Q.Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level $n$ to level $(n-1)$. For large $n$, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The wave number of a spectral line in Brackett series of hydrogen atom is $\dfrac{9}{400}R$. The electron has transmitted from the orbit having quantum number (A) 5 (B) 6 (C) 4 (D) 7

   ›Reveal solutionSolution

   This tests the Rydberg formula for the Brackett series (transitions ending at $n=4$) to find the initial orbit's quantum number from the given wave number. Answer: $n=5$.

   Concept and Intuition

   Each spectral series of hydrogen corresponds to electron transitions ending on a fixed lower energy level: Lyman ($n_f=1$), Balmer ($n_f=2$), Paschen ($n_f=3$), Brackett ($n_f=4$), Pfund ($n_f=5$). The wave number of the emitted photon is given by the Rydberg formula, $\bar\nu=R\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$, where $n_i>n_f$ is the initial (higher) orbit the electron falls from.

   Step-by-Step Solution

   1. For Brackett series, $n_f=4$, so $\bar\nu=R\left(\dfrac1{16}-\dfrac1{n_i^2}\right)$.
   2. Given $\bar\nu=\dfrac{9}{400}R$, so $\dfrac1{16}-\dfrac1{n_i^2}=\dfrac{9}{400}$.
   3. Convert $\dfrac1{16}$ to a denominator of 400: $\dfrac1{16}=\dfrac{25}{400}$.
   4. $\dfrac1{n_i^2}=\dfrac{25}{400}-\dfrac{9}{400}=\dfrac{16}{400}=\dfrac1{25}$.
   5. So $n_i^2=25 \Rightarrow n_i=5$.

   Common Mistakes

   • Using the wrong lower level (e.g. $n_f=2$ for Balmer) instead of $n_f=4$ for Brackett.
   • Arithmetic slip converting $1/16$ to a common denominator with 400.
   • Solving for $n_i^2$ but forgetting to take the square root to get the quantum number itself.

   ✓Final answer

   The correct option is (A) — 5.

   ANSWER: A

2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The atomic number (Z) of a hydrogen like atom whose shortest wavelength of Brackett Series is same as the shortest wavelength of Balmer series of hydrogen atom is (A) $Z = 1$ (B) $Z = 2$ (C) $Z = 3$ (D) $Z = 4$

   ›Reveal solutionSolution

   Matching the series-limit wavelengths of hydrogen's Balmer series and a hydrogen-like atom's Brackett series gives $Z=2$ (i.e. singly-ionized helium).

   Concept and Intuition

   The Rydberg formula for a hydrogen-like ion of atomic number $Z$ is $\dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$. The shortest wavelength of any series (its "series limit") corresponds to the electron falling from $n_2 = \infty$ down to the series' lower level $n_1$, since that transition carries the largest possible energy jump within the series:

   $\frac{1}{\lambda_{min}} = RZ^2\left(\frac{1}{n_1^2}\right).$

   For hydrogen's Balmer series, $n_1 = 2$ and $Z=1$. For the unknown hydrogen-like atom's Brackett series, $n_1 = 4$ with its own $Z$. Setting the two series-limit wavelengths equal lets us solve directly for $Z$.

   Step-by-Step Solution

   1. Hydrogen Balmer series limit ($n_1=2$, $Z=1$): $\dfrac{1}{\lambda_{Balmer}} = R(1)^2\cdot\dfrac{1}{2^2} = \dfrac{R}{4}$.
   2. Hydrogen-like Brackett series limit ($n_1=4$, atomic number $Z$): $\dfrac{1}{\lambda_{Brackett}} = RZ^2\cdot\dfrac{1}{4^2} = \dfrac{RZ^2}{16}$.
   3. Given the two shortest wavelengths are equal, $\dfrac{1}{\lambda_{Balmer}} = \dfrac{1}{\lambda_{Brackett}}$: $\dfrac{R}{4} = \dfrac{RZ^2}{16}$.
   4. Solve: $Z^2 = \dfrac{16}{4} = 4 \Rightarrow Z = 2$.

