Taking the Bohr radius as a₀ = 53pm, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr's model,

NCERT Exemplar · Q1

a_0 = 53\ \text{pm}

(a)

53\ \text{pm}

(b)

27\ \text{pm}

(c)

18\ \text{pm}

(d)

13\ \text{pm}

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Concept understanding — Bohr Model Quantization

Why does an electron not spiral into the nucleus?

Imagine you're pushing a child on a swing. If you push at random moments, the swing jerks and slows down. But if you push exactly in rhythm with the swing's natural motion, each push adds energy smoothly and the swing goes higher and higher. The swing "prefers" to move at specific frequencies — its natural modes.

An electron orbiting a nucleus is similar, but with a crucial twist from the quantum world. Classical physics says an accelerating charge (like an electron going in a circle) must continuously radiate energy. If that were true, the electron would lose energy, spiral into the nucleus, and atoms would collapse in a flash of light. But atoms are stable. So something is fundamentally different.

The radical idea: allowed orbits only

Niels Bohr proposed in 1913 that the electron cannot occupy just any orbit. It can only exist in certain stationary states — orbits where it does not radiate energy. These are like the swing's natural frequencies, but for an electron.

The key condition that picks out these special orbits is called quantization of angular momentum.

L = n \frac{h}{2\pi}, \quad n = 1, 2, 3, \dots

Here $L$ is the orbital angular momentum of the electron, $h$ is Planck's constant, and $n$ is a positive integer called the principal quantum number.

What this means physically

Angular momentum for a circular orbit is $L = m v r$ , where $m$ is the electron mass, $v$ its speed, and $r$ the orbit radius. So the quantization condition becomes:

m v r = n \frac{h}{2\pi}

This is not a formula you derive — it is a postulate, a rule that nature follows. Bohr had no deeper explanation for why this rule works; he simply noticed it gave the right answers for hydrogen's spectrum.

Note

The quantity $\frac{h}{2\pi}$ appears so often that it has its own symbol: $\hbar$ (h-bar). So the condition is often written as $L = n\hbar$ .

What it predicts

Combining this quantization with Newton's law for circular motion (centripetal force = Coulomb attraction) gives:

Radius of the $n$ th orbit: $r_n = n^2 a_0$ , where $a_0 = 0.529 \, \text{Å}$ is the Bohr radius (the smallest orbit, $n=1$ ).
Energy of the $n$ th orbit: $E_n = -\frac{13.6 \, \text{eV}}{n^2}$

The negative sign means the electron is bound to the nucleus. As $n$ increases, the orbit gets larger and the energy becomes less negative (closer to zero).

The key insight for exams

Bohr quantization is not a derivation — it is a condition you apply. When you see a problem about hydrogen-like atoms (one electron), you:

Write $mvr = n\hbar$
Write the force balance: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$ (for nuclear charge $Ze$ )
Solve for $r$ and $v$ in terms of $n$

Watch out

Bohr's model works perfectly only for single-electron systems: H, He $^+$ , Li $^{2+}$ , etc. It fails for multi-electron atoms because it ignores electron-electron repulsion and the wave nature of electrons.

Why "quantization"?

The word comes from the Latin quantus — "how much." In classical physics, angular momentum can take any value. In Bohr's atom, it comes only in discrete packets (quanta) of size $\hbar$ . This is the first hint that at the atomic scale, nature is not continuous but granular.

The electron does not spiral because it cannot lose energy gradually — it can only jump from one allowed orbit to another, emitting or absorbing a photon of exactly the right energy. Between these jumps, it simply exists in a stationary state, defying classical expectations.

Bohr's quantization of angular momentum is one of the defining postulates covered in the NCERT Class 12 Physics Atoms chapter, and students frequently search for "Bohr model quantization condition and derivation" or "Bohr's model important questions" while preparing for CBSE boards and JEE Main/NEET. This concept is also a common launching point for numerical problems on orbital radius and energy levels of hydrogen-like atoms tested across competitive exams.

Why this formula?

