Q.A hypothetical electromagnetic wave is shown in the figure below. Find out the wavelength of the radiation.

Showing the 12 most recent of 71 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.In an atomic spectrum of hydrogen, a series of lines with wavelengths at 656.46, 486.27, $x$ and 410.29 nm was obtained. What is the value of $x$ (in nm)? ($R_H = 1.097 \times 10^7 \, \mathrm{m^{-1}}$) (A) 453.15 (B) 449.32 (C) 434.17 (D) 428.37

   ›Reveal solutionSolution

   This tests the Rydberg formula for the Balmer series of hydrogen to find the missing wavelength ($n_i=5\to n_f=2$ transition) in a listed sequence of visible spectral lines. Answer: 434.17 nm.

   Concept and Intuition

   The Balmer series consists of hydrogen emission lines that end on the $n=2$ level, and these fall in the visible range — they are the most famous hydrogen lines (H-alpha, H-beta, H-gamma, H-delta at 656.3, 486.1, 434.1, 410.2 nm respectively, as commonly tabulated). The listed wavelengths 656.46, 486.27, $x$, 410.29 nm are exactly this sequence for $n_i=3,4,5,6$, so $x$ corresponds to the $n_i=5\to n_f=2$ transition (H-gamma).

   Step-by-Step Solution

   1. Rydberg formula: $\dfrac1\lambda=R_H\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$.
   2. Recognize the series: with $n_f=2$, $n_i=3$ gives 656.46 nm, $n_i=4$ gives 486.27 nm, $n_i=6$ gives 410.29 nm — matching the given data confirms $n_f=2$ (Balmer) and that $x$ is the $n_i=5$ line.
   3. For $n_i=5$: $\dfrac1\lambda=R_H\left(\dfrac14-\dfrac1{25}\right)=R_H\left(\dfrac{25-4}{100}\right)=R_H\times\dfrac{21}{100}=0.21R_H$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Wavelength of a photon emitted during electron transition from $n = 4$ state to $n = 2$ state in the hydrogen atom is $x$ nm. Wavelength of a photon emitted during electron transition from $n = 4$ state to $n = 1$ state in the same atom is $y$ nm. $\dfrac{y}{x}$ is equal to (A) 0.4 (B) 0.2 (C) 0.5 (D) 0.3

   ›Reveal solutionSolution

   This tests the Rydberg/Bohr formula for hydrogen-spectrum wavelengths; the ratio of the two transition wavelengths works out to 0.2.

   Concept and Intuition

   For the hydrogen atom, the energy released (and hence the wavenumber $1/\lambda$ of the emitted photon) during a transition from a higher level $n_2$ to a lower level $n_1$ is given by the Rydberg formula. A bigger energy jump means a shorter wavelength, so a transition all the way down to $n=1$ releases more energy (and gives a shorter wavelength) than one landing at $n=2$.

   Step-by-Step Solution

   1. Rydberg formula: $\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$, with $n_1<n_2$.
   2. For $n=4\to n=2$ (wavelength $x$): $\dfrac{1}{x}=R\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)=R\left(\dfrac14-\dfrac1{16}\right)=R\cdot\dfrac{3}{16}$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The energy of spectral line of lowest frequency in Lyman series of $\mathrm{Li^{2+}}$ spectrum is $x$ J. The energy of second spectral line in Balmer series of $\mathrm{He^{+}}$ spectrum is $y$ J. The ratio of $x$ and $y$ is (A) $3 : 1$ (B) $1 : 3$ (C) $1 : 9$ (D) $9 : 1$

   ›Reveal solutionSolution

   Both are hydrogen-like ions, so the Rydberg-type energy formula with $Z^2$ scaling applies; identifying the correct transitions (lowest-frequency Lyman line, second Balmer line) and computing gives $x:y=9:1$.

