Q.For the ground state, the electron in the H-atom has an angular momentum $= \dfrac{h}{2\pi}$, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.An electron in the ground state of hydrogen atom is revolving in anti-clockwise direction in a circular orbit. The orbital magnetic moment of the electron is given by (A) $\dfrac{h}{2\pi}$ (B) $\dfrac{eh}{2\pi m}$ (C) $\dfrac{eh}{4\pi m}$ (D) $\dfrac{h}{4\pi m}$

   ›Reveal solutionSolution

   This tests the gyromagnetic-ratio relation between orbital magnetic moment and angular momentum for the ground-state hydrogen electron; the answer is $\dfrac{eh}{4\pi m}$.

   Concept and Intuition

   A charge moving in a circular orbit constitutes a tiny current loop, and any current loop has a magnetic moment $\mu = IA$. This can always be re-expressed in terms of the particle's orbital angular momentum $L$ via the classical gyromagnetic ratio $\mu_L = \dfrac{e}{2m}L$, valid regardless of orbit size or speed, as long as we know $L$. For hydrogen's ground state, the Bohr model fixes $L = \dfrac{h}{2\pi}$ (i.e., $n=1$, $L = n\hbar$), so combining the two relations gives the magnetic moment directly.

   Step-by-Step Solution

   1. Orbital magnetic moment in terms of angular momentum: $\mu_L = \dfrac{e}{2m}L$ (standard result for a charge $-e$ orbiting; magnitude used here). …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The energy of an electron in Bohr's hydrogen atom is $-3.4$ eV. The angular momentum of the electron is (A) $\dfrac{2h}{\pi}$ (B) $\dfrac{h}{2\pi}$ (C) $\dfrac{h}{\pi}$ (D) $\dfrac{h}{4\pi}$

   ›Reveal solutionSolution

   Identify the orbit number from the given energy, then use Bohr's quantization rule $L=nh/2\pi$. $E=-3.4$ eV corresponds to $n=2$, giving $L=h/\pi$.

   Concept and Intuition

   In the Bohr model, an electron in the $n$-th orbit of hydrogen has energy $E_n = -13.6/n^2$ eV, and its orbital angular momentum is quantized as $L_n = n\hbar = \dfrac{nh}{2\pi}$ — this quantization condition is the postulate that let Bohr explain the discrete hydrogen spectrum.

   Step-by-Step Solution

   1. Given $E_n=-3.4$ eV. Set $-\dfrac{13.6}{n^2}=-3.4 \Rightarrow n^2=\dfrac{13.6}{3.4}=4 \Rightarrow n=2$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If the ratio of the time periods of the electrons revolving in the first and $n^{th}$ orbits of Hydrogen atom is 1 : 64, then the angular momentum of the electron in the $n^{th}$ excited state of Hydrogen atom is (h – Planck's constant) (A) $\frac{3.5h}{\pi}$ (B) $\frac{5h}{\pi}$ (C) $\frac{2.5h}{\pi}$ (D) $\frac{2h}{\pi}$

   ›Reveal solutionSolution

   $T\propto n^3$ gives orbit $n=4$; the 'nth (fourth) excited state' means principal number 5, so $L = 5\cdot\dfrac{h}{2\pi} = \dfrac{2.5h}{\pi}$.

   Concept and Intuition

   In Bohr's model $r\propto n^2$ and $v\propto 1/n$, so the period $T = \dfrac{2\pi r}{v}\propto n^3$. The period ratio pins the orbit number. The angular momentum is quantised as $L = n\dfrac{h}{2\pi}$.

   Step-by-Step Solution

   1. $\dfrac{T_1}{T_n} = \dfrac{1^3}{n^3} = \dfrac{1}{64} \Rightarrow n^3 = 64 \Rightarrow n = 4$.
   2. The 'nth excited state' with $n=4$ means the fourth excited state; counting ground $=1$, 1st excited $=2$, …, 4th excited corresponds to principal quantum number $5$. …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Suppose an electron is attracted towards the origin by a force $\dfrac{K}{r}$, where K is a constant and r is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the $n^{th}$ orbit of the electron is found to be $r_n$, and the kinetic energy of the electron to be $T_n$, then which of the following is true? (A) $T_n$ is independent of n, $r_n \propto n$ (B) $T_n \propto \dfrac{1}{n}$, $r_n \propto n$ (C) $T_n \propto \dfrac{1}{n}$, $r_n \propto n^2$ (D) $T_n \propto \dfrac{1}{n^2}$, $r_n \propto n^2$

   ›Reveal solutionSolution

   An unusual force law $F=K/r$ makes the orbital speed the same in every Bohr orbit, so kinetic energy doesn't depend on $n$ at all, while quantized angular momentum then forces the radius to grow linearly with $n$. Answer: (A).

