JEE Mains Physics · Atoms

Bohr Orbits: Radius, Speed and Energy

In a hydrogen-like ion the radius goes as n²/Z, the speed as Z/n and the energy as −13.6 Z²/n² eV; every other orbital quantity is built from these three.

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Why this matters

Twenty-three PYQs, seventeen of them multiple choice, and two from 2026. Fifteen scale an orbit: eight work with the radius, four with the speed, and three with a quantity built from both, such as the frequency of revolution, the magnetic moment or the field at the nucleus. Eight are about the energy: three split it into kinetic and potential energy, four find a level's energy from Z²/n², and one replaces the electron with a muon.

Concept 1 of 2: Bohr radius and orbital speed in hydrogen-like ions

Put Bohr's rule mvr = nh/2π together with the Coulomb pull mv²/r = kZe²/r². Two results fall out: the radius grows as n² and shrinks as Z, and the speed falls as 1/n and grows with Z. Outer orbits are big and slow; a heavier nuclear charge pulls every orbit in and speeds it up. Nearly every question on this page is a ratio, so the constants cancel and only the powers of n and Z matter.

Definition

Radius: $r_n = a_0\dfrac{n^2}{Z}$ , with $a_0 = 0.529$ Å (a question may give 0.53 Å or 0.51 Å; use its value).
Speed: $v_n = v_1\dfrac{Z}{n}$ , with $v_1 = 2.19 \times 10^{6}$ m/s, about c/137.
Period: $T = \dfrac{2\pi r}{v} \propto \dfrac{n^3}{Z^2}$ . Frequency of revolution: $f \propto \dfrac{Z^2}{n^3}$ .
Current of the orbiting electron: $I = ef \propto \dfrac{Z^2}{n^3}$ .
Magnetic moment: $\mu = IA = \dfrac{evr}{2} = \dfrac{neh}{4\pi m}$ . It grows as n and does not depend on Z.
Field at the nucleus: $B = \dfrac{\mu_0 I}{2r} \propto \dfrac{Z^2/n^3}{n^2/Z} = \dfrac{Z^3}{n^5}$ .
Radius from the ionisation energy: the bound energy is $E = -\dfrac{kZe^2}{2r}$ , so $r = \dfrac{kZe^2}{2|E|}$ .
Radius to n: divide by $a_0/Z$ and take the square root.

Bohr radius and speed

r_n = 0.529\,\frac{n^2}{Z}\ \text{Å}, \qquad v_n = 2.19 \times 10^{6}\,\frac{Z}{n}\ \text{m/s}

Worked example

For the electron in the third orbit of He⁺, find (a) the orbit radius, (b) its speed, and (c) its frequency of revolution as a fraction of that in the ground state of hydrogen.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q25Moderate

Example 1 · Atoms · Bohr Orbits: Radius, Speed and Energy

The radius of

2^{\text{nd~}}

orbit of

He^{+}

of Bohr's model is

r_{1}

and that of fourth orbit of

Be^{3 +}

is represented as

r_{2}

. Now the ratio

\frac{r_{2}}{r_{1}}

x:1

. The value of

x

is____

Radius goes as n², speed as 1/n

Swapping the powers is the commonest slip. The radius of the third orbit is 9 times the first; the speed is one third of it.

A larger Z makes the orbit smaller

r = a₀n²/Z, so Li²⁺ orbits are a third the size of hydrogen's for the same n. The speed, on the other hand, is three times as large.

Build a derived quantity from r and v

Frequency is v/2πr, current is ef, magnetic moment is evr/2, field at the centre is μ₀I/2r. Write each in terms of n and Z step by step; guessing the power is where marks go.

Drill 14 more on bohr radius and orbital speed in hydrogen-like ions

Concept 2 of 2: Energy levels and the kinetic–potential split

The electron is bound, so its total energy is negative. Its kinetic energy is positive, its potential energy negative and twice as large, so the total equals minus the kinetic energy. Moving to a higher level, the electron slows down (kinetic energy falls) while its potential and total energy rise towards zero. Two levels in different ions sit at the same energy when Z/n is the same.

Definition

$E_n = -13.6\,\dfrac{Z^2}{n^2}$ eV.
$K = \dfrac{kZe^2}{2r} = -E$ , $U = -\dfrac{kZe^2}{r} = 2E$ . So $K : |U| = 1 : 2$ in every orbit of every hydrogen-like ion.
Going up a level: K decreases; U and E increase (towards zero).
Ionisation energy from level n = binding energy = $|E_n| = 13.6\,\dfrac{Z^2}{n^2}$ eV.
Naming levels: n = 1 is the ground state, n = 2 the first excited state, n = 3 the second. The kth excited state is n = k + 1.
Two levels have equal energy when $\dfrac{Z}{n}$ is equal.
The orbiting particle's mass: $E \propto m$ and $r \propto \dfrac{1}{m}$ . A heavier particle in place of the electron sits in deeper, smaller orbits (taking the nucleus as fixed).

Energy of level n

E_n = -13.6\,\frac{Z^2}{n^2}\ \text{eV}, \qquad K = -E, \quad U = 2E

Worked example

For the electron in the fourth orbit of Li²⁺, find its total energy, its kinetic and potential energy, and the energy needed to remove it from the ion.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q15Moderate

Example 2 · Atoms · Bohr Orbits: Radius, Speed and Energy

Considering Bohr's atomic model for hydrogen atom : (A) the energy of H atom in ground state is same as energy of

{He}^{+}

ion in its first excited state. (B) the energy of H atom in ground state is same as that for

{Li}^{+ +}

ion in its second excited state. (C) the energy of H atom in its ground state is same as that of

{He}^{+}

ion for its ground state. (D) the energy of

{He}^{+}

ion in its first excited state is same as that for

{Li}^{+ +}

ion in its ground state Choose the correct answer from the options given below :

Z is squared in the energy

E = −13.6 Z²/n² eV. Using Z instead of Z² gives He⁺ half its true ground-state energy. The radius has Z to the first power; the energy has Z².

Excited-state numbers are one behind n

The first excited state is n = 2 and the second excited state is n = 3. Reading second excited as n = 2 is wrong in every such question.

Kinetic energy falls as the electron moves out

Total and potential energy rise towards zero in a higher orbit, but the kinetic energy falls, because K = −E. Saying all three increase is a standard wrong option.

Drill 7 more on energy levels and the kinetic–potential split

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 (2)

Bohr radius and orbital speed in hydrogen-like ions
Bohr radius and speed
$r_n = 0.529\,\frac{n^2}{Z}\ \text{Å}, \qquad v_n = 2.19 \times 10^{6}\,\frac{Z}{n}\ \text{m/s}$
Energy levels and the kinetic–potential split
Energy of level n
$E_n = -13.6\,\frac{Z^2}{n^2}\ \text{eV}, \qquad K = -E, \quad U = 2E$

Watch out for (6)

Radius goes as n², speed as 1/n
→ Bohr radius and orbital speed in hydrogen-like ions
A larger Z makes the orbit smaller
→ Bohr radius and orbital speed in hydrogen-like ions
Build a derived quantity from r and v
→ Bohr radius and orbital speed in hydrogen-like ions
Z is squared in the energy
→ Energy levels and the kinetic–potential split
Excited-state numbers are one behind n
→ Energy levels and the kinetic–potential split
Kinetic energy falls as the electron moves out
→ Energy levels and the kinetic–potential split

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