   Concept and Intuition

   For any hydrogen-like species (single electron, nuclear charge $Ze$), the energy released in a transition from level $n_2$ to $n_1$ ($n_2>n_1$) is

   $E = 13.6\,Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \text{eV}.$

   Within a series (fixed lower level $n_1$), the lowest-frequency (least energetic) line corresponds to the smallest jump, i.e. the transition from the level immediately above $n_1$. The Lyman series has $n_1=1$, so its lowest-frequency line is $2\to1$. The Balmer series has $n_1=2$; its lines in increasing energy (and frequency) order are $3\to2$ (first/weakest), $4\to2$ (second), $5\to2$ (third), etc. — so the second Balmer line is $4\to2$.

   Step-by-Step Solution

   1. Lyman, lowest-frequency line of $\mathrm{Li^{2+}}$ ($Z=3$): transition $2\to1$.

   $x = 13.6\times3^2\left(\frac1{1^2}-\frac1{2^2}\right) = 13.6\times9\times\frac34 = 13.6\times6.75 = 91.8\ \text{eV (in energy units, up to a common constant)}$

   2. Balmer, second line of $\mathrm{He^+}$ ($Z=2$): second line means $4\to2$ (first is $3\to2$). …

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4. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Quantum number sets of four electrons I, II, III, IV are given below. The correct order of the energy of these electrons is I. $n = 3,\ l = 1,\ m_l = -1,\ m_s = +\frac{1}{2}$ II. $n = 4,\ l = 1,\ m_l = 0,\ m_s = +\frac{1}{2}$ III. $n = 4,\ l = 2,\ m_l = -2,\ m_s = +\frac{1}{2}$ IV. $n = 3,\ l = 2,\ m_l = -1,\ m_s = -\frac{1}{2}$ The correct answer is (A) I > IV > II > III (B) III > II > I > IV (C) III > IV > II > I (D) III > II > IV > I

   ›Reveal solutionSolution

   Applying the $(n+l)$ (Aufbau) rule with the n+l tie-breaker gives the energy order III > II > IV > I — testing whether the tie-break (lower $n$ wins for equal $n+l$) is applied correctly.

   Concept and Intuition

   In multi-electron atoms, orbital energy is not determined by $n$ alone (as in hydrogen) but follows the empirical $(n+l)$ rule: orbitals with a lower value of $n+l$ have lower energy. When two orbitals share the same $n+l$ value, the one with the smaller $n$ (and hence larger $l$) has lower energy — because a larger $l$ means the electron is, on average, farther from the nucleus in angular terms but the radial penetration effects work out so that lower-$n$/higher-$l$ combinations of equal $n+l$ sit lower in energy (e.g. $3d$ fills before $4p$... more precisely $4s$ before $3d$, but the classic comparison here is between $4p$ ($n+l=5$) and $3d$ ($n+l=5)$, where $3d$ is lower).

   Step-by-Step Solution

   1. Compute $n+l$ for each electron (the $m_l,m_s$ values don't affect orbital energy, only $n,l$ do):
      • I: $n=3,l=1$ (3p) $\Rightarrow n+l=4$
      • II: $n=4,l=1$ (4p) $\Rightarrow n+l=5$
      • III: $n=4,l=2$ (4d) $\Rightarrow n+l=6$
      • IV: $n=3,l=2$ (3d) $\Rightarrow n+l=5$
   2. Order by $n+l$ ascending (lower = lower energy): I(4) $<$ {II, IV}(5) $<$ III(6). …

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5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The wavelength of spectral line (X) of hydrogen spectrum is same as that of spectral line of $He^+$ spectrum corresponding to $n = 4 \to n = 2$ electron transition. The correct electron transition corresponding to X is (A) $n = 2 \to n = 1$ (B) $n = 3 \to n = 2$ (C) $n = 4 \to n = 2$ (D) $n = 3 \to n = 1$

   ›Reveal solutionSolution

   This tests the Rydberg formula scaled by $Z^2$ and matching two different hydrogenic spectra to the same photon energy/wavelength. The transition is $n=2\to n=1$.

