Hydrogen Spectrum - Meaning, Definition, Relation, FAQs

For chemistry as a science, the Line Spectrum of Hydrogen-Like Atoms occupies a prominent place in the consideration of chemical principles of atomic structure and action. Developed from some principles established by Neil Bohr and Werner Heisenberg, it is one of the most characteristic phenomena illustrating the interactions between particles inside atoms. When analyzing these atoms, scholars investigate interactions of individual isolated quanta of light produced or absorbed by the atoms when changing their energy levels, which leads to exploring the quantum reality in which electrons orbit nuclei. It also reveals significant information concerning the quantized aspect of energy, as well as the foundation for interpretations of the quanta make-up, stability and behaviours of the various elements. From hydrogen as the cleaner fluid, to its counterparts, the Line Spectrum is rich in finding the mechanics of atomic physics, to this date continuing the challenge in place for chemistries.

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In this article, we will cover the concept of the Line spectrum of hydrogen. This concept falls under the broader category of Classification of elements and the periodic table, which is a crucial chapter in Class 11 chemistry. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Let us study the Line spectrum of hydrogen in detail to gain insights into this topic and solve a few related problems.

Line Spectrum of Hydrogen-Like Atoms

When an electric discharge is passed through gaseous hydrogen, the H₂ molecules dissociate and the energetically excited hydrogen atoms produced emit electromagnetic radiation of discrete frequencies. These radiations are emitted because of electronic transitions upon de-excitation to different energy levels and based on the final energy level of transition, the hydrogen spectrum consists of several series of lines named after their discoverers like the Lyman series, Balmer Series, Paschen Series, Bracket Series, Pfund Series.

Related Topics,

• Atomic number mass number
• Quantum Numbers
• Bohrs Model
• Isotope and Isobars
• Proton Neutron Discovery
  
• NMR Spectroscopy

Hydrogen Spectrum Wavelength

$
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
$

Where $\mathrm{R}$ is called the $\underline{\text { Rydberg }}$ constant, $\mathrm{R}=109677 \mathrm{~cm}^{-1}, \mathrm{Z}$ is the atomic number $n_1=$ final orbit occupied after de-excitation $=1,2,3 \ldots$.
$\mathrm{n}_2=$ initial orbit occupied before de-excitation

Lyman Series spectrum:

Transition of electrons from higher orbits to n=1 result in the Lyman Series

n₁= 1 and n₂= 2, 3, 4....

For H atom, this lies in Ultraviolet region. For elements with higher Z, the Balmer lines lie in the Ultraviolet region

Balmer Series Spectrum:

Transition of electrons from higher orbits to n=2 result in the Balmer Series

Where n₁= 2 and n₂= 3, 4, 5, 6....

For H atom, this generally lies in visible region.

Paschen, Bracket and Pfund Series spectrums:

Transition of electrons from higher orbits to n=3, 4 and 5 respectively result in the Paschen, Bracket and the Pfund Series

These lines lie in the Infrared Region for H atom.

Recommended topic video on(hydrogen atom)

Solved Examples Based On-Line spectrum of hydrogen

Example1: For the emission line of atomic hydrogen from $n_i=8$ to $n_f=n$, the plot of wave number $(\bar{v})$ against $\left(\frac{1}{n^2}\right)$ will be (The Rydberg constant is in wave number unit)

1) Linear with intercept -R_H

2) Non-linear

3) Linear with slope R_H

4) (correct) Linear with slope -R_H

Solution:

As We Know The Formula

$
\begin{aligned}
& \frac{1}{\lambda}=\bar{\nu}=R_H Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\
& n_1 \rightarrow 8 \text { to } n_2 \rightarrow n \text { and } Z=1 \\
& \bar{\nu}=R_H(1)^2\left(\frac{1}{8^2}-\frac{1}{n^2}\right) \\
& \bar{\nu}=\frac{R_H}{64}-\frac{R_H}{n^2}
\end{aligned}
$

Comparing $y=m x+c$
So, Slope $(m)=-R_H$

Hence, the answer is the option (4).

Example2:The transition of electrons from higher orbits to $n=2$ results in the Balmer Series Where $n_1=2$ and $n_2=3,4,5,6 \ldots$ For the $\mathrm{H}$ atom, this generally lies in the visible region. For any given series of spectral lines of atomic hydrogen, let $\Delta \bar{v}=\bar{v}_{\max }-\bar{v}_{\min }$ be the difference in maximum and minimum frequencies in $\mathrm{cm}^{-1}$.

