Explain emission and absorption spectra. (Diagram not necessary)

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Q. Explain emission and absorption spectra. (Diagram not necessary)

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 10mImportance★★★★★

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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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Emission spectra are produced by atoms/molecules that have been excited and re-emit energy as characteristic radiation; absorption spectra are produced when a continuous spectrum passes through a cooler gas that absorbs specific wavelengths, leaving dark lines at exactly those wavelengths.

Emission spectrum

When a substance is excited — by heating to incandescence, by an electric discharge, or by other means — its atoms/molecules absorb energy and their electrons jump to higher energy states. These excited electrons are unstable and fall back to lower energy states, emitting the excess energy as electromagnetic radiation (photons) of definite frequencies. Viewed through a spectroscope, this radiation forms the emission spectrum, of three kinds:

Continuous emission spectrum: emitted by incandescent solids, liquids, or gases under high pressure (e.g. a hot filament, molten metal, the Sun's photosphere); it contains all wavelengths merging into one another with no gaps.
Line emission spectrum: emitted by atoms of an excited rarefied (low-pressure) gas; consists of sharp, discrete bright lines at wavelengths characteristic of that element only, because electrons in an isolated atom occupy only specific discrete energy levels, so only specific photon energies (hence wavelengths) are emitted. Each element has its own unique set of lines, so this spectrum is used to identify elements.
Band emission spectrum: emitted by excited molecules; consists of groups of closely spaced lines (bands) arising from the more complex rotational/vibrational energy levels of molecules in addition to electronic levels.

Absorption spectrum

When white light (a continuous spectrum) from a hot source is passed through a comparatively cooler gas or vapour before entering the spectroscope, the atoms of the cooler gas absorb photons of exactly those wavelengths that they are capable of emitting (i.e. those matching the energy differences between their allowed energy levels), promoting their electrons to higher states. The rest of the continuous spectrum passes through unaffected. The result, seen on the spectroscope, is a continuous bright background crossed by dark lines at the absorbed wavelengths — the absorption spectrum.

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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.Explain emission and absorption spectra. (Diagram not necessary)

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