Q.In a hydrogen atom, the electron revolving in the fourth orbit has angular momentum equal to

Showing the 12 most recent of 29 on this concept.

1. CBSE 2026Set 55/2/11 markMCQ
   Q.In Bohr model of hydrogen atom, for large values of $n$, the distance between the consecutive orbits is proportional to (A) $\sqrt{n}$ (B) $n$ (C) $n^2$ (D) $n^3$

   ›Reveal solutionSolution

   In the Bohr model, the orbital radius scales as $r_n \propto n^2$. For large $n$, the gap between consecutive orbits $\Delta r = r_{n+1} - r_n$ behaves like $2n + 1$, which is proportional to $n$ — so the answer is (B).

   The Bohr model gives a beautifully simple picture of the hydrogen atom: electrons orbit the nucleus in fixed circular paths, with quantised angular momentum. The radius of the $n$-th orbit is

   $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \quad \text{or, more compactly,} \quad r_n = a_0 n^2,$

   where $a_0 \approx 0.529 \, \text{Å}$ is the Bohr radius. So the radius grows as $n^2$.

   The question asks about the distance between consecutive orbits — that is, the difference $r_{n+1} - r_n$ — for large $n$. Many students instinctively think this difference is constant, or that it grows like $n^2$ because the radii themselves do. But a difference between two quadratic terms behaves differently from either term alone. Let’s work it out.

   1. Write the radii for two neighbouring orbits:

   $r_n = a_0 n^2, \qquad r_{n+1} = a_0 (n+1)^2.$

   2. The gap between them is

   $\Delta r = r_{n+1} - r_n = a_0 \left[ (n+1)^2 - n^2 \right].$

   3. Expand $(n+1)^2 = n^2 + 2n + 1$. Then

   $\Delta r = a_0 \left( n^2 + 2n + 1 - n^2 \right) = a_0 (2n + 1).$

   4. For large $n$, the constant $1$ becomes negligible compared to $2n$. So

   $\Delta r \approx 2 a_0 n.$

   Thus the spacing between consecutive orbits is proportional to $n$ itself — not $n^2$, not $\sqrt{n}$, not $n^3$.

   Watch out

   A common mistake is to think that because $r_n \propto n^2$, the difference must also be $\propto n^2$. But the difference of two squares is linear in $n$: $(n+1)^2 - n^2 = 2n+1$. Always check the actual difference, not just the scaling of the individual term.

   Tip

   This is a neat example of how a discrete difference approximates a derivative: $\frac{dr}{dn} \propto 2n$, which is linear in $n$. For large $n$, the Bohr atom’s orbits get farther apart linearly — a fact that connects to the correspondence principle, where classical behaviour emerges.

   ✓Final answer

   The distance between consecutive orbits is proportional to $n$, so the correct option is (B).

2. CBSE 2026Set A1 markMCQ
   Q.The radius of the lowest Bohr's orbit in hydrogen atom is r0. The radius of Bohr's second orbit is (A) r0 (B) 2r0 (C) 4r0 (D) r0/2

   ›Reveal solutionSolution

   Bohr radii scale as n², so r₂ = 4 r₀.

   In Bohr's model of hydrogen the radius of the $n$-th orbit is:

   $r_n=n^2 r_0$

   where $r_0$ is the radius of the lowest (ground-state, $n=1$) orbit. For the second orbit, $n=2$:

   $r_2=2^2 r_0=4r_0.$

   ✓Final answer

   (C) 4r₀.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the first Bohr radius of hydrogen atom be R, then the radius of the third orbit is

   (a) 9R

   (b) R/3

   (c) 3R

   (d) R/9

   ›Reveal solutionSolution

   Bohr radius of the nth orbit scales as $n^2$, so the third orbit's radius is $9R$.

   In the Bohr model, the radius of the $n$th orbit of the hydrogen atom is

   $r_n = n^2 r_1$

   where $r_1$ is the radius of the first orbit (given as $R$ here). For $n=3$:

   $r_3 = 3^2 \times R = 9R$

   ✓Final answer

   (a) 9R.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Case study: Bohr's model addressed the instability of the Rutherford model by introducing quantization. The model is based on three postulates, which successfully explained the discrete line spectrum of hydrogen. According to the model, the radius of the nth stationary orbit is r_n ∝ n², and the total energy is E_n = −13.6 eV / n², when an electron jumps from a higher energy level (E_i) to a lower one (E_f), a photon of energy hν = E_i − E_f is emitted. Transitions ending at the n = 1 level form the Lyman series. According to Bohr's second postulate, which quantity is quantized?

