Question

A spectroscopic instrument can resolve two nearby wavelengths $\lambda $ and $\lambda + \Delta \lambda $ if $\lambda /\Delta \lambda $ is smaller than $8000$ . This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument?

Answer 1

Hint: In this problem, to determine the number of spectral lines of the Balmer series of hydrogen, resolved by the spectroscopic instrument; we will apply Rydberg's formula for the hydrogen atom and then differentiate it in order to compute the ratio of $\lambda /\Delta \lambda $ which will ultimately help us in finding the correct solution.

Formula used:
The formula used in this problem is Rydberg’s formula which is defined as: -
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
where, ${R_H} = $ Rydberg constant for the hydrogen atom, ${n_1}{\text{ }}and{\text{ }}{n_2}$ are the numbers for low energy level and high energy level respectively.

Complete step by step solution:
We know that the expression for Rydberg’s formula can be stated as: -
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(for\,hydrogen\,atom)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(1)$
This formula is used for computing the wavelength of light that an atom's electron emits during its transition between different energy levels. Now it is given that the spectroscopic instrument is used to study the spectral lines of the Balmer series of hydrogen. Therefore, for Balmer’s series of lines ${n_1} = 2$
From the equation $(1)$, we get
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{n_2^2}}} \right)\, = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{{n_2^2}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(2)$

On differentiating equation $(2)$, we get
$\dfrac{{ - d\lambda }}{{{\lambda ^2}}} = {R_H}\left( {0 - \dfrac{{\left( { - 2} \right)}}{{n_2^3}}d({n_2})} \right) \\ $
$\dfrac{{ - d\lambda }}{\lambda } = \lambda {R_H} \cdot \dfrac{2}{{n_2^3}}d{n_2}$
From equation $(2)$, substitute the value of $\lambda \, = \dfrac{1}{{{R_H}}}\left( {\dfrac{{4n_2^2}}{{n_2^2 - 4}}} \right)\,$ in the above expression, we get
$\dfrac{{ - d\lambda }}{\lambda } = \dfrac{1}{{{R_H}}}\left( {\dfrac{{4n_2^2}}{{n_2^2 - 4}}} \right)\,{R_H} \cdot \dfrac{2}{{n_2^3}}d{n_2} \\ $
By simplifying it, we get
$ \Rightarrow \dfrac{{ - d\lambda }}{\lambda } = \dfrac{8}{{{n_2}(n_2^2 - 4)}}d{n_2} \\ $
Since, ${n_1} < {n_2}$ therefore, $d\lambda $ will always be negative and as a result, $ - d\lambda > 0$ (always positive).
$ \Rightarrow \dfrac{{\Delta \lambda }}{\lambda } = \dfrac{{8 \cdot \Delta {n_2}}}{{{n_2}(n_2^2 - 4)}} \\ $
But also, it is given that $\dfrac{\lambda }{{\Delta \lambda }} < 8000$ i.e., $\dfrac{{\Delta \lambda }}{\lambda } > \dfrac{1}{{8000}} \\ $
$ \Rightarrow \dfrac{1}{{8000}} < \dfrac{{8 \cdot \Delta {n_2}}}{{{n_2}(n_2^2 - 4)}} \\ $
For two nearby wavelengths, we can use $\Delta {n_2} = 1$
$\dfrac{1}{{8000}} < \dfrac{{8 \cdot \left( 1 \right)}}{{{n_2}^3 - 4{n_2}}} \Leftrightarrow {n_2}^3 - 4{n_2} < 64000 \\ $
$ \therefore {n_2}^3 - 4{n_2} - 64000 < 0 \\ $
This equation holds true up to ${n_2} = 40$ , the maximum value of ${n_2}$.
Thus, the total number of lines resolved will be $ = {n_2} - {n_1} = 40 - 2 = 38$.

Hence, approximately $38$ spectral lines of the Balmer series of hydrogen will be resolved by the given spectroscopic instrument.

Note: In this problem of Bohr’s model, the study of spectral lines of the Balmer series of hydrogen is done using a spectroscopic instrument. Therefore, to determine the number of lines resolved by this instrument, Rydberg’s formula for hydrogen atom must be used and the key points like $ - d\lambda > 0$ and $\Delta {n_2} = 1$ (for two nearby wavelengths) must be kept in mind while doing calculation part.