   Common Mistakes

   • Using the longest wavelength (first line) transition instead of the series limit ($n_2\to n_1+1$ rather than $n_2\to\infty$) — the "shortest wavelength" phrase specifically signals the series-limit transition.
   • Mixing up which series' $n_1$ value belongs to which atom (Balmer $n_1=2$ for hydrogen, Brackett $n_1=4$ for the unknown atom).

   ✓Final answer

   The correct option is (B) — $Z = 2$.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The minimum frequency of light which can ionize a hydrogen atom is approximately (A) $3.3 \times 10^{15}$ Hz (B) $5 \times 10^{15}$ Hz (C) 91.1 Hz (D) 30.5 Hz

   ›Reveal solutionSolution

   Ionizing a ground-state hydrogen atom requires supplying at least its binding energy, 13.6 eV. Converting this energy to a photon frequency via $E=hf$ gives about $3.3\times10^{15}$ Hz.

   Concept and Intuition

   The electron in a hydrogen atom's ground state is bound with energy $-13.6$ eV. To ionize it (remove it to infinity with zero kinetic energy, the minimum condition), a photon must supply exactly this much energy. The minimum frequency is the one where the photon energy just equals the ionization energy.

   Step-by-Step Solution

   1. Ionization energy of hydrogen: $E = 13.6$ eV $= 13.6 \times 1.6\times10^{-19}\text{ J} = 2.176\times10^{-18}$ J.
   2. Using $E = hf$: $f = \dfrac{E}{h} = \dfrac{2.176\times10^{-18}}{6.63\times10^{-34}}$.
   3. $f \approx 3.28\times10^{15}$ Hz $\approx 3.3\times10^{15}$ Hz.

   Common Mistakes

   • Using an excited-state energy instead of the ground-state (n=1) binding energy for the minimum ionizing frequency.
   • Forgetting to convert eV to joules before dividing by $h$.

   ✓Final answer

   The correct option is (A) — $3.3\times10^{15}$ Hz.

   ANSWER: A

4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.In hydrogen spectrum, the ratio of the shortest wavelength of Balmer series and longest wavelength of pfund series is (A) 36:125 (B) 11:225 (C) 11:125 (D) 36:11

   ›Reveal solutionSolution

   Comparing the shortest Balmer wavelength (series limit) with the longest Pfund wavelength ($n=6\to5$) gives a ratio of 11:225.

   Concept and Intuition

   For hydrogen, $\dfrac1\lambda=R\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)$. A series's shortest wavelength corresponds to the transition from $n_2=\infty$ (series limit, the maximum possible energy gap for that series). A series's longest wavelength corresponds to the transition from the very next higher level (the minimum energy gap).

   Step-by-Step Solution

   1. Balmer series ($n_1=2$) shortest $\lambda$: transition from $n_2=\infty$. $\dfrac1{\lambda_{Balmer}}=R\left(\dfrac14-0\right)=\dfrac R4 \Rightarrow \lambda_{Balmer}=\dfrac4R$.
   2. Pfund series ($n_1=5$) longest $\lambda$: transition from $n_2=6$ (smallest possible jump). $\dfrac1{\lambda_{Pfund}}=R\left(\dfrac1{25}-\dfrac1{36}\right)=R\cdot\dfrac{36-25}{900}=\dfrac{11R}{900} \Rightarrow \lambda_{Pfund}=\dfrac{900}{11R}$.
   3. Ratio: $\dfrac{\lambda_{Balmer}}{\lambda_{Pfund}}=\dfrac{4/R}{900/(11R)}=\dfrac{4\times11}{900}=\dfrac{44}{900}=\dfrac{11}{225}$.

   Common Mistakes

   • Mixing up which transition ($\infty$ vs. adjacent level) gives the shortest vs. longest wavelength.
   • Using the wrong base level for the Pfund series (it's $n_1=5$, not 4 or 6).

   ✓Final answer

   The correct option is (B) — 11:225.

   ANSWER: B

5. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.The energy required to excite an electron from $1^{st}$ to $3^{rd}$ Bohr orbit in $Li^{2+}$ ion is (A) 212.1 eV (B) 136.3 eV (C) 108.8 eV (D) 122.4 eV

   ›Reveal solutionSolution

   Bohr model energy levels scale as $Z^2/n^2$; for a hydrogen-like ion, the excitation energy between two orbits is just the difference of these level energies.

