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The wavelength of radiation emitted when an electron in hydrogen atom undergoes a transition from $n=2$ to $n=1$ is $\lambda_1$ and that for $n=4$ to $n=2$ is $\lambda_2$. The ratio $\lambda_1/\lambda_2$ is:
Options
1
$1/4$
2
$1/2$
3
$2$
4
$4$
Correct Answer
$\lambda_1/\lambda_2 = 1/4$
Solution
1

Rydberg formula: $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$

$$\frac{1}{\lambda_1} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$$$$\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$$
2
$$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{3R/16}{3R/4} = \boxed{\frac{1}{4}}$$
Key: $\lambda_1/\lambda_2 = (1/\lambda_2) \div (1/\lambda_1)$ — take reciprocal of wavenumber ratio!
Answer = $1/4$
Theory: Atomic Structure
1. Bohr Model and Energy Levels

Bohr (1913) proposed three postulates: electrons move in circular orbits without radiating; only orbits where $L = n\hbar$ are allowed; radiation emitted/absorbed only during transitions. Results for hydrogen: $r_n = n^2 a_0$ (Bohr radius $a_0 = 0.529$ Å), $E_n = -13.6/n^2$ eV. Ground state ($n=1$): $E = -13.6$ eV. Second level ($n=2$): $E = -3.4$ eV. Third level: $-1.51$ eV. Fourth: $-0.85$ eV. Infinity: $0$ eV (free electron). For hydrogen-like ions (He⁺, Li²⁺, etc.): $E_n = -13.6Z^2/n^2$ eV. The quantization of energy levels directly explains the discrete line spectrum of hydrogen — electrons can only have specific energies, so photons of only specific frequencies are emitted or absorbed.

2. Rydberg Formula and Spectral Series

The Rydberg formula: $1/\lambda = R_H(1/n_1^2 - 1/n_2^2)$ where $n_2 > n_1$ (emission). $R_H = 1.097 \times 10^7$ m⁻¹ = 109677 cm⁻¹. Spectral series: Lyman ($n_1=1$): UV, 91.2–121.6 nm. First line (2→1): 121.6 nm (Lyman alpha). Balmer ($n_1=2$): UV+visible, 364.6–656.3 nm. Lines: H-α=656 nm (red), H-β=486 nm (blue-green), H-γ=434 nm (violet), H-δ=410 nm (violet). Paschen ($n_1=3$): near IR, 820–1875 nm. Brackett ($n_1=4$): IR. Pfund ($n_1=5$): far IR. Important rule: maximum wavelength (minimum energy) in any series = transition to adjacent level (e.g., 3→2 for Balmer). Series limit (minimum wavelength) = $n_2=\infty$. The Balmer series is historically important because it falls in visible range and was discovered first (Balmer, 1885) — before Bohr's theory explained it.

3. Calculating Spectral Lines — Key Shortcuts

Energy of transition: $\Delta E = 13.6(1/n_1^2 - 1/n_2^2)$ eV. Wavelength: $\lambda = 1240/\Delta E$ (nm) where $\Delta E$ in eV. For ratio problems: since $1/\lambda \propto (1/n_1^2 - 1/n_2^2)$, taking ratios is straightforward. Number of lines when electron drops from level $n$: $n(n-1)/2$. For $n=4$: $4(3)/2 = 6$ lines. For $n=5$: $5(4)/2 = 10$ lines. Series limit wavelength: $\lambda_{limit} = n_1^2/R_H$ for series with lower level $n_1$. Lyman limit: $1/R_H = 91.2$ nm. Balmer limit: $4/R_H = 364.8$ nm. Common NEET questions: find wavelength of specific transition, identify series, calculate ratio of wavelengths, find number of spectral lines. Always draw energy level diagram and identify $n_1$ (lower) and $n_2$ (upper) correctly.