Why Angular Momentum is Quantised in the Bohr Model

The Bohr model's most famous result — that angular momentum comes only in integer multiples of $\frac{h}{2\pi}$ — is not an arbitrary assumption. It follows directly from a single, elegant idea: the electron's wave must close on itself.

The core problem Bohr faced

By 1913, physicists knew two things that seemed contradictory:

Rutherford's nuclear model showed electrons orbiting the nucleus.
Maxwell's equations predicted that any accelerating charge (like an orbiting electron) must radiate energy, spiral inward, and collapse in about $10^{-11}$ seconds.

Atoms are stable. Something was missing.

Bohr's breakthrough was to combine the newly discovered quantum idea (Planck's constant $h$ ) with classical mechanics, but only for allowed orbits. The key constraint came from thinking of the electron not as a tiny planet, but as a standing wave.

The de Broglie wavelength argument (the cleanest derivation)

A few years after Bohr, de Broglie proposed that every moving particle has a wavelength:

\lambda = \frac{h}{p} = \frac{h}{mv}

For an electron in a circular orbit of radius $r$ , the circumference is $2\pi r$ . For the wave to be stable — not cancelling itself out — the circumference must contain an integer number of wavelengths:

2\pi r = n\lambda, \quad n = 1, 2, 3, \dots

Substitute $\lambda = h/(mv)$ :

2\pi r = n \cdot \frac{h}{mv}

Rearrange:

mvr = n \cdot \frac{h}{2\pi}

That's it. The left side $mvr$ is the angular momentum $L$ . So:

L = n\hbar, \quad \text{where } \hbar = \frac{h}{2\pi}

Note

This is not a separate postulate — it is a consequence of requiring the electron wave to be a standing wave. If the wave doesn't close on itself, it interferes destructively and the orbit cannot exist.

Why this fixes the energy levels

Once angular momentum is quantised, the rest follows from classical physics. For a circular orbit, the centripetal force is provided by the Coulomb attraction:

\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}

Combine this with $mvr = n\hbar$ and solve for $r$ and $E$ :

r_n = \frac{4\pi\epsilon_0 \hbar^2}{m e^2} \cdot n^2 = a_0 n^2

E_n = -\frac{m e^4}{8\epsilon_0^2 h^2} \cdot \frac{1}{n^2} = -\frac{13.6 \text{ eV}}{n^2}

Watch out

A common mistake is to think Bohr derived the quantization rule from first principles. He didn't — he postulated it. The de Broglie standing-wave argument came later and provides the physical reason for the postulate, but it is still a postulate in the full quantum theory.

The deeper reason: it's not really about orbits

The Bohr model is ultimately wrong — electrons don't orbit in neat circles. But the quantization of angular momentum survives in the full quantum mechanical treatment (Schrödinger equation) as the condition that the wavefunction must be single-valued. For the hydrogen atom, the angular momentum quantum number $l$ can take values $0, 1, 2, \dots, n-1$ , and the magnitude is $\sqrt{l(l+1)}\hbar$ , not $n\hbar$ .

Yet the Bohr model's key insight — that only certain discrete states are allowed — remains the foundation of atomic physics. The formula $L = n\hbar$ is the simplest example of a quantum number, and it correctly predicts the hydrogen spectrum to within fine-structure corrections.

Important

The Bohr quantization condition $L = n\hbar$ is a boundary condition on the electron wave, not a dynamical law. It says: for the electron to exist in a stable state, its wave must fit perfectly around the nucleus. This is the same principle that governs standing waves on a string or in an organ pipe — only certain wavelengths survive.

The Bohr radius scales as $r_n \propto n^2 / Z$ . For $\text{Li}^{++}$ (Z=3) in the ground state (n=1), the radius is $a_0 / 3 \approx 18\ \text{pm}$ , so the correct option is (C).

The Bohr model gives us a beautifully simple way to think about atomic radii: the electron orbits the nucleus in quantized circular paths, and the radius of the $n$ -th orbit depends on two things — the principal quantum number $n$ (which tells you the "size" of the orbit) and the nuclear charge $Z$ (which pulls the electron inward more strongly as $Z$ increases).

