Question

One of the lines in the emission spectrum of $L{i^{2 + }}$ has the same wavelength as that of the ${2^{nd}}$ line of the Balmer series in the hydrogen spectrum. The electronic transition corresponding to this line is $n = 12 \to n = x$. Find the value of $x$.
A. $8$
B. $6$
C. $7$
D. $5$

Answer 1

Hint: We already know that an element's absorption spectrum, which contains dark lines in the same locations as its emission spectrum's bright lines, is defined by the number of spectral lines. To get the wavelength of the Balmer series in this problem, we shall use Rydberg's formula. In the Balmer series, the electron transition on the second line, which occurs at the least energy transition, yields the greatest wavelength.

Formula used:
The Rydberg’s Formula is as follows:
$\dfrac{1}{\lambda } = R \times {Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right){m^{ - 1}}$
Here, $\lambda $ is the wavelength, $R$ is the Rydberg constant and $n_1,n_2$ are the energy states.

Complete step by step solution:
In the question, it is given that the electronic transition corresponding to this line is $n = 12 \to n = x$ and the emission spectrum $L{i^{2 + }}$ has a line with the same wavelength as the Balmer series ${2^{nd}}$ line for hydrogen. Use Rydberg’s Formula to determine the value of $x$,
$\dfrac{1}{\lambda } = R \times {Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right){m^{ - 1}}$
As an atomic number of hydrogen is $1$ and as the ground state of the Balmer series is $2$, then we have for ${2^{nd}}$ line of the Balmer series in the hydrogen atom:
$Z = 1 \\$
$\Rightarrow {n_1} = 2 \\$
Now, the transition of the electron must occur in a way that releases the least amount of energy in order for the maximal wavelength to be emitted. Additionally, the transfer of electrons between the two closest energy states should occur for the least energy. In order to know that, the ground state for the Balmer series is ${n_1} = 2$, the just-next-energy state, which is ${n_2} = 4$, would be the nearest energy state for least energy,

Now, substituting the given values in the above formula, then we obtain:
$\dfrac{1}{\lambda } = R(1)\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right) \\$ $\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right) \\$ $\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{3}{{16}}} \right) \\$
Similarly, as an atomic number of lithium is $3$and as the ground state of the Balmer series is $x$, then we have for $x$line of the Balmer series in the lithium atom:
$Z = 3 \\$
${n_1} = x \\$
In order to know that, the ground state for the Balmer series is ${n_1} = 3$, the just-next-energy state, which is ${n_2} = 12$, would be the nearest energy state for least energy. Substituting the given values in the above formula, then we obtain:
$\dfrac{1}{\lambda } = R{(3)^2}\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{{12}^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{\lambda } = 9R\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{144}}} \right) \\$
From the given information, as we know that the emission spectrum of lithium and hydrogen has same wavelength, so:
$9R\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{144}}} \right) = R\left( {\dfrac{3}{{16}}} \right) \\$
$\Rightarrow 3\left( {\dfrac{{144 - {x^2}}}{{144{x^2}}}} \right) = \left( {\dfrac{1}{{16}}} \right) \\$
$\Rightarrow 6912 - 48{x^2} = 144{x^2} \\$
$\therefore x = \sqrt {\dfrac{{6912}}{{192}}} = 6 $

Hence, the correct option is B.

Note: Due to the abundance of hydrogen in the universe, the Balmer lines arise in many celestial objects and are therefore frequently observed and quite strong in comparison to lines from other elements. This makes the Balmer series very helpful in the field of astronomy.