   Concept and Intuition

   Every hydrogen-like ion (H, He$^+$, Li$^{2+}$, …) has energy levels $E_n = -\dfrac{13.6\,Z^2}{n^2}\ \text{eV}$. Two hydrogenic transitions emit the same wavelength exactly when their $Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ values are equal — the $Z^2$ scaling is what lets a lower-$Z$ atom's low-lying transition mimic a higher-$Z$ ion's higher-lying one.

   Step-by-Step Solution

   1. For $He^+$ ($Z=2$), transition $n=4\to n=2$:

   $\frac{1}{\lambda_{He^+}} = R(2)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right) = 4R\left(\frac14-\frac1{16}\right)=4R\cdot\frac{3}{16}=0.75R$

   2. For hydrogen ($Z=1$), we need a transition $n_1\to n_2$ with

   $\frac{1}{\lambda_H} = R(1)^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) = 0.75R$

   3. Test $n=2\to n=1$: $R\left(1-\frac14\right) = 0.75R$. This matches exactly. …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If the energy required to remove an electron from the ground state of $He^+$ is x J, the energy (in J) required to remove an electron from the ground state of $Li^{2+}$ is (A) $\frac{3}{2}x$ (B) $\frac{2}{3}x$ (C) $\frac{9}{4}x$ (D) $\frac{4}{9}x$

   ›Reveal solutionSolution

   Both $He^+$ and $Li^{2+}$ are hydrogen-like (one electron); their ground-state ionization energies scale purely as $Z^2$, giving a factor of $9/4$. Answer: (C).

   Concept and Intuition

   $He^+$ ($Z=2$) and $Li^{2+}$ ($Z=3$) are both single-electron (hydrogen-like) species, so the Bohr-model ionization energy formula $E_n=13.6\,\dfrac{Z^2}{n^2}$ eV applies directly to each. For the ground state ($n=1$), the energy is simply proportional to $Z^2$.

   Step-by-Step Solution

   1. Ionization energy of $He^+$ ($Z=2$, $n=1$): $E_{He^+}=13.6\times2^2=13.6\times4=x$ (given).
   2. Ionization energy of $Li^{2+}$ ($Z=3$, $n=1$): $E_{Li^{2+}}=13.6\times3^2=13.6\times9$.
   3. Ratio: $\dfrac{E_{Li^{2+}}}{E_{He^+}}=\dfrac{13.6\times9}{13.6\times4}=\dfrac{9}{4}$.
   4. So $E_{Li^{2+}}=\dfrac{9}{4}x$. …

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7. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Which of the following set of quantum numbers represent the electron with highest energy? (A) $n=3,\ l=0,\ m=0,\ s=+\frac{1}{2}$ (B) $n=3,\ l=1,\ m=1,\ s=-\frac{1}{2}$ (C) $n=3,\ l=2,\ m=1,\ s=+\frac{1}{2}$ (D) $n=4,\ l=0,\ m=0,\ s=-\frac{1}{2}$

   ›Reveal solutionSolution

   Comparing orbital energies from quantum numbers uses the $(n+l)$ rule (Aufbau/Madelung ordering): the set with the largest $(n+l)$ sum (and, on a tie, the larger $n$) is highest in energy.

   Concept and Intuition

   For multi-electron atoms, orbital energy isn't decided by $n$ alone — the $(n+l)$ rule says orbitals with a lower $(n+l)$ sum fill first (are lower in energy); when two orbitals share the same $(n+l)$, the one with smaller $n$ is lower. This is why 4s (n+l=4) fills before 3d (n+l=5), even though 3d has a smaller $n$.