For a hydrogen-like ion (one electron around a nucleus of charge $+Ze$ ), the radius of the $n$ -th orbit is:

r_n = \frac{n^2}{Z} a_0

where $a_0 = 53\ \text{pm}$ is the Bohr radius for hydrogen ( $Z=1$ , $n=1$ ).

The key insight: higher $Z$ shrinks the orbit because the stronger Coulomb attraction pulls the electron closer. For $\text{Li}^{++}$ , the nucleus has $Z=3$ and there is only one electron left (it's a hydrogen-like ion). In its ground state, $n=1$ .

Let's work through it step by step.

Identify the ion and its parameters.

$\text{Li}^{++}$ means a lithium atom that has lost two electrons, leaving just one electron. So it's a hydrogen-like ion with nuclear charge $Z=3$ . The ground state means the electron is in the lowest energy orbit, $n=1$ .
Recall the Bohr radius formula for hydrogen-like atoms.

The general expression for the radius of the $n$ -th orbit is:

r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \cdot \frac{1}{Z}

The constant factor $\frac{h^2 \varepsilon_0}{\pi m e^2}$ is exactly $a_0$ , the Bohr radius for hydrogen. So:

r_n = \frac{n^2}{Z} a_0

Plug in the numbers. For $\text{Li}^{++}$ in ground state: $n=1$ , $Z=3$ , $a_0 = 53\ \text{pm}$ .

r_1 = \frac{1^2}{3} \times 53\ \text{pm} = \frac{53}{3}\ \text{pm} \approx 17.67\ \text{pm}

Round to the nearest option. $17.67\ \text{pm}$ is about $18\ \text{pm}$ .

Watch out

A common mistake is to forget that $\text{Li}^{++}$ has $Z=3$ , not $Z=1$ (neutral lithium) or $Z=2$ (if you mistakenly think it's like helium). Always check the ionic charge: $\text{Li}^{++}$ means two electrons removed, so the remaining electron sees a full $+3e$ nucleus.

Tip

You can think of it this way: the radius scales inversely with $Z$ , so a $Z=3$ ion has one-third the radius of hydrogen. No need to memorize the full formula — just remember $r \propto n^2/Z$ and that $a_0$ is the reference for $n=1, Z=1$ .

✓Final answer

18\ \text{pm}

Method: Scale Directly From the Known Hydrogen Radius (Ratio Shortcut)

This method solves any "radius/energy/velocity of a hydrogen-like ion" problem without re-deriving the Bohr equations from Coulomb's law each time -- you scale a known reference value using the $n$ and $Z$ dependence alone.

Steps

Step 1: Write down how the quantity scales with $n$ and $Z$

From the Bohr model, every orbit quantity for a one-electron ion depends on $n$ (orbit number) and $Z$ (nuclear charge) in a fixed way:

r_n \propto \frac{n^2}{Z}, \qquad E_n \propto -\frac{Z^2}{n^2}, \qquad v_n \propto \frac{Z}{n}

You don't need to re-derive these from the centripetal-force balance every time -- memorise the proportionality and use the hydrogen value ( $n=1,Z=1$ ) as your anchor.

Step 2: Identify $n$ and $Z$ for the ion in question

Determine the principal quantum number of the state asked about, and the nuclear charge $Z$ seen by the single remaining electron (equal to the atomic number, since all other electrons have been stripped away).

Step 3: Form the ratio against the hydrogen reference

\frac{r_n(\text{ion})}{r_1(\text{H})} = \frac{n^2}{Z}

so $r_n(\text{ion}) = \dfrac{n^2}{Z}\, a_0$ , where $a_0 = 53\ \text{pm}$ is the known hydrogen ground-state radius.