The ratio $\Delta \bar{v}_{\text {Lyman }} / \Delta \bar{v}_{\text {Balmer }}$ is :
1) $4: 1$
2) (correct) $9: 4$
3) $5: 4$
4) $27: 5$

Solution:

For Lyman Series : $\bar{\nu}_{\max } \Rightarrow \infty \rightarrow 1 \quad$ (electron jump)

\begin{aligned}
& \bar{\nu}_{\max }=R(1)^2\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]=R \\
& \bar{\nu}_{\min } \Rightarrow 2 \rightarrow 1 \\
& \bar{\nu}_{\min }=R(1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3 R}{4} \\
& \left(\bar{\nu}_{\text {max }}-\bar{\nu}_{\text {min }}\right)=R-\frac{3 R}{4}=\frac{R}{4}
\end{aligned}

For Balmer Series :
$
\begin{gathered}
\bar{\nu}_{\max } \Rightarrow \infty \rightarrow 2 \\
\bar{\nu}_{\max }=R(1)^2\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=\frac{R}{4} \\
\bar{\nu}_{\min } \Rightarrow 3 \rightarrow 2 \\
\bar{\nu}_{\max }=R(1)^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{9-4}{36} R=\frac{5 R}{36} \\
\bar{\nu}_{\max }-\bar{\nu}_{\min }=\frac{R}{4}-\frac{5 R}{36}=\frac{4 R}{36}=\frac{R}{9}
\end{gathered}
$

So,

Ratio $=\frac{\Delta \bar{\nu}_{\text {Lyman }}}{\Delta \bar{\nu}_{\text {Balmer }}}=\frac{R}{4} \times \frac{9}{R}=9: 4$

Hence, the answer is the option (2).

Example 3:Heat treatment of muscular pain involves radiation of a wavelength of about $900 \mathrm{~nm}$. Which spectral line of $\mathrm{H}$-atom is suitable for this purpose?
$
\left[R_H=1 \times 10^5 \mathrm{~cm}^{-1}, h=6.6 \times 10^{-34} \mathrm{Js}, c=3 \times 10^8 \mathrm{~ms}^{-1}\right]
$
1) Paschen, $5 \rightarrow 3$
2) Lyman, $\infty \rightarrow 1$
3) Balmer, $\infty \rightarrow 2$
4) (correct) Paschen, $\infty \rightarrow 3$

Solution:

Line Spectrum of Hydrogen atoms -

$\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

- wherein

Where $R$ is called the Rydberg constant, $R=1.097 \times 10^7, \mathrm{Z}$ is the atomic number

\begin{aligned}
& n_1=1,2,3 \ldots \\
& n_2=n_1+1, n_1+2 \ldots \ldots
\end{aligned}

Balmer Series Spectrum -
$
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
$

Where $n_1=2$ and $n_2=3,4,5,6 \ldots$.
It lies in the visible region
Paschen, $\infty \rightarrow 3$ is correct
$
\begin{aligned}
& \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=10^7\left(\frac{1}{(3) 2}-\frac{1}{\infty}\right) \\
& \lambda=9 \times 10^{-7} \mathrm{~m} \\
& \lambda=900 \mathrm{~nm} .
\end{aligned}
$

Hence, the answer is the option (4).

Example 4: The pfund series wavelength of H-atom lies in which part of EM waves:

1) UV region

2) Visible region

3) (correct) Infrared region

4) Both 1 and 3

Solution:

For Pfund Series
$
\begin{aligned}
& \frac{1}{\lambda^{\max }}=R\left(\frac{1}{5^2}-\frac{1}{6^2}\right)=R\left(\frac{11}{25 * 36}\right) \\
& \lambda^{\max }=\frac{900}{11 R} \\
& \frac{1}{\lambda^{\min }}=R\left(\frac{1}{5^2}-\frac{1}{\infty^2}\right)=\frac{R}{25} \\
& \lambda^{\min }=\frac{25}{R} \\
& \text { when } \frac{1}{R}=912 A^{\circ}
\end{aligned}
$

Wavelength lies in infrared regions.

Hence, the answer is the option (3).

Also Read:

• NCERT solutions for Class 11 Chemistry Chapter 2 Structure of Atom
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Conclusion

Therefore, while considering the case of the Line Spectrum of Hydrogen-Like Atoms, one can only be amazed by the complex nature of the phenomena taking place at the atomic level. studying detailed behavior and using conceptual analysis, it has been found that atoms either emitting or taking in light follow strange mechanical patterns of the quantum world that control our existence. Returning to the historical perspective in the global transition from Bohr and Heisenberg to today’s developments in spectroscopy, one can confidently conclude that the Line Spectrum is a fundamental platform for defining the very nature of matter. Hence, its importance is not limited to chemistry but can be seen crossing over into physics, astronomy, and more. Thus, by examining the sharp contours that write the spectral constitution on the body of its culture and the lines of the constellations in the universe that represent our limbs, we do not merely analyze how atoms communicate but decipher the alphabet of the cosmos. Thus, the Line Spectrum of Hydrogen-Like Atoms stays as the shining logo of this free spirit which will challenge us eternally to journey forward into the realms of the unknown.

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