   Step-by-Step Solution

   1. (A) $n=3,l=0$ (3s): $n+l=3$.
   2. (B) $n=3,l=1$ (3p): $n+l=4$.
   3. (C) $n=3,l=2$ (3d): $n+l=5$.
   4. (D) $n=4,l=0$ (4s): $n+l=4$. …

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8. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Wavelength of a particular line in Balmer series of atomic spectrum of hydrogen is 656.4 nm. What is the wavelength (in nm) of corresponding line in the spectrum of $\mathrm{He^+}$? (A) 328.2 (B) 164.1 (C) 492.3 (D) 246.1

   ›Reveal solutionSolution

   The same Balmer transition in the hydrogen-like He$^+$ ion has wavelength scaled down by $Z^2=4$ from hydrogen's, giving $164.1\ nm$.

   Concept and Intuition

   The Rydberg formula for any hydrogen-like (single-electron) species is $\dfrac1\lambda = RZ^2\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)$. For the same pair of energy levels $n_1,n_2$ (the "corresponding line"), everything is identical between hydrogen and He$^+$ except the nuclear charge $Z$. Since $\dfrac1\lambda \propto Z^2$, the wavelength itself is inversely proportional to $Z^2$: a higher-charge nucleus pulls electrons in more tightly, so transition energies are larger and wavelengths shorter.

   Step-by-Step Solution

   1. For H ($Z=1$): $\dfrac{1}{\lambda_H} = R\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)$.
   2. For He$^+$ ($Z=2$): $\dfrac{1}{\lambda_{He^+}} = R(2)^2\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right) = 4\times\dfrac{1}{\lambda_H}$. …

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9. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.The difference between the radii of M and N shells of $He^+$ is $\Delta R_1 (nm)$. The difference between the radii of L and N shells of $Li^{2+}$ is $\Delta R_2 (nm)$. The ratio of $\Delta R_1$ to $\Delta R_2$ is (A) 8 : 7 (B) 7 : 8 (C) 3 : 4 (D) 4 : 5

   ›Reveal solutionSolution

   This uses the Bohr-model radius formula for hydrogen-like ions to compute two shell-radius differences and compare them.

   Concept and Intuition

   For any hydrogen-like species, the radius of the $n$-th Bohr orbit is $r_n=\dfrac{a_0 n^2}{Z}$, where $a_0$ is the Bohr radius and $Z$ is the atomic number. Shell letters map to principal quantum numbers: K=1, L=2, M=3, N=4.

   Step-by-Step Solution

   1. For $He^+$ ($Z=2$): $r_M=\dfrac{a_0\cdot3^2}{2}=4.5a_0$; $r_N=\dfrac{a_0\cdot4^2}{2}=8a_0$. $\Delta R_1=r_N-r_M=8a_0-4.5a_0=3.5a_0$.
   2. For $Li^{2+}$ ($Z=3$): $r_L=\dfrac{a_0\cdot2^2}{3}=\dfrac{4}{3}a_0$; $r_N=\dfrac{a_0\cdot4^2}{3}=\dfrac{16}{3}a_0$. $\Delta R_2=r_N-r_L=\dfrac{16}{3}a_0-\dfrac{4}{3}a_0=4a_0$. …

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10. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
    Q.If the velocity of electron in first Bohr's orbit of H-atom is $x\ ms^{-1}$, then the velocity (in $ms^{-1}$) of electron in fourth Bohr's orbit of same atom is ($m_e = 9\times10^{-31}\ kg$) (A) $x/2$ (B) $x/4$ (C) $x/6$ (D) $x/5$

    ›Reveal solutionSolution

    A direct application of the Bohr-model scaling law that orbital electron speed is inversely proportional to the orbit number.

    Concept and Intuition

    In the Bohr model, the electron's orbital speed in the $n$-th orbit of a hydrogen-like atom is $v_n=\dfrac{Ze^2}{2\epsilon_0 nh}\propto\dfrac{1}{n}$ for fixed $Z$. So as the orbit number increases, the electron moves proportionally slower.