   Concept and Intuition

   Li²⁺ is hydrogen-like (one electron, nuclear charge $Z=3$), so its orbit energies follow the same Bohr formula as hydrogen but scaled by $Z^2$: $E_n=-13.6\dfrac{Z^2}{n^2}$ eV. Because the charge is 3 times that of hydrogen's, all the level spacings are $9\times$ larger — a scaling factor that's often forgotten.

   Step-by-Step Solution

   1. $E_n=-13.6\times\dfrac{3^2}{n^2}=-\dfrac{122.4}{n^2}$ eV.
   2. $E_1=-122.4$ eV.
   3. $E_3=-122.4/9=-13.6$ eV.
   4. Energy absorbed to excite from $n=1$ to $n=3$: $\Delta E=E_3-E_1=-13.6-(-122.4)=108.8$ eV.

   Common Mistakes

   • Forgetting to square $Z=3$ (using $Z=3$ instead of $Z^2=9$ in the formula).
   • Using hydrogen's own level energies ($-13.6$ eV, $-1.51$ eV) without the ionic scaling.

   ✓Final answer

   The correct option is (C) — 108.8 eV.

   ANSWER: C

6. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.In hydrogen atom, if the kinetic energy of an electron in an orbit having angular momentum $\dfrac{2h}{\pi}$ is E, then the potential energy of the electron in the first orbit of hydrogen atom is (h – Planck's constant) (A) $-32E$ (B) $-8E$ (C) $-4E$ (D) $-16E$

   ›Reveal solutionSolution

   This uses Bohr's quantization of angular momentum to identify the orbit number, then the $1/n^2$ scaling of kinetic energy and the virial-theorem relation between KE and PE.

   Concept and Intuition

   In the Bohr model, $L_n=\dfrac{nh}{2\pi}$ identifies the orbit. Kinetic energy scales as $KE_n\propto Z^2/n^2$, and for the Coulomb potential the virial theorem gives $PE_n=-2\,KE_n$ (total energy $E_n=KE_n+PE_n=-KE_n$).

   Step-by-Step Solution

   1. Given $L=\dfrac{2h}{\pi}$; setting $\dfrac{nh}{2\pi}=\dfrac{2h}{\pi}$ gives $n=4$.
   2. So $KE_4=E$ (as given). Since $KE_n\propto\dfrac{1}{n^2}$: $\dfrac{KE_1}{KE_4}=\dfrac{4^2}{1^2}=16\Rightarrow KE_1=16E$.
   3. By the virial theorem for the Coulomb (1/r) potential: $PE_n=-2\,KE_n$.
   4. So $PE_1=-2\times16E=-32E$.

   Common Mistakes

   • Mixing up which orbit is "first" vs the orbit identified by the given angular momentum (here $n=4$, not the orbit whose energy is asked).
   • Using $PE_n=-KE_n$ (total energy relation) instead of the correct $PE_n=-2KE_n$.

   ✓Final answer

   The correct option is (A) — $-32E$.

   ANSWER: A

7. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.In the pfund series of hydrogen spectrum, the wavelength of first spectral line is (Rydberg constant $= 1.097 \times 10^7 m^{-1}$) (A) 8547 nm (B) 6574 nm (C) 3729 nm (D) 7458 nm

   ›Reveal solutionSolution

   The Pfund series (transitions to $n=5$) has its first line at $n=6\to n=5$; the Rydberg formula gives $\lambda\approx7458$ nm, in the infrared.

   Concept and Intuition

   The hydrogen spectral series are named by the lower energy level $n_1$ the electron falls to: Lyman ($n_1=1$), Balmer ($n_1=2$), Paschen ($n_1=3$), Brackett ($n_1=4$), Pfund ($n_1=5$). The "first line" (longest wavelength, smallest energy gap) of any series is the transition from the very next higher level, $n_2=n_1+1$.

