PhysicsCBSE Class 12 · Chapter 12

Atoms Class 12: Notes, Bohr Model and Solved Numericals

Updated 14 October 2026 · Fact-checked

Atoms is Chapter 12 of CBSE Class 12 Physics. It covers the alpha-particle scattering experiment and Rutherford's nuclear model, its failure to explain stability and line spectra, Bohr's postulates, the expressions for the radius, speed and energy of the electron in the nth orbit of hydrogen, and the hydrogen line spectrum qualitatively.

What Atoms is about

What does an atom look like inside? Thomson's 'plum pudding' model spread positive charge evenly through the atom. Geiger and Marsden, working with Rutherford, tested this by firing alpha particles at thin gold foil. Most went straight through, but a few bounced back almost the way they came. Rutherford concluded that all the positive charge and almost all the mass are packed into a tiny nucleus at the centre, with electrons moving around it at relatively large distances.

Rutherford's model had two problems: classical physics says an orbiting electron should radiate energy and spiral into the nucleus, and it predicts a continuous spectrum instead of the sharp lines that hydrogen actually emits. Bohr fixed this with three postulates: electrons move only in certain stationary orbits without radiating, the angular momentum in these orbits is a whole multiple of $\frac{h}{2 π}$ , and light is emitted or absorbed only when an electron jumps between orbits, with $h ν = E_{i} - E_{f}$ . This gives the orbit radius $r_{n} \propto n^{2}$ , speed $v_{n} \propto \frac{1}{n}$ and energy $E_{n} = - \frac{13.6}{n ^{2}} e V$ , and explains the hydrogen line spectrum.

For 2026-27 you need the expressions (not the derivation of the radius) and the line spectra qualitatively; the spectral series formulae and de Broglie's explanation of Bohr's second postulate are not in the syllabus. In our question bank this chapter appears in all eight board papers from 2023 to 2026, mostly as 1-mark MCQs and assertion-reason items on energy levels and Rutherford scattering.

Weightage: This chapter is part of Units VII and VIII, Dual Nature of Radiation and Matter; Atoms and Nuclei, which carry 12 of the 70 theory marks together with Dual Nature of Radiation and Matter and Nuclei.Source: CBSE Physics syllabus 2026-27, Units VII and VIII Dual Nature of Radiation and Matter; Atoms and Nuclei

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 4.4 marks per paper. See the questions

Key concepts

Alpha-particle scattering experiment

NCERT section 12.2

Alpha particles (helium nuclei, charge +2e) from a radioactive source were collimated into a narrow beam and aimed at a very thin gold foil (about $2.1 \times 1 0^{- 7} m$ thick). A movable zinc-sulphide screen with a microscope counted the flashes made by scattered particles at different angles.

Observations:

Most alpha particles passed straight through or were deflected by less than 1°.
A very small fraction (NCERT reports about 1 in 8000) was deflected by more than 90°, and a few came almost straight back.

Conclusions: most of the atom is empty space. A large deflection needs a very strong repulsive force, which can come only from positive charge concentrated in a very small region: the nucleus, about $1 0^{- 15}$ to $1 0^{- 14} m$ across, roughly 10 000 to 100 000 times smaller than the atom. The nucleus holds almost all the mass of the atom. Gold was chosen because it can be beaten into very thin foils and has a large nuclear charge (Z = 79).

Worked example

In a scattering experiment, very few alpha particles bounce back from a gold foil. What two conclusions does this support about the structure of the atom? [2 marks]

Try it first, then show the solutionSolution

Most particles pass through undeflected, so most of the atom is empty space.
Bouncing back needs a very strong repulsion from a heavy, positively charged object: the positive charge and nearly all the mass are concentrated in a tiny central nucleus.

Answer: The atom is mostly empty; its positive charge and mass are concentrated in a tiny nucleus.

Alpha-particle trajectory: impact parameter and closest approach

NCERT section 12.2.1

The path of an alpha particle depends on its impact parameter b: the perpendicular distance of its initial line of motion from the centre of the nucleus.