   (a) Energy of electron

   (b) Orbital angular momentum

   (c) Linear momentum

   (d) Frequency of revolution

   ›Reveal solutionSolution

   Bohr's second postulate quantizes the electron's orbital angular momentum in integer multiples of $h/2\pi$.

   Bohr's second postulate states that an electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of $\hbar = h/2\pi$:

   $mvr = \dfrac{nh}{2\pi}, \quad n = 1, 2, 3, \dots$

   ✓Final answer

   Orbital angular momentum is the quantized quantity (option b).

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.According to Bohr's hypothesis the following physical quantity is quantised

   (a) angular momentum

   (b) angular velocity

   (c) potential energy

   (d) momentum

   ›Reveal solutionSolution

   Bohr's key postulate (beyond classical mechanics) was that only orbits where the electron's angular momentum is an integer multiple of $h/2\pi$ are allowed.

   Bohr's second postulate states that the electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of $h/2\pi$:

   $L = mvr = \dfrac{nh}{2\pi}, \quad n = 1, 2, 3, \ldots$

   This quantisation condition (not present in classical Rutherford's planetary model) is what restricts the electron to discrete stable orbits and discrete energy levels.

   ✓Final answer

   (a) angular momentum.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The radius of Bohr's stable orbit for hydrogen is r. The radius of Bohr's second orbit is

   (a) r

   (b) r/2

   (c) 2r

   (d) 4r

   ›Reveal solutionSolution

   In the Bohr model, the radius of the nth orbit of hydrogen scales as n^2, so if the (first, ground-state) orbit has radius r, the second orbit has radius 4r.

   Bohr's model gives the radius of the nth stationary orbit of hydrogen as:

   r_n = n^2 r_1

   where r_1 is the radius of the first (n=1) orbit. Taking the given "stable orbit" radius r to be r_1, the second orbit (n=2) has radius:

   r_2 = (2)^2 r_1 = 4r

   This n^2 dependence comes from combining Bohr's quantization of angular momentum (mvr = nh/2-pi) with the Coulomb force providing the centripetal force.

   ✓Final answer

   (d) 4r.

7. CBSE 2025Set ANNUAL1 mark
   Q.If radius of first electron orbit of hydrogen is r0, radius of second electron orbit of hydrogen is ______.

   ›Reveal solutionSolution

   In the Bohr model, the radius of the nth orbit of hydrogen scales as $r_n = n^2 r_0$, where $r_0$ is the first-orbit (Bohr) radius.

   Bohr's model gives the radius of the nth stationary orbit of the hydrogen atom as

   $r_n = n^2 r_0$

   where $r_0 \approx 0.529\ \text{\AA}$ is the radius of the first orbit (n = 1). This follows from equating the Coulomb force to the centripetal force together with Bohr's quantization condition on angular momentum, $mvr = n\dfrac{h}{2\pi}$.

   For the second orbit, n = 2:

   $r_2 = 2^2 r_0 = 4r_0$

   ✓Final answer

   $r_2 = 4r_0$.

8. CBSE 2024Set ANNUAL1 mark
   Q.If the radius of first orbit of hydrogen atom is 0.5 x 10^-10 m, then the radius of its second orbit will be __________ m.

   ›Reveal solutionSolution

   Bohr's model gives orbit radius proportional to n², so the second orbit's radius is 2² = 4 times the first orbit's radius.

   In Bohr's model of the hydrogen atom, the radius of the nth orbit is:

   $r_n = n^2 r_1$

   where $r_1$ is the radius of the first (ground state) orbit. Given $r_1 = 0.5 \times 10^{-10}\ \text{m}$, for the second orbit (n=2):

   $r_2 = 2^2 \times r_1 = 4 \times 0.5\times 10^{-10} = 2 \times 10^{-10}\ \text{m}$

   ✓Final answer

   2 × 10^-10 m.

9. CBSE 2024Set ANNUAL1 mark
   Q.What is Bohr's quantisation condition for the angular momentum of an electron in the second orbit ?

   ›Reveal solutionSolution

   Bohr's second postulate: angular momentum is quantised as $L=\dfrac{nh}{2\pi}$; for the second orbit, n=2.

   Bohr's quantisation condition states that an electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of $h/2\pi$:

   $L = m v r = \dfrac{nh}{2\pi}, \quad n = 1,2,3,\dots$

   For the second orbit, $n=2$:

   $L = \dfrac{2h}{2\pi} = \dfrac{h}{\pi}$

   ✓Final answer

   $L=\dfrac{2h}{2\pi}=\dfrac{h}{\pi}$.

