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.

JKBOSE Class 12 Annual Regular Examination 2024Set SZ3 marks
$Q. On the basis of Bohr's atomic model, find an expression for radius of nth orbit of a hydrogen atom.$
›Reveal solutionSolution

Equating the Coulomb attraction to the required centripetal force, and combining it with Bohr's quantization of angular momentum, gives the radius of the nth orbit as r_n = n²h²ε0/(πme²) — increasing as the square of the orbit number n.

Step 1 — Centripetal force condition: In Bohr's model, an electron of mass m and charge −e moves in a circular orbit of radius r around the nucleus (charge +e for hydrogen), held in orbit by the electrostatic (Coulomb) force acting as the centripetal force:

(1/4πε0) × e²/r² = mv²/r

⟹ mv² = e² / (4πε0 r) ... (1)

Step 2 — Bohr's quantization postulate: The angular momentum of the electron is quantized in integral multiples of h/2π:

mvr = nh/2π ⟹ v = nh / (2πmr) ... (2)

Step 3 — Combine (1) and (2): Substitute v from (2) into (1):

m × [nh/(2πmr)]² = e²/(4πε0 r)

n²h² / (4π²mr²) = e² / (4πε0 r)

Multiplying both sides by r and rearranging for r:

n²h² / (4π²mr) = e² / (4πε0)

r = n²h² × 4πε0 / (4π²me²)

r_n = n²h²ε0 / (πme²)

This shows the orbit radius increases as the square of the quantum number n (r_n ∝ n²) — for hydrogen, r1 works out to about 0.53 Å (the Bohr radius), r2 = 4 × 0.53 Å, and so on.

✓Final answer
$r_n = n²h²ε0 / (πme²) — derived from equating the Coulomb force to the centripetal force and using Bohr's angular-momentum quantization mvr = nh/2π.$
JKBOSE Class 12 Annual Regular Examination 2023Set ANNUAL3 marks
$Q. Write the postulates of Bohr's modal of hydrogen atom.$
›Reveal solutionSolution
Bohr's model of the hydrogen atom rests on three postulates: stable non-radiating orbits, quantized angular momentum (mvr = nh/2pi), and photon emission/absorption only during transitions between orbits (h*nu = delta-E).

Niels Bohr proposed the following postulates to explain the stability of atoms and the discrete (line) spectrum of hydrogen, combining classical mechanics with early quantum ideas:

Postulate of stationary (stable) orbits: An electron in an atom revolves around the nucleus only in certain specific, permitted circular orbits, called stationary orbits, without radiating energy - even though it is accelerating (contrary to classical electromagnetic theory, which would predict continuous energy loss and the electron spiralling into the nucleus). In these orbits, the necessary centripetal force is provided by the electrostatic (Coulomb) attraction between the electron and the nucleus.

Postulate of quantization of angular momentum: Only those orbits are permitted (stable) for which the angular momentum of the electron is an integral multiple of h/(2pi): L = mvr = nh/(2*pi), where n = 1, 2, 3, ... is the principal quantum number, m is electron mass, v its orbital speed, r the orbit radius, and h is Planck's constant.

Postulate of frequency condition (radiation postulate): An atom radiates (emits) energy only when an electron makes a transition from a higher-energy stationary orbit to a lower-energy one, and it absorbs energy when the electron jumps from a lower to a higher orbit. The energy of the photon emitted or absorbed equals the difference in energy of the two orbits: h*nu = E2 - E1 (or E_i - E_f), where nu is the frequency of the emitted/absorbed radiation and h is Planck's constant. While the electron remains in a given stationary orbit, no radiation occurs regardless of its acceleration.

These three postulates together successfully explained the stability of the hydrogen atom and correctly predicted its observed line spectrum (Balmer, Lyman series, etc.).

✓Final answer
$Bohr's postulates: (i) electrons move in stable, non-radiating circular orbits; (ii) angular momentum is quantized, L = n h/(2 pi); (iii) radiation is emitted/absorbed only during a transition between orbits, with photon energy h*nu = E2 - E1.$
JKBOSE Class 12 Annual Regular Examination 2022Set SZ3 marks
$Q. Write down the postulates of Bohr's model of hydrogen atom.$
›Reveal solutionSolution
Bohr postulated stationary orbits with quantised angular momentum, and that spectral lines arise from photon emission/absorption during transitions between these orbits.