   Concept and Intuition

   Bohr's model has two ingredients regardless of the force law: (i) the given force supplies the centripetal force for circular motion, and (ii) angular momentum is quantized, $mvr_n=n\hbar$. Normally (Coulomb force $\propto 1/r^2$) both $v$ and $r$ depend on $n$ in specific ways; here the unusual $1/r$ force makes the speed itself independent of $r$ (hence of $n$), which is the key simplifying feature of this problem.

   Step-by-Step Solution

   1. Centripetal condition: $\dfrac{mv^2}{r}=\dfrac{K}{r}\Rightarrow mv^2=K$, i.e. $v=\sqrt{K/m}$ — a constant, the same for every orbit (independent of $r$ or $n$).
   2. Kinetic energy: $T_n=\tfrac12 mv^2=\dfrac{K}{2}$ — a constant, independent of $n$.
   3. Bohr's angular-momentum quantization: $mvr_n=n\hbar$. …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.An electron has an angular momentum of $90\hbar\ J-S$ while orbiting with a linear velocity of $\pi \times 10^5\ ms^{-1}$ then the radius of the orbit is (mass of electron $= 9 \times 10^{-3}\ Kg$ and Planks Constant $= 6.6 \times 10^{-34}\ J\ s$) (A) $66 \times 10^{-15}\ m$ (B) $33 \times 10^{-15}\ m$ (C) $66 \times 10^{-10}\ m$ (D) $33 \times 10^{-8}\ m$

   ›Reveal solutionSolution

   Angular momentum of a circulating particle is $L=mvr$, so the orbit radius is $r=L/(mv)$, with $L$ given as a multiple of $\hbar=h/2\pi$. Answer: $\approx3.3\times10^{-8}\ m$.

   Concept and Intuition

   For a particle moving in a circular orbit with speed $v$ at radius $r$, its angular momentum about the center is simply $L = mvr$ (mass times linear momentum times the lever arm, which here is the radius itself since velocity is tangential/perpendicular to the radius vector). Given the angular momentum in units of the reduced Planck constant ($L=n\hbar$), we can invert this relation to solve directly for the radius.

   Step-by-Step Solution

   1. Angular momentum: $L = n\hbar = n\dfrac{h}{2\pi}$, with $n=90$, $h=6.6\times10^{-34}\ Js$. $L = 90\times\dfrac{6.6\times10^{-34}}{2\pi} = 90\times1.05\times10^{-34} \approx 9.45\times10^{-33}\ J\,s$.
   2. Since $L = mvr$, the radius is $r = \dfrac{L}{mv}$.
   3. Momentum: $mv = (9\times10^{-31})\times(\pi\times10^{5}) \approx 2.83\times10^{-25}\ kg\,m/s$. …

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6. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.The ratio of areas of $2^{nd}$ and $3^{rd}$ Bohr's orbits in a doubly ionized Lithium atom is (A) 16 : 81 (B) 4 : 5 (C) 4 : 9 (D) 2 : 3

   ›Reveal solutionSolution

   Bohr orbit radius scales as $n^2$, so orbit area scales as $n^4$; for $n=2$ and $n=3$ this gives an area ratio of $16:81$.

   Concept and Intuition

   In the Bohr model, $r_n = \dfrac{n^2 a_0}{Z}$. Since $Z$ (here, for doubly-ionised lithium, $Z=3$) is the same for both orbits being compared, it cancels out in any ratio of radii for the same atom/ion. The area of a circular orbit scales as the square of the radius, so it scales as $n^4$.