Step 4: Applying to this problem

For the electron remaining in $\text{Li}^{++}$ : this is a hydrogen-like ion with $Z=3$ , and the question asks about the ground state, $n=1$ . The ratio gives $r = \frac{1}{3}\times 53\ \text{pm} \approx 17.7\ \text{pm}$ , which rounds to the listed option $18\ \text{pm}$ . The same ratio approach works instantly for energy ( $E_n = -13.6\,Z^2/n^2\ \text{eV}$ ) or speed in any hydrogen-like ion, without redoing the force-balance derivation.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

GUJCET 2026Set x1 markMCQ
$5.3 \times 10^{-11}$
›Reveal solutionSolution

$r_n=n^2 r_1\Rightarrow r_3=9(5.3\times10^{-11})=4.77\times10^{-10}\text{ m}$ .

In the Bohr model the orbit radius scales as the square of the principal quantum number:
$r_n=n^2\,r_1$
With $r_1=5.3\times10^{-11}\text{ m}$ and $n=3$ :
$r_3=3^2\times5.3\times10^{-11}=9\times5.3\times10^{-11}=4.77\times10^{-10}\text{ m}$
✓Final answer
$4.77\times10^{-10}$

ANSWER: (D)
GUJCET 2025Set 031 markMCQ
$h = 6.63 \times 10^{-34}$
›Reveal solutionSolution

Third excited state means $n=4$ ; Bohr angular momentum $L=\dfrac{nh}{2\pi}$ .

Ground $n=1$ , so third excited $\Rightarrow n=4$ .
$L=\frac{4h}{2\pi}=\frac{4\times6.63\times10^{-34}}{2\pi}=\frac{26.52\times10^{-34}}{6.283}\approx4.2\times10^{-34}\ \text{kg m}^2\text{s}^{-1}.$
✓Final answer
$4.2\times10^{-34}$

ANSWER: (A)
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. The ratio of radius of third and second orbits of a Hydrogen atom is ______.$
(a) 2/3
(b) 4/9
(c) 3/2
(d) 9/4
›Reveal solutionSolution

In the Bohr model, the radius of the nth orbit scales as rn ∝ n².

r3/r2 = 3²/2² = 9/4.

✓Final answer
$(d) 9/4.$
GUJCET 2024Set 131 markMCQ
$4 : 9$
›Reveal solutionSolution

Bohr orbit radius $r_n\propto n^2$ .

Steps. For hydrogen $r_n \propto n^2$ , so
$\frac{r_2}{r_3} = \frac{2^2}{3^2} = \frac{4}{9}.$
✓Final answer
$4 : 9$

ANSWER: (A)
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. If the radius of hydrogen atom in its first orbit is a0, then its radius in third excited state is ___.$
(a) 3a0
(b) 4a0
(c) 9a0
(d) 16a0
›Reveal solutionSolution

Bohr radius formula: rn = n²a0. Ground state is n=1; first excited n=2; second excited n=3; third excited state is n=4.

r4 = 4² a0 = 16a0.

✓Final answer
$(d) 16a0.$
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. In accordance with the Bohr's model, the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5 x 10^11 m with orbit speed 3 x 10^4 m/s is ___. (Mass of earth is 6.0 x 10^24 kg, h = 6.625 x 10^-34 Js)$
(a) 2.6 x 10^72
(b) 2.6 x 10^74
(c) 2.6 x 10^39
(d) 2.6 x 10^73
›Reveal solutionSolution

Bohr's angular momentum quantization: L = Mvr = n(h/2π), so n = 2πMvr/h.

M = 6.0 × 10²⁴ kg, v = 3 × 10⁴ m/s, r = 1.5 × 10¹¹ m, h = 6.625 × 10⁻³⁴ Js.

Mvr = (6.0 × 10²⁴)(3 × 10⁴)(1.5 × 10¹¹) = 2.7 × 10⁴⁰

2π × 2.7 × 10⁴⁰ = 1.696 × 10⁴¹

n = 1.696 × 10⁴¹ / 6.625 × 10⁻³⁴ ≈ 2.56 × 10⁷⁴ ≈ 2.6 × 10⁷⁴.

