MHT-CET Physics · Structure of Atoms and Nuclei

Bohr's Model: How Each Orbit Scales With n

In Bohr's model the electron's angular momentum is a whole number of h/2π, which fixes each orbit: the radius grows as n², the speed falls as 1/n, and the energy is −13.6 Z²/n² eV, so every other orbital quantity is a power of n.

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Why this matters

39 PYQs, 11 of them HARD — the largest page in the chapter. Twenty-two ask how an orbit property scales with n: radius, speed, period, acceleration, the field at the nucleus, the de Broglie wavelength. Twelve are the energy levels and the photon between two of them, and five apply the quantum condition outside the atom — to a rotating molecule and to a charge circling in a magnetic field. Three cards.

Concept 1 of 3: Radius, Speed and Everything Built From Them

Two equations fix an orbit: the Coulomb pull supplies the centripetal force, and mvr = nh/2π. Solving them gives r ∝ n²/Z and v ∝ Z/n. Every other quantity is built from these two, so its power of n follows by counting: the period 2πr/v goes as n³, the revolution frequency as n⁻³, the acceleration v²/r and the force as n⁻⁴, the current e·f as n⁻³, and the field at the nucleus μ₀I/2r as n⁻⁵. The de Broglie wavelength h/mv grows as n, and exactly n wavelengths fit round the orbit. Read 'first excited state' as n = 2 and 'second excited state' as n = 3.

Definition

$r_n = \dfrac{\varepsilon_0 n^2 h^2}{\pi m Z e^2} = 0.53\,\dfrac{n^2}{Z}$ Å; $v_n = \dfrac{Ze^2}{2\varepsilon_0 n h} \approx 2.2\times10^6\,\dfrac{Z}{n}$ m/s.
$L = \dfrac{nh}{2\pi}$ : one step in n changes L by $\dfrac{h}{2\pi} \approx 1.05\times10^{-34}$ J s.
Period $\propto n^3$ ( $T_n = \dfrac{4\varepsilon_0^2 n^3 h^3}{me^4}$ ); frequency $\propto n^{-3}$ ; acceleration and force $\propto n^{-4}$ .
Current $\propto n^{-3}$ ; field at the nucleus $\dfrac{\mu_0 I}{2r} \propto n^{-5}$ .
de Broglie: $\lambda_n = \dfrac{2\pi r_n}{n} \propto n$ (third orbit: $6\pi a_0$ ).
Area of the orbit $\propto r^2 \propto n^4$ ; moment of inertia $mr^2 \propto n^4$ .

Bohr orbit

r_n \propto \frac{n^2}{Z}, \qquad v_n \propto \frac{Z}{n}, \qquad mvr = \frac{nh}{2\pi}

Worked example

The electron in hydrogen moves from n = 1 to n = 3. By what factor do its speed and orbital period change?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q13Hard

Example 1 · Structure of Atoms and Nuclei · Bohr Model and Atomic Properties

Magnetic field at the centre of the hydrogen atom due to motion of electrons in

n

th orbit is proportional to

Numbering the excited states from 1

The ground state is n = 1, so the FIRST excited state is n = 2 and the third excited state is n = 4. Counting from 1 puts every ratio one orbit out.

Stopping at the current for the field at the nucleus

The current goes as n⁻³, but the field is μ₀I/2r, and r grows as n². The field therefore goes as n⁻⁵, not n⁻³.

Drill 21 more on radius, speed and everything built from them

Concept 2 of 3: Energy Levels and the Photon Between Them

The total energy of the nth orbit is E_n = −13.6 Z²/n² eV. It is negative because the electron is bound, and it rises towards zero as n grows. The kinetic energy is −E and the potential energy is 2E, so moving outward raises the potential energy (less negative) and lowers the kinetic energy. A jump between two levels emits or absorbs one photon carrying the difference, hν = E_i − E_f. Exciting hydrogen from the ground state to n = 2 needs 10.2 eV; removing the electron entirely needs 13.6 eV. The photon also carries momentum h/λ, so the atom recoils when it emits.

