Question

Which transition in \[{\rm{L}}{{\rm{i}}^{{\rm{2 + }}}}\] would have the same wavelength as the \[2 \to 4\] transition in \[{\rm{H}}{{\rm{e}}^ + }\] ion?
A. 3 to 6
B. 4 to 3
C. 5 to 6
D. 5 to 4

Answer 1

Hint: The emission of spectral lines takes place by the hydrogen atoms in the visible part of the spectrum. The wavelengths of the spectral lines are given by the formula of \[\dfrac{1}{\lambda } = {R_H}{Z^2}\left[ {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right]\]. Here, \[\lambda \] stands for wavelength, \[{R_H}\] is Rydberg's constant, Z is for atomic number, \[{n_i}\]stands for initial shell and the \[{n_f}\] stands for final shell.

Complete Step by Step Answer:
Here, we need to identify the transition of \[{\rm{L}}{{\rm{i}}^{{\rm{2 + }}}}\]in which wavelength is equal to the wavelength in the transition of \[2 \to 4\]in a helium ion.

Let's first calculate the Helium ion's wavelength using the above equation.
We know that the Z for He atom is 2, and given that transition of helium ion is given as 2 to 4.
So, we put the above values in the wavelength formula.
\[\dfrac{1}{\lambda } = {R_H}{2^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right]\]
\[\dfrac{1}{\lambda } = {R_H}4\left[ {\dfrac{{4 - 1}}{{16}}} \right]\]
\[\dfrac{1}{\lambda } = {R_H}\dfrac{{4 \times 3}}{{16}}\]
\[\dfrac{1}{\lambda } = {R_H}\dfrac{3}{4}\] …… (1)

Given that,
\[{\lambda _{{\rm{L}}{{\rm{i}}^{2 + }}}} = {\lambda _{{\rm{H}}{{\rm{e}}^ + }}}(2 \to 4)\]
Therefore, we can put the above value of wavelength in the equation of wavelength for the \[{\rm{L}}{{\rm{i}}^{{\rm{2 + }}}}\]ion. And we know, Z for lithium is 3.

So, the wavelength equation for \[{\rm{L}}{{\rm{i}}^{{\rm{2 + }}}}\]is,
\[\dfrac{1}{\lambda } = {R_H}{3^2}\left[ {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right]\]
\[{R_H}\dfrac{3}{4} = {R_H}{3^2}\left[ {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right]\]
\[\dfrac{3}{4} = {3^2}\left[ {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right]\]
\[\dfrac{3}{{4 \times 9}} = \left[ {\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}}} \right]\]
\[\dfrac{1}{{{n_i}^2}} - \dfrac{1}{{{n_f}^2}} = \dfrac{1}{{12}}\]
On solving the above quadratic equation, we get the values of \[{n_i} = 3\]and \[{n_f} = 6\] .
Therefore, option A is right.

Note: The phenomenon of emission of spectral lines by hydrogen atoms kept in a discharge lamp is due to the absorption of energy and the excitation of the electrons to a higher state of energy. When the electron comes back to its original state, the emission of energy in the form of radiation happens. This series was discovered by Balmer and the series got named after him.