    Step-by-Step Solution

    1. $v_n\propto\dfrac{1}{n}$, so $\dfrac{v_4}{v_1}=\dfrac{n_1}{n_4}=\dfrac{1}{4}$.
    2. Given $v_1=x$: $v_4=\dfrac{x}{4}$.

    Common Mistakes …

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11. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
    Q.In hydrogen spectrum, the frequency of the spectral line corresponding to electron transition $n_2 = 3$ to $n_1 = 2$ is $x$ Hz. What is the frequency (in Hz) of the spectral line corresponding to electron transition $n_2 = 4$ to $n_1 = 3$ of $He^+$ spectrum? (A) $\dfrac{5x}{7}$ (B) $\dfrac{7x}{5}$ (C) $\dfrac{20x}{7}$ (D) $\dfrac{7x}{20}$

    ›Reveal solutionSolution

    Both frequencies come from the same Rydberg formula (with $Z^2$ scaling for He$^+$); dividing the two expressions gives $\nu_{He^+}=\dfrac{7x}{5}$.

    Concept and Intuition

    The Rydberg formula for the frequency of a spectral line in a hydrogen-like ion of nuclear charge $Z$ is $\nu=RcZ^2\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)$. For He$^+$ ($Z=2$), every transition frequency is scaled up by a factor of $Z^2=4$ compared to the equivalent transition in hydrogen (same $n_1,n_2$), on top of whatever the specific $n$-dependence contributes.

    Step-by-Step Solution

    1. Hydrogen, $n_2=3\to n_1=2$: $\nu_H=Rc(1)^2\left(\dfrac1{2^2}-\dfrac1{3^2}\right)=Rc\left(\dfrac14-\dfrac19\right)=Rc\cdot\dfrac{9-4}{36}=Rc\cdot\dfrac{5}{36}=x$. …

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12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.Which of the following statements are correct? I) The energy of hydrogen atom in its ground state is $-13.6$ eV II) On the basis of Bohr's model, the radius of the $3^{rd}$ orbit of hydrogen atom is 158.7 pm III) The order of radius of the first orbit of H, $\mathrm{He^+}$, $\mathrm{Li^{2+}}$ and $\mathrm{Be^{3+}}$ is $\mathrm{H > He^+ > Li^{2+} > Be^{3+}}$ (A) II & III only (B) I & III only (C) I & II only (D) I, II, III

    ›Reveal solutionSolution

    This tests the Bohr model formulas for orbit energy and radius, and their $Z$-dependence across isoelectronic hydrogen-like species. Statements I and III are correct; II misapplies the radius formula.

    Concept and Intuition

    In the Bohr model of the hydrogen atom (and hydrogen-like ions), the electron's energy in the $n^{th}$ orbit is $E_n = -\dfrac{13.6\,Z^2}{n^2}$ eV, and its radius is $r_n = 0.529\,\dfrac{n^2}{Z}$ Å. For hydrogen ($Z=1$), the ground state ($n=1$) has $E_1=-13.6$ eV — this is a foundational number worth memorizing. The radius grows as $n^2$ (not linearly with $n$), which is a common trap. For isoelectronic one-electron species (H, $He^+$, $Li^{2+}$, $Be^{3+}$), all have one electron but increasing nuclear charge $Z=1,2,3,4$; since radius $\propto 1/Z$, a larger nuclear pull contracts the orbit.

    Step-by-Step Solution

    1. Statement I: For hydrogen, ground state ($n=1$), $E_1 = -13.6\times 1^2/1^2 = -13.6$ eV. True.
    2. Statement II: For hydrogen, $n=3$: $r_3 = 0.529 \times 3^2/1\ \text{Å} = 0.529\times 9 = 4.761\ \text{Å} = 476.1$ pm. The claimed value of 158.7 pm is actually $0.529\times 3\times 100$ pm — that is, someone used $n$ instead of $n^2$. So Statement II is False. …

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