   Step-by-Step Solution

   1. $r_n \propto n^2/Z$; since $Z$ is fixed (same ion, $Z=3$), $r_n \propto n^2$.
   2. Area $A_n = \pi r_n^2 \propto n^4$.
   3. Ratio of areas of $2^{nd}$ and $3^{rd}$ orbits: $\dfrac{A_2}{A_3} = \dfrac{2^4}{3^4} = \dfrac{16}{81}$. …

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7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.An electron is moving in an orbit of hydrogen atom in which there can be a maximum of six transitions. Another electron is moving in an another orbit of hydrogen atom in which there can be maximum of three transitions. The ratio of the velocity of electrons in these two orbits is (A) $3/4$ (B) $5/4$ (C) $2/1$ (D) $1/2$

   ›Reveal solutionSolution

   Maximum-transitions count fixes the principal quantum numbers as $n=4$ and $n=3$; since $v_n\propto 1/n$, their speed ratio is $3/4$.

   Concept and Intuition

   In the Bohr model, an electron in level $n$ can make a transition to any of the $(n-1)$ lower levels, and the total number of distinct spectral lines obtainable from levels $1$ to $n$ (or equivalently, the number of possible transitions starting from the topmost level $n$ when all electrons are in it) is $\binom{n}{2}=\dfrac{n(n-1)}{2}$. Once we know $n$ for each orbit, we use the Bohr result that orbital speed $v_n = \dfrac{v_1}{n} \propto \dfrac1n$ — higher orbits move slower.

   Step-by-Step Solution

   1. First orbit: $\dfrac{n(n-1)}{2} = 6 \Rightarrow n(n-1)=12 \Rightarrow n=4$ (since $4\times3=12$).
   2. Second orbit: $\dfrac{n(n-1)}{2} = 3 \Rightarrow n(n-1)=6 \Rightarrow n=3$ (since $3\times2=6$).
   3. Bohr orbital speed: $v_n \propto \dfrac1n$, so $v_4 \propto \dfrac14$ and $v_3\propto \dfrac13$. …

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8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.If the angular momenta of electrons in two orbits of hydrogen atom are $\dfrac{h}{\pi}$ and $\dfrac{1.5h}{\pi}$, then the ratio of velocities of electrons in these two orbits is (h - Planck's constant) (A) 3 : 2 (B) 3 : 4 (C) 9 : 4 (D) 1 : 3

   ›Reveal solutionSolution

   Converting the given angular momenta to orbit numbers ($n_1=2,\,n_2=3$) and using $v_n\propto 1/n$ gives $v_1:v_2 = 3:2$.

   Concept and Intuition

   Bohr's quantization condition states the angular momentum of an electron in the $n$-th orbit is $L_n = n\dfrac{h}{2\pi}$. Also, from the Bohr model, the orbital speed is $v_n = \dfrac{Ze^2}{2\varepsilon_0 n h}$, i.e. $v_n \propto \dfrac{1}{n}$ — electrons in higher (larger) orbits move slower. So once we identify which orbit numbers correspond to the given angular momenta, the velocity ratio follows immediately by inverting the $n$ ratio.

   Step-by-Step Solution

   1. First orbit: $L_1 = \dfrac{h}{\pi} = n_1\dfrac{h}{2\pi} \implies n_1 = 2$.
   2. Second orbit: $L_2 = \dfrac{1.5h}{\pi} = n_2\dfrac{h}{2\pi} \implies n_2 = 3$. …

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9. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.The ratio of the time periods of the revolution of the electrons in the second and third excited states of hydrogen atom is (A) $9:16$ (B) $27:64$ (C) $4:9$ (D) $8:27$

   ›Reveal solutionSolution

   Using Bohr's model, the orbital period scales as $n^3$; for $n=3$ (second excited) and $n=4$ (third excited) the ratio is $27:64$.

   Concept and Intuition

   In the Bohr model, orbit radius grows as $r_n\propto n^2$ while orbital speed falls as $v_n\propto 1/n$. The time period is $T_n=\dfrac{2\pi r_n}{v_n}$, so combining these gives $T_n\propto n^2\times n=n^3$ — outer orbits take dramatically longer to complete one revolution.

   Step-by-Step Solution

   1. Identify the principal quantum numbers: ground state is $n=1$, so first excited $=n=2$, second excited $=n=3$, third excited $=n=4$.
   2. Since $T_n\propto n^3$: $\dfrac{T_3}{T_4}=\dfrac{3^3}{4^3}=\dfrac{27}{64}$. …

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10. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
    Q.If the angular momentum of an electron in the second orbit of hydrogen atom is J, then the angular momentum of an electron in the third excited state of hydrogen atom is (A) 2 J (B) 3 J (C) 4 J (D) 6 J

    ›Reveal solutionSolution

    Bohr's angular momentum quantization ($L_n \propto n$) turns the given $L_2=J$ into $L_4=2J$ for the third excited state ($n=4$).

