Q.In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron.) Assuming the nucleus to be massive, calculate the kinetic energy of an $n = 4$ Auger electron emitted by Chromium by absorbing the energy from a $n = 2$ to $n = 1$ transition.

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: $\dfrac1{16}=\dfrac{25}{400}$. …

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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 $Z$): $\dfrac{1}{\lambda_{Brackett}} = RZ^2\cdot\dfrac{1}{4^2} = \dfrac{RZ^2}{16}$. …

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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 $E = hf$: $f = \dfrac{E}{h} = \dfrac{2.176\times10^{-18}}{6.63\times10^{-34}}$. …

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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 $n_2=\infty$. $\dfrac1{\lambda_{Balmer}}=R\left(\dfrac14-0\right)=\dfrac R4 \Rightarrow \lambda_{Balmer}=\dfrac4R$. …

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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. …

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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: $PE_n=-2\,KE_n$. …

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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. $\dfrac1{25}-\dfrac1{36}=\dfrac{36-25}{900}=\dfrac{11}{900}$. …

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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: $\nu_2-\nu_1 = \dfrac{8Rc}{9}-\dfrac{3Rc}{4}$. …

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

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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 $\dfrac{7}{144}=\dfrac{175}{3600}$. …

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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. …

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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 $n=2$. …

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