Definition

$E_n = -\dfrac{13.6\,Z^2}{n^2}$ eV: $-13.6, -3.4, -1.51, -0.85$ eV for hydrogen.
$KE = -E$ , $PE = 2E$ : at $E = -3.4$ eV, KE = 3.4 eV, PE = −6.8 eV.
Photon: $h\nu = E_i - E_f$ ; $1 \to 2$ needs 10.2 eV; ionisation from the ground state 13.6 eV.
Between atoms: $E \propto \dfrac{Z^2}{n^2}$ (H in n = 3 has E; He⁺ in n = 5 has $\dfrac{36E}{25}$ ).
Recoil of the emitting atom: $Mv = \dfrac{h}{\lambda}$ .

Energy of the nth level

E_n = -\frac{13.6\,Z^2}{n^2}\ \text{eV}, \qquad h\nu = E_i - E_f

Worked example

An electron in hydrogen has total energy −1.51 eV. Which orbit is it in, and what are its kinetic and potential energies?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q27Hard

Example 2 · Structure of Atoms and Nuclei · Bohr Model and Atomic Properties

In the third orbit of hydrogen atom the energy of an electron is

E

. In the fifth orbit of helium (

Z=2

) the energy of an electron will be

Saying the potential energy falls as the electron moves out

The potential energy is negative and becomes LESS negative outward: −6.8 eV at n = 2 is greater than −27.2 eV at n = 1. It increases, while the kinetic energy decreases.

Forgetting Z² when changing the atom

He⁺ and Li²⁺ are hydrogen-like but their levels are Z² deeper. Compare two atoms with E ∝ Z²/n², both factors at once.

Drill 11 more on energy levels and the photon between them

Concept 3 of 3: The Quantum Condition Outside the Atom

The papers apply L = nh/2π to two other systems. A diatomic molecule of moment of inertia I rotating with L = nh/2π has energy L²/2I = n²h²/8π²I. A charge q moving in a circle in a magnetic field B has qBR = mv; combined with mvR = nh/2π, vR = nh/2πm, and the energy ½mv² = ½qB·vR = nqBh/4πm, which grows as n, not n².

Definition

Rotor: $E_n = \dfrac{L^2}{2I} = \dfrac{n^2 h^2}{8\pi^2 I}$ (n = 2: $\dfrac{h^2}{2\pi^2 I}$ ; n = 3: $\dfrac{9h^2}{8\pi^2 I}$ ).
Charge in a field: $qBR = mv$ and $mvR = \dfrac{nh}{2\pi}$ give $E_n = \dfrac{nqBh}{4\pi m}$ .

Rotor and charge in a field

E_{\text{rot}} = \frac{n^2 h^2}{8\pi^2 I}, \qquad E_{B} = \frac{n\,qBh}{4\pi m}

Worked example

Rotational energy of a diatomic molecule of moment of inertia I in the level n = 1?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q36Hard

Example 3 · Structure of Atoms and Nuclei · Bohr Model and Atomic Properties

A diatomic molecule has moment of inertia '

I

'. By applying Bohr's quantization condition, its rotational energy in the

n

th level is

[n\geq1]

[h = Planck's constant]

Writing L = nh

Bohr's condition is L = nh/2π. Dropping the 2π puts a factor 4π² into every energy, and the options include that wrong answer.

Drill 4 more on the quantum condition outside the atom

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Radius, Speed and Everything Built From Them
Bohr orbit
$r_n \propto \frac{n^2}{Z}, \qquad v_n \propto \frac{Z}{n}, \qquad mvr = \frac{nh}{2\pi}$
Energy Levels and the Photon Between Them
Energy of the nth level
$E_n = -\frac{13.6\,Z^2}{n^2}\ \text{eV}, \qquad h\nu = E_i - E_f$
The Quantum Condition Outside the Atom
Rotor and charge in a field
$E_{\text{rot}} = \frac{n^2 h^2}{8\pi^2 I}, \qquad E_{B} = \frac{n\,qBh}{4\pi m}$

Watch out for (5)

Numbering the excited states from 1
→ Radius, Speed and Everything Built From Them
Stopping at the current for the field at the nucleus
→ Radius, Speed and Everything Built From Them
Saying the potential energy falls as the electron moves out
→ Energy Levels and the Photon Between Them
Forgetting Z² when changing the atom
→ Energy Levels and the Photon Between Them
Writing L = nh
→ The Quantum Condition Outside the Atom

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