Niels Bohr proposed the following postulates for the hydrogen atom, combining classical mechanics with quantum ideas:

Stationary (stable) orbits: An electron in an atom revolves around the nucleus in certain fixed, stable circular orbits without radiating energy, contrary to what classical electromagnetic theory (an accelerating charge should continuously radiate energy) would predict. These allowed orbits are called stationary states.

Quantisation of angular momentum: The electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of $h/2\pi$ :

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

where $n$ is called the principal quantum number.

Emission/absorption on transition: An electron can transition ("jump") from one stationary orbit to another. When it jumps from a higher energy orbit ( $E_2$ ) to a lower one ( $E_1$ ), it emits a photon of energy equal to the energy difference; conversely, it absorbs a photon of that energy to jump from the lower to the higher orbit:

$h\nu = E_2 - E_1$

These postulates successfully explained the discrete line spectrum of hydrogen (the various spectral series such as Lyman, Balmer, Paschen) and gave correct values for the energy levels and radius of the hydrogen atom.

✓Final answer
$mvr=nh/2\pi$
JKBOSE Class 12 Annual Regular Examination 2021Set SZ3 marks
$Q. State the basic postulates of Bohr's model of atom.$
›Reveal solutionSolution

Bohr combined classical mechanics with quantum ideas: electrons orbit without radiating, only in orbits with quantised angular momentum, and photons are emitted/absorbed only during a jump between orbits.

Niels Bohr proposed three basic postulates to explain the stability of atoms and the discrete line spectra observed (overcoming the problem that a classically accelerating orbiting electron should radiate energy continuously and spiral into the nucleus):

1. Postulate of stationary orbits: An electron in an atom can revolve only in certain special, discrete circular orbits, called stationary states or orbits, without radiating energy, even though it is undergoing centripetal acceleration. In these orbits, the electrostatic force of attraction between the nucleus and electron provides the necessary centripetal force:

$\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2} = \frac{mv^2}{r}$

2. Postulate of quantisation of angular momentum: Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of $h/2\pi$ :

$L = mvr = \frac{nh}{2\pi}, \qquad n = 1,2,3,\ldots$

where $n$ is called the principal quantum number.

3. Postulate of quantum jumps (frequency condition): An atom radiates (or absorbs) energy only when an electron jumps from a higher energy orbit to a lower one (or vice versa). The energy of the emitted/absorbed photon equals the difference in energy between the two orbits:

$h\nu = E_i - E_f$

where $E_i$ and $E_f$ are the energies of the initial and final orbits.

These three postulates together successfully explained the stability of the hydrogen atom and the observed discrete spectral lines (Lyman, Balmer, Paschen series, etc.).

✓Final answer
$L=nh/2\pi$
JKBOSE Class 12 Annual Regular Examination 2020Set SZ3 marks
$Q. State postulates of Bohr's theory of Hydrogen atom.$
›Reveal solutionSolution
Bohr's three postulates fix stable non-radiating orbits with quantised angular momentum, and explain spectral lines as photon emission/absorption during orbit jumps.

Stationary orbits postulate: An electron in an atom revolves around the nucleus in certain fixed circular orbits, called stationary orbits, without radiating any energy, even though it is accelerating (which classically should cause continuous radiation and spiral collapse into the nucleus).

Quantisation of angular momentum: Only those orbits are permitted for which the angular momentum of the electron is an integral multiple of h/2π (h being Planck's constant):

mvr = nh/2π, n = 1, 2, 3, …

where n is called the principal quantum number.

Frequency condition (transition postulate): An electron can jump from one stationary orbit to another. When it jumps from a higher energy orbit E₂ to a lower energy orbit E₁, the energy difference is emitted as a photon of frequency ν given by

hν = E₂ − E₁

Conversely, the atom absorbs a photon of exactly this energy to jump from E₁ to E₂.

✓Final answer
$(1) Electrons move in fixed non-radiating stationary orbits. (2) Angular momentum is quantised, mvr = nh/2π. (3) Energy is emitted/absorbed only on transition between orbits, hν = E₂ - E₁.$

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