    Concept and Intuition

    Bohr's model postulates that the angular momentum of an electron in the $n$-th orbit is quantized as $L_n=\dfrac{nh}{2\pi}$ — it simply scales linearly with the orbit number n. The key subtlety here is counting orbits correctly: the ground state is $n=1$, so the "first excited state" is $n=2$, the "second excited state" is $n=3$, and the "third excited state" is $n=4$.

    Step-by-Step Solution

    1. Given: angular momentum in the second orbit ($n=2$) is $J$. So $L_2=\dfrac{2h}{2\pi}=J$, giving $\dfrac{h}{2\pi}=\dfrac{J}{2}$.
    2. Identify the third excited state: ground state $n=1$ → 1st excited $n=2$ → 2nd excited $n=3$ → 3rd excited $n=4$. …

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11. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.μ – meson of charge 'e', mass 208 me moves in a circular orbit around a heavy nucleus having charge +3e. The quantum state 'n' for which the radius of the orbit is same as that of the first Bohr orbit for hydrogen atom is [approximately] (A) $n \approx 20$ (B) $n \approx 25$ (C) $n \approx 28$ (D) $n \approx 29$

    ›Reveal solutionSolution

    The Bohr radius scales as $n^2/(Z\cdot m)$; matching the muon's orbit radius (Z=3, m=208

    electron masses) to hydrogen's first Bohr radius gives $n^2=624$, i.e. $n\approx25$.

    Concept and Intuition

    The Bohr model radius for a hydrogen-like system is $r_n = \dfrac{n^2\hbar^2}{Z m e^2/(4\pi \epsilon_0)} \propto \dfrac{n^2}{Zm}$ (for fixed fundamental constants), where m is the orbiting

    particle's mass. A heavier, more charge-attracted particle (the muon, with $m=208m_e$ around

    $Z=3$) needs a much larger quantum number n to reach the same radius as the electron's smallest

    hydrogen orbit.

    Step-by-Step Solution

    1. Write $r_n(\text{muon}) = \dfrac{n^2}{Z}\cdot\dfrac{m_e}{m_\mu}\,a_0$, where $a_0$ is hydrogen's first Bohr radius (with $n=1,Z=1,m=m_e$).
    2. We want $r_n(\text{muon}) = a_0$ (matches the first Bohr orbit of hydrogen). …

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12. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.The speed of the electron is a hydrogen atom in the $n = 3$ level is (Plank constant $= 6.6 \times 10^{-34}\ \mathrm{Js}$) (A) $6.2 \times 10^5\ \mathrm{ms^{-1}}$ (B) $3.7 \times 10^5\ \mathrm{ms^{-1}}$ (C) $7.3 \times 10^5\ \mathrm{ms^{-1}}$ (D) $1.6 \times 10^5\ \mathrm{ms^{-1}}$

    ›Reveal solutionSolution

    The Bohr-model orbital speed scales as $1/n$; using the standard first-orbit speed of hydrogen, the $n=3$ speed comes out to about $7.3\times10^5$ m/s.

    Concept and Intuition

    In Bohr's model, angular momentum is quantized ($mvr = nh/2\pi$) and the Coulomb force provides centripetal force. Solving these together gives an orbital speed $v_n = \dfrac{e^2}{2\varepsilon_0 n h} = \dfrac{v_1}{n}$ for hydrogen, where $v_1 \approx 2.18\times10^6\ \mathrm{m\,s^{-1}}$ is the speed in the ground state ($n=1$). Higher orbits therefore have progressively lower orbital speeds.

    Step-by-Step Solution

    1. Use $v_n = v_1/n$ with $v_1 = 2.18\times10^6\ \mathrm{m\,s^{-1}}$ (standard hydrogen ground-state speed, consistent with $e^2/2\varepsilon_0 h$ using the given Planck constant).
    2. For $n=3$: $v_3 = 2.18\times10^6/3 = 7.27\times10^5\ \mathrm{m\,s^{-1}}$. …

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