   Step-by-Step Solution

   1. Pfund series: $n_1=5$. First line: $n_2=6\to n_1=5$.
   2. Rydberg formula: $\dfrac1\lambda=R\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)=R\left(\dfrac1{25}-\dfrac1{36}\right)$.
   3. $\dfrac1{25}-\dfrac1{36}=\dfrac{36-25}{900}=\dfrac{11}{900}$.
   4. $\dfrac1\lambda=1.097\times10^7\times\dfrac{11}{900}=1.097\times10^7\times0.012222\approx1.341\times10^5\ \text{m}^{-1}$.
   5. $\lambda=\dfrac1{1.341\times10^5}\approx7.458\times10^{-6}\ \text{m}=7458$ nm.

   Common Mistakes

   • Using $n_1=4,n_2=5$ (that would be Brackett) instead of $n_1=5,n_2=6$ for Pfund.
   • Forgetting to invert $1/\lambda$ to get $\lambda$ at the end.

   ✓Final answer

   The correct option is (D) — 7458 nm.

   ANSWER: D

8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The difference between the frequencies of the first and second Lyman lines of hydrogen atom is (R - Rydberg constant and c - speed of light in vacuum) (A) $\dfrac{9Rc}{28}$ (B) $\dfrac{7Rc}{12}$ (C) $\dfrac{3Rc}{8}$ (D) $\dfrac{5Rc}{36}$

   ›Reveal solutionSolution

   The Lyman series frequencies follow $\nu=Rc(1-1/n^2)$ for transitions to $n=1$; the difference between the first two lines is $\dfrac{5Rc}{36}$.

   Concept and Intuition

   The Lyman series corresponds to transitions from higher levels $n=2,3,4,\dots$ down to the ground state $n=1$. Using the Rydberg formula for wavenumber, $\bar\nu = R\left(\dfrac{1}{1^2}-\dfrac{1}{n^2}\right)$, and multiplying by $c$ gives frequency $\nu = Rc\left(1-\dfrac{1}{n^2}\right)$. The "first" Lyman line is the $n=2\to1$ transition and the "second" is $n=3\to1$.

   Step-by-Step Solution

   1. First Lyman line ($n=2\to1$): $\nu_1 = Rc\left(1-\dfrac{1}{4}\right) = \dfrac{3Rc}{4}$.
   2. Second Lyman line ($n=3\to1$): $\nu_2 = Rc\left(1-\dfrac{1}{9}\right) = \dfrac{8Rc}{9}$.
   3. Difference: $\nu_2-\nu_1 = \dfrac{8Rc}{9}-\dfrac{3Rc}{4}$.
   4. Common denominator 36: $= \dfrac{32Rc}{36}-\dfrac{27Rc}{36} = \dfrac{5Rc}{36}$.

   Common Mistakes

   • Confusing "first" and "second" line ordering (second line has HIGHER frequency, so subtract the smaller from the larger).
   • Forgetting to convert wavenumber to frequency by multiplying by $c$ (some derivations stop at $R$ alone).

   ✓Final answer

   The correct option is (D) — $\dfrac{5Rc}{36}$.

   ANSWER: D

9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The ratio of the wavelengths of the first Lyman line and the second Balmer line of hydrogen atom is (A) 3 : 4 (B) 1 : 4 (C) 2 : 3 (D) 1 : 3

   ›Reveal solutionSolution

   This tests the Rydberg formula applied to specific named spectral lines. The ratio of wavelengths (Lyman-first : Balmer-second) is 1 : 4.

   Concept and Intuition

   Each spectral series of hydrogen corresponds to transitions ending on a fixed lower level ($n=1$ for Lyman, $n=2$ for Balmer), and the "first", "second", etc. lines within a series correspond to the electron starting from successively higher levels. The Rydberg formula converts each specific transition into a wavenumber $1/\lambda$, and comparing two lines is just comparing these wavenumbers.

   Step-by-Step Solution

   1. Rydberg formula: $\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$.
   2. First Lyman line: transition $n=2\to n=1$:

   $\frac{1}{\lambda_{L1}}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R\left(1-\frac14\right)=\frac{3R}{4}$

   3. Second Balmer line: transition $n=4\to n=2$:

   $\frac{1}{\lambda_{B2}}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R\left(\frac14-\frac1{16}\right)=\frac{3R}{16}$

   4. Ratio of wavelengths:

   $\frac{\lambda_{L1}}{\lambda_{B2}}=\frac{1/\lambda_{L1}^{-1}}{1}=\frac{3R/16}{3R/4}=\frac{4}{16}=\frac14$

   So $\lambda_{L1}:\lambda_{B2}=1:4$.