10. CBSE 2023Set 55/1/11 markMCQ
    Q.The radius of the $n^{\text{th}}$ orbit in Bohr model of hydrogen atom is proportional to :

    (a) $\dfrac{1}{n^2}$

    (b) $\dfrac{1}{n}$

    (c) $n^2$

    (d) $n$

    ›Reveal solutionSolution

    In the Bohr model, the electron's orbit radius grows with the principal quantum number because higher orbits require more angular momentum and lower electrostatic attraction. The radius is proportional to $n^2$.

    The Bohr model treats the hydrogen atom as a miniature solar system where the electron orbits the nucleus in circular paths. But unlike planets, the electron can only occupy certain allowed orbits, determined by quantum conditions. The question asks how the orbital radius scales with the quantum number $n$.

    The key insight is that two forces govern the electron's motion: the electrostatic attraction pulling it inward and the requirement that its angular momentum be quantized. Let me show you how these constraints lead to the radius formula.

    The physics behind the orbit

    For a stable circular orbit, the centripetal force must equal the electrostatic force:

    $\frac{mv^2}{r} = \frac{ke^2}{r^2}$

    where $m$ is the electron mass, $v$ its speed, $k$ is Coulomb's constant, and $e$ the electron charge. This gives us one equation relating $r$ and $v$.

    Bohr's quantum condition provides the second equation. He postulated that angular momentum is quantized:

    $mvr = n\hbar$

    where $\hbar = \frac{h}{2\pi}$ and $n = 1, 2, 3, \ldots$ is the principal quantum number.

    Deriving the radius dependence

    1. From the angular momentum condition, solve for $v$:

    $v = \frac{n\hbar}{mr}$

    2. Substitute this into the force balance equation:

    $\frac{m}{r}\left(\frac{n\hbar}{mr}\right)^2 = \frac{ke^2}{r^2}$

    3. Simplify the left side:

    $\frac{m \cdot n^2\hbar^2}{m^2r^3} = \frac{n^2\hbar^2}{mr^3} = \frac{ke^2}{r^2}$

    4. Multiply both sides by $r^3$:

    $\frac{n^2\hbar^2}{m} = ke^2 r$

    5. Solve for $r$:

    $r = \frac{n^2\hbar^2}{mke^2}$

    Everything on the right except $n^2$ is a constant (fundamental constants of nature). Therefore:

    $r_n \propto n^2$

    The proportionality constant is the Bohr radius $a_0 = \frac{\hbar^2}{mke^2} \approx 0.529$ Å, so $r_n = n^2 a_0$.

    Physical interpretation

    Why does the radius grow as $n^2$ and not linearly? As $n$ increases, the electron must carry more angular momentum ($\propto n$). But angular momentum is $mvr$, and the velocity actually decreases in higher orbits (the electron is less tightly bound). To compensate for the slower speed while maintaining higher angular momentum, the radius must increase faster than $n$ — specifically, as $n^2$.

    Watch out

    Don't confuse this with the energy levels, which go as $E_n \propto -\frac{1}{n^2}$. The radius grows while the (negative) energy increases toward zero.

    ✓Final answer

    The correct option is (c) $n^2$.

11. CBSE 2023Set ANNUAL1 mark
    Q.Write the mathematical form of Bohr's postulate regarding angular momentum of electron in atom. (Write the answer only)

    ›Reveal solutionSolution

    Bohr's second postulate states that the angular momentum of the electron in a stationary orbit is an integral multiple of $h/2\pi$.

    Bohr postulated that an electron can revolve only in those orbits for which its orbital angular momentum is quantized:

    $L = mvr = \dfrac{nh}{2\pi}, \quad n = 1, 2, 3, \ldots$

    where $m$ is the electron mass, $v$ its orbital speed, $r$ the orbit radius, $h$ Planck's constant, and $n$ the principal quantum number. Only orbits satisfying this condition are 'allowed' (stable, non-radiating) orbits.

    ✓Final answer

    $mvr = nh/2\pi$.

12. CBSE 2023Set ANNUAL1 markMCQ
    Q.The Bohr model of atoms

    (1) assumes that the angular momentum of electrons is quantized

    (2) uses Einstein's photoelectric equation

    (3) predicts continuous emission spectra for atoms

    (4) predicts the same emission spectra for all types of atoms

    ›Reveal solutionSolution

    Bohr's central postulate was that electrons can only occupy orbits where angular momentum is an integer multiple of h/2π.

    Bohr postulated that an electron revolves only in those orbits for which its angular momentum is quantized: $L=mvr=\dfrac{nh}{2\pi}$, $n=1,2,3,\dots$. This quantization condition (not option b, which is unrelated to Bohr's model; and not options c/d, since Bohr's model correctly predicts DISCRETE, atom-specific line spectra, not continuous or universal ones) is the defining feature of the Bohr model.

    ✓Final answer

    (1) assumes that the angular momentum of electrons is quantized.