• For the first line of the Balmer series : $n_1=2and{n}_2=3$.
• The atomic number of Hydrogen($\mathrm{H}$) and Lithium-ion(${\mathrm{Li}}^{2+}$) are 1 and 3 respectively.
• Lithium-ion(${\mathrm{Li}}^{2+}$) is a Hydrogen-like element due to having an electron, hence, it exhibits these spectrums.

Step 2: Using the Rydberg formula:

• The wave number of the given series in Hydrogen($\mathrm{H}$) with atomic number Z = 1 is given as

$$\begin{gathered} \overline{{\nu }_H}=\frac{1}{{\lambda }_H}=R_H\times 1^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ \Rightarrow \overline{{\nu }_H}=15200c{m}^{-1}\left(Given\right) \end{gathered}$$

• In the case of Lithium-ion(${\mathrm{Li}}^{2+}$), the atomic number Z = 3. Hence, the following relation can be expressed:

$$\begin{gathered} \overline{{\nu }_{L{i}^{2+}}}=\frac{1}{{\lambda }_{L{i}^{2+}}}=R_H\times 3^2\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=9\times R_H\times \left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ \Rightarrow \overline{{\nu }_{L{i}^{2+}}}=9\times \overline{{\nu }_H} \\ \Rightarrow \overline{{\nu }_{L{i}^{2+}}}=9\times 15200{\mathrm{cm}}^{-1} \\ \Rightarrow \overline{{\nu }_{L{i}^{2+}}}=136800{\mathrm{cm}}^{-1} \end{gathered}$$