Question

What electronic transition in $L{i^{2 + }}$ ​produces the radiation of the same wavelength as the first line in the Balmer's series of hydrogen spectrum?
A. ${n_2} = 3\,to\,{n_1} = 2$
B. ${n_2} = 6\,to\,{n_1} = 3$
C. ${n_2} = 9\,to\,{n_1} = 6$
D. ${n_2} = 9\,to\,{n_1} = 8$

Answer 1

When electrons in a molecule are excited from one energy level to a higher energy level, this is known as a molecular electronic transition. The energy change associated with this transition reveals a molecule's structure and influences a variety of chemical attributes, including colour.

Complete step by step answer:
The energy of an electron in the nth orbit, according to Bohr's Theory, is:
${E_n} = - {R_H}\left( {\dfrac{{{Z^2}}}{{{n^2}}}} \right)\,J$
where ${R_H} = 2.18 \times {10^{ - 18}}$ and is $Z$ the atomic number of an atom or ion having one electron.

With the emission of radiation, the electron jumps from higher orbit/energy level ${n_2}$ to lower energy state ${n_1}$ in the atomic emission spectra of Hydrogen. The difference in energy between the two states involved equals the energy of the emitted radiation:
$\Delta E = {E_{{n_1}}} - {E_{{n_2}}}$

As a result of solving for the first line of the Balmer series, ${n_1} = 2$ and ${n_2} - {n_1} = 1$ for Hydrogen, $Z = 1$ emitted radiation energy is:
$\Delta E = - {R_H}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)$F
For $L{i^{2 + }}\left( {Z = 3} \right)$ , energy of radiation is:
$\Delta E = - {R_H} \cdot {3^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
The energy difference between transition stages should be the same for the two radiations to have the same wavelength. Hence,
${1^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) = {3^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \\\ \Rightarrow \dfrac{1}{{{2^2}}} = \dfrac{{{3^2}}}{{n_1^2}}\,and \,\dfrac{1}{{{3^2}}} = \dfrac{{{3^2}}}{{n_2^2}}$
Hence, ${n_1} = 6$ and ${n_2} = 9$ transition is the answer.

So, the correct option is C.

Note: The Balmer series is particularly valuable in astronomy because, due to the abundance of hydrogen in the universe, the Balmer lines arise in many celestial objects and are thus widely detected and relatively strong compared to lines from other elements. The relative strength of spectral lines is particularly essential in the spectral classification of stars, which is primarily a determination of surface temperature. The Balmer series in particular is very important.