Q.The number of radial nodes for 3p orbital is __________.

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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