4. Electronic Configuration — Rules

Three rules govern electron filling: Aufbau principle: electrons occupy lowest available energy orbital first. Order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p... (Madelung/diagonal rule). Pauli exclusion: maximum 2 electrons per orbital, with opposite spins. Hund rule: in degenerate orbitals (same energy), electrons fill singly with parallel spins before pairing. This minimises electron repulsion and is the ground state configuration. Important exceptions to Aufbau: Cr (Z=24): [Ar]3d⁵4s¹ (not 3d⁴4s²) — half-filled d is extra stable. Cu (Z=29): [Ar]3d¹⁰4s¹ (not 3d⁹4s²) — fully-filled d is extra stable. These exceptions are regularly asked in NEET. Similar exceptions occur for Mo, Ag, Au, Pd.

5. Quantum Numbers in Detail

Principal quantum number $n$ (1,2,3,...): determines energy (∝ $-1/n^2$) and average distance from nucleus (∝ $n^2$). Azimuthal $l$ (0 to $n-1$): s-orbital ($l=0$, spherical), p ($l=1$, dumbbell), d ($l=2$, cloverleaf), f ($l=3$, complex). Number of orbitals: $2l+1$. Max electrons: $2(2l+1)$. Magnetic $m_l$ ($-l$ to $+l$): orientation. p has 3 orbitals ($p_x, p_y, p_z$); d has 5; f has 7. Spin $m_s$ ($±1/2$): intrinsic angular momentum. Nodes: radial nodes = $n-l-1$; angular nodes = $l$; total nodes = $n-1$. The 1s orbital has 0 nodes (spherically symmetric, maximum density at nucleus for hydrogen). 2s has 1 radial node (spherical shell of zero density). 2p has 1 angular node (nodal plane through nucleus). Understanding node patterns helps visualize orbital shapes and predict reactivity.

6. Atomic Orbitals and Shapes

s-orbitals: spherically symmetric, one per subshell. 1s: smallest, most tightly bound. 2s: larger, one spherical node. Electron density has a maximum outside the node as well as at small $r$. p-orbitals: dumbbell shaped, along x, y, or z axis. Three degenerate p-orbitals ($p_x, p_y, p_z$). Nodal plane through nucleus. d-orbitals: five degenerate. $d_{xy}, d_{xz}, d_{yz}$: lobes between axes. $d_{x^2-y^2}$: lobes along x and y axes. $d_{z^2}$: unique shape with lobes along z-axis and ring (torus) in xy-plane. f-orbitals: seven degenerate, complex shapes. The directionality of p and d orbitals is responsible for directional bonding and geometry in molecules and coordination compounds. The different spatial orientations allow for σ, π, and δ bonds formed by different orbital overlaps.

7. Ionisation Energy Trends

Ionisation energy (IE): energy needed to remove outermost electron from gaseous atom. IE increases across period (left to right): increasing nuclear charge with same shielding → electrons held more tightly. IE decreases down group: increasing atomic radius → outer electron farther from nucleus → less attracted. Exceptions across period: Be (IE > B): 2s is more stable than 2p. N (IE > O): N has half-filled 2p (extra stability from exchange energy). Mg (IE > Al): 3s vs 3p. Successive ionisation energies: large jump when removing electron from inner shell. Used to determine valence electrons. Example: Na: 1st IE = 496 kJ/mol (easy, removes 3s). 2nd IE = 4562 kJ/mol (large jump — removing 2p electron, inner shell). Confirms Na has 1 valence electron. IE₁ < IE₂ < IE₃... always. Electron affinity: energy released when atom accepts electron. Cl has highest EA (349 kJ/mol).

8. Atomic Radius and Periodic Trends

Atomic radius decreases across period (left to right): increasing nuclear charge pulls electrons closer (same shell, increasing Z). Atomic radius increases down group: new shells added, shielding increases. Covalent radius: half the distance between two identical atoms bonded by single bond. Van der Waals radius: half the distance between non-bonded atoms in contact. Ionic radius: radius of ion. Cation: smaller than parent atom (electron removed, proton-to-electron ratio increases). Anion: larger than parent atom (electron added, repulsion increases radius). Isoelectronic species: same electrons, different nuclear charge. Example: N³⁻, O²⁻, F⁻, Ne, Na⁺, Mg²⁺, Al³⁺ all have 10 electrons. Radius decreases with increasing atomic number: N³⁻ > O²⁻ > F⁻ > Ne > Na⁺ > Mg²⁺ > Al³⁺. Lanthanide contraction: filling of 4f orbitals in lanthanides → poor shielding → radius contraction → 5th period d-block similar size to 4th period.

