Discuss the spectral series of the hydrogen atom.

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Q. Discuss the spectral series of the hydrogen atom.

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Concept understanding — Hydrogen Spectral Series

When an individual gaseous element such as hydrogen is excited (by heating or an electrical discharge) and its emitted light is examined through a spectrometer, the result is not a smooth, continuous rainbow of colours but rather a set of sharp, discrete bright lines at very specific wavelengths - a line spectrum, unique to that particular element, functioning almost like a fingerprint that lets scientists identify a gas's composition from its light alone (including the composition of distant stars and the Sun).

For hydrogen specifically, every observed spectral-line wavelength is given by a single Rydberg-type formula,

\frac{1}{\lambda}=R\left(\frac{1}{n^2}-\frac{1}{m^2}\right),\qquad m>n\ \text{positive integers}

where $R=1.09737\times10^7\ \text{m}^{-1}$ is the Rydberg constant. Fixing the lower level $n$ and letting $m$ range over every larger integer generates one complete named series of spectral lines: $n=1$ gives the Lyman series (ultraviolet region); $n=2$ gives the Balmer series (visible region - the only series bright enough to be seen directly with the eye); $n=3$ gives the Paschen series (near infrared); $n=4$ gives the Brackett series (mid infrared); and $n=5$ gives the Pfund series (far infrared). Within any one series, successive lines (as $m$ increases) crowd progressively closer together, approaching a limiting wavelength - the series limit - as $m\to\infty$ .

This same formula also lets the total number of distinct spectral lines an excited hydrogen sample can emit be counted: if an electron has been excited up to level $n$ , it can subsequently make a transition down to any lower level, and the total number of distinct transitions (and hence distinct emitted wavelengths) possible works out to $n(n-1)/2$ . …

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Step 1. The governing formula. Every hydrogen spectral-line wavelength satisfies $\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)$ , with $R=1.09737\times10^7\ \text{m}^{-1}$ the Rydberg constant and $m>n$ positive integers, where the transition is from level $m$ down to level $n$ (emission) or the reverse (absorption).

Step 2. Lyman series. Setting $n=1$ and letting $m=2,3,4,\dots$ generates the Lyman series, lying in the ultraviolet region.

Step 3. Balmer series. Setting $n=2$ and $m=3,4,5,\dots$ generates the Balmer series, lying in the visible region - the only series directly observable with the naked eye.

Step 4. Paschen, Brackett and Pfund series. Setting $n=3,4,5$ respectively (with $m$ one greater and beyond, in each case) generates the Paschen, Brackett and Pfund series in turn, all lying in the infrared region (near, middle and far IR respectively). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set 55/3/11 markMCQ
$Q. In Bohr's model of the hydrogen atom, the ratio of the maximum frequency to the minimum frequency of light emitted in the Balmer series of the hydrogen spectrum is :$
(a) $\dfrac{11}{9}$
(b) $\dfrac{9}{5}$
(c) $\dfrac{11}{7}$
(d) $\dfrac{16}{7}$
›Reveal solutionSolution

Frequency is inversely proportional to wavelength, so the frequency ratio equals the wavelength ratio $\lambda_{\max}/\lambda_{\min}$ . Taking the visible band as $400$ nm (violet) to $720$ nm (red) gives $\dfrac{720}{400} = \dfrac{9}{5}$ — option (b).

Since $f = c/\lambda$ , the highest frequency corresponds to the shortest wavelength and the lowest frequency to the longest wavelength:
$\frac{f_{\max}}{f_{\min}} = \frac{c/\lambda_{\min}}{c/\lambda_{\max}} = \frac{\lambda_{\max}}{\lambda_{\min}}.$
The options fix the intended wavelength band. Keeping the violet edge at the standard $\lambda_{\min} = 400$ nm, only $\lambda_{\max} = 720$ nm gives a value that is still in the visible (red) region: …
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CBSE 2017Set ANNUAL1 markMCQ
$Q. In hydrogen atom, which of the following transitions produce a spectral line of maximum frequency ?$
(a) 6 → 2
(b) 2 → 1
(c) 4 → 3
(d) 5 → 2
›Reveal solutionSolution
Comparing the actual Bohr energy-level gaps for all four listed transitions shows $2\to1$ releases the most energy (10.2 eV), and hence has the maximum frequency.

In the Bohr model of the hydrogen atom, the energy of the $n$ -th orbit is $E_n = -\dfrac{13.6}{n^2}$ eV, and a spectral line's photon energy (and hence frequency, via $E=h\nu$ ) equals the magnitude of the energy released when an electron falls from a higher level $n_i$ to a lower level $n_f$ : $h\nu = E_{n_i}-E_{n_f}$ in magnitude. A larger energy gap directly means a higher-frequency (shorter-wavelength) spectral line, so the question reduces to finding which of the four listed transitions has the largest $|\Delta E|$ .

Computing each level's energy: $E_1=-13.6$ eV, $E_2=-3.4$ eV, $E_3\approx-1.51$ eV, $E_4=-0.85$ eV, $E_5\approx-0.544$ eV, $E_6\approx-0.378$ eV.

$6\to2$ : $|E_2-E_6| = |-3.4-(-0.378)| \approx 3.02$ eV.

$2\to1$ : $|E_1-E_2| = |-13.6-(-3.4)| = 10.2$ eV.

$4\to3$ : $|E_3-E_4| = |-1.51-(-0.85)| \approx 0.66$ eV.

$5\to2$ : $|E_2-E_5| = |-3.4-(-0.544)| \approx 2.86$ eV. …
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Q.Discuss the spectral series of the hydrogen atom.

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