Question

What transition of ${{L}}{{{i}}^{{{ + 2}}}}$ spectrum will have same wavelength as that of second line of Balmer series in ${{H}}{{{e}}^{{ + }}}$ spectrum?

Answer 1

The wavelengths of spectral lines are given by the wavelength equation. We can find out the transition from which level by knowing the Rydberg constant and the value of ’n’.

Complete step by step answer:
Hydrogen atoms in a discharge lamp emits a series of lines in the visible part of the spectrum. This occurs when the atoms absorb energy and the electrons get excited to a higher energy state. When this electron comes back to its original state, it emits the energy in the form of radiations. This series was discovered by a Swiss teacher Johann Balmer, after whom the series was named as Balmer series. The equation for this series is given by;
$\dfrac{{{1}}}{{{\lambda }}}{{ = R[}}\dfrac{{{1}}}{{{{{2}}^{{2}}}}}{{ - }}\dfrac{{{1}}}{{{{{n}}^{{2}}}}}{{]}}$
where ${{\lambda }}$is the wavelength, ${{n = 1,2,3}}...$ and ${{R}}$ is a constant called Rydberg constant whose value is ${{R = 1}}{{.097 \times 1}}{{{0}}^{{7}}}{{{m}}^{{{ - 1}}{{.}}}}$.
Then in 1889, a scientist named Johannes Rydberg found several series of spectra and a formula that fits to the empirical formula of Balmer. This general relationship can be written as follows;
$\dfrac{{{1}}}{{{\lambda }}}{{ = R[}}\dfrac{{{1}}}{{{{n}}_{{f}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{i}}^{{2}}}}{{]}}\;{{{n}}_{{i}}}{{ > }}{{{n}}_{{f}}}$, where ${{{n}}_{{i}}}$stands for the initial level of the atom and ${{{n}}_{{f}}}$stands for the final level of the atom.
Here we have to find out the transition of ${{L}}{{{i}}^{{{ + 2}}}}$spectrum which has the same wavelength of the second line of the Balmer series of ${{H}}{{{e}}^{{ + }}}$spectrum. We will use the above equation to find out the value of ${{n}}$ in this situation. Then the equation will be as follows;
$\dfrac{{{1}}}{{{\lambda }}}{{ = R}}{{{Z}}^{{2}}}{{[}}\dfrac{{{1}}}{{{{n}}_{{f}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{i}}^{{2}}}}{{]}}$. Here we have the new term Z which is the atomic number of the element.
We have the wavelength of ${{H}}{{{e}}^{{ + }}}$ spectrum = $\dfrac{{{1}}}{{{{{\lambda }}_{{{H}}{{{e}}^{{ + }}}}}}}{{ = }}{{{2}}^{{2}}}{{[}}\dfrac{{{1}}}{{{{n}}_{{2}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{4}}^{{2}}}}{{]}}$ and
the wavelength of ${{L}}{{{i}}^{{{ + 2}}}}$spectrum = $\dfrac{{{1}}}{{{{{\lambda }}_{{{L}}{{{i}}^{{{ + 2}}}}}}}}{{ = }}{{{3}}^{{2}}}{{[}}\dfrac{{{1}}}{{{{n}}_{{f}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{i}}^{{2}}}}{{]}}$
Equating these two equations we will get,
${{{2}}^{{2}}}{{[}}\dfrac{{{1}}}{{{{n}}_{{2}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{4}}^{{2}}}}{{]}}$=${{{3}}^{{2}}}{{[}}\dfrac{{{1}}}{{{{n}}_{{f}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{i}}^{{2}}}}{{]}}$
${{{2}}^{{2}}}{{[}}\dfrac{{{1}}}{{{{{2}}^{{2}}}}}{{ - }}\dfrac{{{1}}}{{{{{4}}^{{2}}}}}{{]}}$=${{9[}}\dfrac{{{1}}}{{{{n}}_{{f}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{i}}^{{2}}}}{{]}}$
${{4[}}\dfrac{{{1}}}{{{4}}}{{ - }}\dfrac{{{1}}}{{{{16}}}}{{]}}\;{{ = 9[}}\dfrac{{{1}}}{{{{n}}_{{f}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{i}}^{{2}}}}{{]}} \\\ \Rightarrow \dfrac{{{4}}}{{{9}}}{{ \times }}\dfrac{{{3}}}{{{{16}}}}\;{{ = [}}\dfrac{{{1}}}{{{{n}}_{{f}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{i}}^{{2}}}}{{]}} \\\ \Rightarrow \dfrac{{{1}}}{{{{12}}}}\;{{ = }}\dfrac{{{1}}}{{{{n}}_{{f}}^{{2}}}}{{ - }}\dfrac{{{1}}}{{{{n}}_{{i}}^{{2}}}} \\\$
Which is a quadratic equation and on solving it we get the values of ${{{n}}_{{f}}}{{ = 3}} \;{{and }}\;{{{n}}_{{i}}}{{ = 6}}$.

Therefore, the transition from ${{n = 3}}\;{{to\; n\; = 6}}$ in ${{L}}{{{i}}^{{{ + 2}}}}$ N will have the same wavelength of the second line of Balmer series of ${{H}}{{{e}}^{{ + }}}$ spectrum.

Note: The emission of energy by atoms when they absorb and emit energy is the form of light is characteristic of an element. This characteristic of certain elements is used for the identification of those elements when they are in a compound. The wavelength equation can be used to find out the wavelength and the initial and final levels of an electron in an atom who have absorbed and emitted energy. The name of the series corresponding to different wavelengths are named after the scientists who discovered it.