   Common Mistakes

   • Confusing "second Balmer line" ($n=4\to2$) with "second line overall" or with $n=3\to2$ (which is the first Balmer line).
   • Inverting the ratio (mixing up wavenumber ratio with wavelength ratio).

   ✓Final answer

   The correct option is (B) — 1 : 4.

   ANSWER: B

10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.The difference between the frequencies of second and first Paschen lines of hydrogen atom is (R - Rydberg constant and c - speed of light in vacuum) (A) $\dfrac{9Rc}{16}$ (B) $\dfrac{16Rc}{25}$ (C) $\dfrac{9Rc}{400}$ (D) $\dfrac{3Rc}{200}$

    ›Reveal solutionSolution

    The Paschen series lines all terminate on $n=3$; working out the frequencies of the $4\to3$ and $5\to3$ transitions and subtracting gives $\dfrac{9Rc}{400}$.

    Concept and Intuition

    The hydrogen spectral series are named by the final (lower) energy level the electron falls to: Lyman ($n_f=1$), Balmer ($n_f=2$), Paschen ($n_f=3$), and so on. Within a series, lines are numbered by increasing initial level: the first Paschen line is the transition from the next level up ($n=4\to3$), the second Paschen line is from one level further ($n=5\to3$), etc. The Rydberg formula gives the frequency of any transition as $f = Rc\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)$, so once we know which two transitions are meant, this is purely an algebra exercise in fractions.

    Step-by-Step Solution

    1. First Paschen line ($4\to3$): $f_1 = Rc\left(\dfrac{1}{9}-\dfrac{1}{16}\right) = Rc\cdot\dfrac{16-9}{144} = \dfrac{7Rc}{144}$.
    2. Second Paschen line ($5\to3$): $f_2 = Rc\left(\dfrac{1}{9}-\dfrac{1}{25}\right) = Rc\cdot\dfrac{25-9}{225} = \dfrac{16Rc}{225}$.
    3. Find a common denominator for $\dfrac{16}{225}$ and $\dfrac{7}{144}$: since $225=9\times25$ and $144=16\times9$, the LCM is $3600$.
    4. $\dfrac{16}{225} = \dfrac{256}{3600}$, and $\dfrac{7}{144}=\dfrac{175}{3600}$.
    5. Difference: $f_2 - f_1 = Rc\cdot\dfrac{256-175}{3600} = Rc\cdot\dfrac{81}{3600} = Rc\cdot\dfrac{9}{400}$ (dividing numerator and denominator by 9).

    Common Mistakes

    • Mixing up "first" and "second" line (i.e., swapping $4\to3$ and $5\to3$) — the difference is the same magnitude either way here since we just subtract, but it's easy to instead compute $f_1-f_2$ (a sign slip) or use the wrong pair of levels (e.g. $3\to2$, which belongs to the Balmer series, not Paschen).
    • Arithmetic error in combining the two fractions — reducing $81/3600$ to $9/400$ is a common place to make an error.

    ✓Final answer

    The correct option is (C) — $\dfrac{9Rc}{400}$.

    ANSWER: C

11. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.Of the following, Bohr's atomic model is applicable to (A) explain relative intensities of spectral lines emitted by hydrogen atoms (B) helium atom (C) lithium atom (D) hydrogenic atoms

    ›Reveal solutionSolution

    Bohr's model is a single-electron model; it applies to hydrogen and hydrogen-like (hydrogenic) ions, not to multi-electron neutral atoms, and it does not predict spectral line intensities.