Frequently Asked Questions
1. How to avoid the common mistake in wavelength ratio problems?
The most common mistake: students calculate $(1/n_1^2 - 1/n_2^2)$ for each transition and then directly take the ratio. This gives the WAVENUMBER ratio ($\bar{\nu}_1/\bar{\nu}_2$), not the wavelength ratio. Remember: wavenumber $\bar{\nu} = 1/\lambda$. So $\lambda_1/\lambda_2 = \bar{\nu}_2/\bar{\nu}_1$ (RECIPROCAL of wavenumber ratio). In this problem: $\bar{\nu}_1 : \bar{\nu}_2 = 3R/4 : 3R/16 = 4:1$. Therefore $\lambda_1/\lambda_2 = 1:4 = 1/4$. Always be careful: high wavenumber = short wavelength = high energy. Low wavenumber = long wavelength = low energy.
2. Why is n=4→2 transition in the Balmer series?
The Balmer series consists of all transitions where the electron ends up in the $n=2$ level (final state $n_f = 2$). Since $n=4$ is the initial state and $n=2$ is the final state (lower level), this is indeed a Balmer series transition. Specifically: $n=3\to2$ = H-α (656 nm, red), $n=4\to2$ = H-β (486 nm, blue-green), $n=5\to2$ = H-γ (434 nm, violet), $n=6\to2$ = H-δ (410 nm, violet). The H-β line is commonly seen in emission nebulae and stellar spectra. All Balmer lines are more energetic (shorter wavelength) than the H-α line.
3. How do you identify which series a transition belongs to?
Look at the LOWER energy level (final level for emission, initial for absorption): $n_f = 1$: Lyman series (UV). $n_f = 2$: Balmer series (visible/near UV). $n_f = 3$: Paschen series (near IR). $n_f = 4$: Brackett series (IR). $n_f = 5$: Pfund series (far IR). For this problem: transition $n=2\to1$ has $n_f=1$ → Lyman. Transition $n=4\to2$ has $n_f=2$ → Balmer. The Balmer series is the only one with lines in the visible region (wavelengths 400-700 nm), which is why it was discovered first (1885) and is most commonly seen in astrophysical spectra.
4. What are the exact wavelengths for common hydrogen transitions?
Important wavelengths to remember: Lyman-α (2→1): 121.6 nm (UV). Balmer-α/H-α (3→2): 656.3 nm (red). Balmer-β/H-β (4→2): 486.1 nm (blue-green). Balmer-γ/H-γ (5→2): 434.1 nm (violet). Lyman series limit: 91.2 nm. Balmer series limit: 364.6 nm. For calculation: $\lambda = 1240/E(eV)$ nm. H-α: $E = 13.6(1/4-1/9) = 13.6 \times 5/36 = 1.89$ eV. $\lambda = 1240/1.89 = 656$ nm ✓. These wavelengths are used as standard references in spectroscopy and astronomy for measuring redshift of distant galaxies (cosmological redshift tells us the universe is expanding).
5. What is the significance of the hydrogen spectrum in astronomy?
Hydrogen is the most abundant element in the universe (~75% by mass). Hydrogen spectral lines appear in: stellar spectra (absorption lines in stellar atmospheres), emission nebulae (H-II regions with excited hydrogen), quasar spectra, 21 cm radio line (hyperfine transition — used to map spiral arm structure of Milky Way). Cosmological redshift: all hydrogen lines shifted to longer wavelengths in distant galaxies → $z = \Delta\lambda/\lambda_0 = v/c$ (Hubble law). Edwin Hubble (1929) used Balmer lines to show galaxies are receding → expanding universe → Big Bang theory. The measurement of redshift from hydrogen lines is now the primary method for determining distances to galaxies and the Hubble constant ($H_0 \approx 70$ km/s/Mpc).
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