✓Final answer
$(b) 2.6 \times 10⁷⁴.$
GUJCET 2023Set 091 markMCQ
$5^{th}$
›Reveal solutionSolution
[!TLDR]

Using $L=\dfrac{nh}{2\pi}$ , the change in angular momentum from the 5th to the 3rd orbit is $\dfrac{h}{\pi}$ .

Concept

In Bohr's model of the hydrogen atom, the orbital angular momentum is quantised: $L = \dfrac{nh}{2\pi}$ , where $n$ is the orbit number.

Solution

For the two orbits:
$L_5 = \frac{5h}{2\pi}, \qquad L_3 = \frac{3h}{2\pi}.$
The magnitude of the change is
$\Delta L = L_5 - L_3 = \frac{(5-3)h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi}.$
[!ANSWER]

(A) $\dfrac{h}{\pi}$
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. What is the angular momentum of electron of Be^+3 ion in n = 5 orbit?$
(a) 3.3 x 10^-34 Js
(b) 6.6 x 10^-34 Js
(c) 5.3 x 10^-34 Js
(d) 1.3 x 10^-34 Js
›Reveal solutionSolution

Bohr's quantisation gives L = n h/(2*pi) for any hydrogen-like ion; for n = 5, L = 5 x 1.055 x 10^-34 = 5.3 x 10^-34 Js.

Bohr's postulate for angular momentum applies to any single-electron (hydrogen-like) system such as Be^3+:

L = n h/(2*pi) = n (h-bar).

It depends only on n, not on the nuclear charge Z. For n = 5:

L = 5 x (6.63 x 10^-34)/(2*pi) = 5 x 1.055 x 10^-34 = 5.27 x 10^-34 approximately 5.3 x 10^-34 Js.

✓Final answer
$(c) 5.3 x 10^-34 Js.$
GUJCET 2022Set 171 markMCQ
$5.3 \times 10^{-11}$
›Reveal solutionSolution
Bohr radius scales as $n^2$ : $r_n=r_1 n^2$ .

Concept. In the Bohr model the orbit radius is $r_n = r_1 n^2$ , with $r_1=5.3\times10^{-11}$ m.

Steps.

$r_3 = 5.3\times10^{-11}\times 3^2 = 5.3\times10^{-11}\times 9$ .

$r_3 = 4.77\times10^{-10}\ \text{m}$ .

✓Final answer
$4.77 \times 10^{-10}$

ANSWER: (B)
GUJCET 2022Set 171 markMCQ
$1.5 \times 10^{11}$
›Reveal solutionSolution
Bohr quantisation of angular momentum: $mvr = \dfrac{nh}{2\pi}$ .

Steps.

$mvr = (6\times10^{24})(3\times10^4)(1.5\times10^{11}) = 2.7\times10^{40}\ \text{kg m}^2/\text{s}$ .

$n=\dfrac{2\pi\,mvr}{h}=\dfrac{6.283\times2.7\times10^{40}}{6.625\times10^{-34}}$ .

$n\approx 2.6\times10^{74}$ .

✓Final answer
$2.6 \times 10^{74}$

ANSWER: (C)
GUJCET 2021Set 151 markMCQ
$\frac{1}{R}$
›Reveal solutionSolution

The Balmer series limit (n to infinity, to n=2) gives the shortest wavelength.

Concept. $\dfrac{1}{\lambda}=R\left(\dfrac{1}{2^2}-\dfrac{1}{\infty}\right)=\dfrac{R}{4}$ .

Solution. $\lambda=\dfrac{4}{R}$ .

✓Final answer
$\frac{4}{R}$

ANSWER: (D)
GUJCET 2021Set 151 markMCQ
$5.3 \times 10^{-11}$
›Reveal solutionSolution

Bohr radii scale as n^2: r_n = n^2 r1.

Concept. $r_n=n^2 r_1$ .

Solution. $r_4=4^2\times5.3\times10^{-11}=16\times5.3\times10^{-11}=8.48\times10^{-10}$ m.

✓Final answer
$8.48 \times 10^{-10}$

ANSWER: (B)

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