    Concept and Intuition

    Bohr's postulates were built specifically around one electron orbiting a nucleus of charge $+Ze$ under a pure Coulomb force, with angular momentum quantized as $n\hbar$. This works cleanly whenever there is exactly one electron in the system — hydrogen ($Z=1$) and "hydrogenic" ions like $\mathrm{He}^+$ ($Z=2$, one electron), $\mathrm{Li}^{2+}$ ($Z=3$, one electron), etc. As soon as more than one electron is present (neutral helium, neutral lithium), electron-electron repulsion and screening effects make the simple circular-orbit picture invalid, so Bohr's model does not extend to those. Also, Bohr's theory predicts energy levels and hence spectral line frequencies, but says nothing about intensities of the lines, which require quantum-mechanical transition probabilities beyond the scope of the model.

    Step-by-Step Solution

    1. Recall Bohr's derivation assumes a single electron orbiting a central charge — the model's validity is tied to being "one-electron" in nature.
    2. Option (A), relative intensities of spectral lines, is outside Bohr's theory's predictive power (it only gives energy levels/frequencies, not transition probabilities) — reject.
    3. Option (B), helium atom, is a two-electron neutral atom — Bohr's simple model fails here due to electron-electron interaction — reject.
    4. Option (C), lithium atom, is a three-electron neutral atom — same issue, reject.
    5. Option (D), hydrogenic atoms (one-electron species like H, He⁺, Li²⁺), is exactly the class of systems Bohr's model correctly describes — accept.

    Common Mistakes

    • Confusing "helium atom" (2 electrons, Bohr model fails) with "He⁺ ion" (1 electron, a hydrogenic species Bohr's model handles well).
    • Assuming Bohr's model, because it explains spectral lines' positions, also explains their intensities.

    ✓Final answer

    The correct option is (D) — hydrogenic atoms.

    ANSWER: D

12. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
    Q.When an electron beam of energy $10.2\ \mathrm{eV}$ is used to excite hydrogen gas, then the possible spectral line is (A) first Balmer line (B) first Lyman line (C) second Balmer line (D) second Lyman line

    ›Reveal solutionSolution

    An electron beam of exactly $10.2\,\mathrm{eV}$ can only excite a hydrogen atom from $n=1$ to $n=2$ (since higher-level excitations need more energy). The de-excitation $n=2\to n=1$ produces the first Lyman line.

    Concept and Intuition

    Bohr's hydrogen energy levels are $E_n = -\dfrac{13.6}{n^2}\,\mathrm{eV}$. Electron-impact excitation can only promote the atom to a level if the beam energy equals (or in an inelastic collision, at least equals) the energy gap to that level — but if there's exactly enough energy for one specific transition and not the next, only that transition occurs. Here $10.2\,\mathrm{eV}$ exactly matches $E_1 \to E_2$, so the atom is excited only up to $n=2$. When it relaxes back to the ground state ($n=2\to n=1$), it emits a Lyman-series photon — specifically the first (lowest energy) line of that series, called Lyman-$\alpha$.

    Step-by-Step Solution

    1. $E_1 = -13.6\,\mathrm{eV}$ (ground state), $E_2 = -\dfrac{13.6}{4} = -3.4\,\mathrm{eV}$.
    2. Energy gap $E_2 - E_1 = -3.4-(-13.6) = 10.2\,\mathrm{eV}$ — exactly the beam energy given.
    3. Check $E_3 = -\dfrac{13.6}{9} \approx -1.51\,\mathrm{eV}$; gap $E_3-E_1 \approx 12.09\,\mathrm{eV}$, which is more than $10.2\,\mathrm{eV}$ — so $n=3$ cannot be reached.
    4. So the atom is excited only to $n=2$.
    5. On de-excitation $n=2\to n=1$, the emitted photon belongs to the Lyman series (all transitions ending at $n=1$); since it is the smallest-gap transition possible in that series ($n=2\to1$), it is the first Lyman line.

    Common Mistakes

    • Confusing Balmer series (transitions ending at $n=2$) with Lyman series (transitions ending at $n=1$) — since the atom is excited to $n=2$, the emitted photon on relaxation is a Lyman line, not a Balmer line.
    • Assuming any energy at least 10.2 eV could reach higher levels — but the atom absorbs the beam energy in a single quantum jump matching an allowed transition; here it matches only the $n=1\to2$ gap.

    ✓Final answer

    The correct option is (B) — first Lyman line.

    ANSWER: B