A large b means a glancing encounter and a small deflection.
A small b means a nearly head-on encounter and a large deflection.
b = 0 (head-on): the particle comes to rest momentarily and bounces straight back (scattering angle 180°).

For a head-on collision, at the distance of closest approach d all the kinetic energy K has become electric potential energy:

$K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{d} \Rightarrow d = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{K}$

This gives an upper limit for the size of the nucleus. A handy value: $\frac{e ^{2}}{4 π ε _{0}} = 1.44 M e V f m$ (1 fm = $1 0^{- 15}$ m).

Worked example

An alpha particle of kinetic energy 5.0 MeV approaches a gold nucleus (Z = 79) head-on. Find the distance of closest approach. ( $\frac{e ^{2}}{4 π ε _{0}} = 1.44 M e V f m$ ) [2 marks]

Try it first, then show the solutionSolution

$d = \frac{2 Z}{K} \times \frac{e ^{2}}{4 π ε _{0}} = \frac{2 \times 79 \times 1.44 M e V f m}{5.0 M e V}$
$d = \frac{227.5}{5.0} \approx 45.5 f m = 4.55 \times 1 0^{- 14} m$

Answer: About $4.6 \times 1 0^{- 14} m$ (45.5 fm)

Rutherford's nuclear model and its limitations

NCERT section 12.2.2 and 12.3

In Rutherford's model, electrons revolve round the nucleus in orbits, the electrostatic attraction providing the centripetal force: $\frac{m v ^{2}}{r} = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}}$ for hydrogen. The total energy of the electron is then $E = K + U = \frac{e ^{2}}{8 π ε _{0} r} - \frac{e ^{2}}{4 π ε _{0} r} = - \frac{e ^{2}}{8 π ε _{0} r}$ : negative, showing the electron is bound.

Limitations:
1. Stability: an orbiting electron is accelerating, and classical electromagnetism says an accelerating charge radiates. The electron would lose energy, spiral inward and fall into the nucleus in a tiny fraction of a second. Real atoms are stable.
2. Line spectra: a spiralling electron would emit a continuous range of frequencies. Real atoms emit only certain sharp wavelengths, a line spectrum characteristic of each element (hydrogen, for example, has lines in the ultraviolet, visible and infrared).

Worked example

In Rutherford's model of hydrogen, the electron's kinetic energy is K. Write its potential energy and total energy in terms of K. What does the sign of the total energy mean? [2 marks]

Try it first, then show the solutionSolution

$K = \frac{e ^{2}}{8 π ε _{0} r}$ and $U = - \frac{e ^{2}}{4 π ε _{0} r} = - 2 K$
$E = K + U = - K$
A negative total energy means the electron is bound to the nucleus: energy must be supplied to free it.

Answer: U = −2K, E = −K; negative means bound.

Bohr's postulates

NCERT section 12.4

Bohr combined Rutherford's nucleus with quantum ideas in three postulates:

1. Stationary orbits: an electron can revolve in certain stable orbits without radiating energy. Each such orbit has a definite total energy.
2. Quantisation of angular momentum: the allowed orbits are those for which the angular momentum is a whole-number multiple of $\frac{h}{2 π}$ :
$L = m v r = \frac{nh}{2 π}, n = 1, 2, 3, \dots$ (n is the principal quantum number).
3. Frequency condition: an electron radiates or absorbs energy only when it jumps from one stationary orbit to another. The photon's energy equals the energy difference:
$h ν = E_{i} - E_{f}$

These postulates explain why atoms are stable and why they emit only certain frequencies.

Worked example

Find the angular momentum of the electron in the third orbit of hydrogen according to Bohr. ( $h = 6.63 \times 1 0^{- 34} J s$ ) [1 mark]

Try it first, then show the solutionSolution

$L = \frac{nh}{2 π} = \frac{3 \times 6.63 \times 1 0 ^{- 34}}{2 \times 3.14}$
$L \approx 3.17 \times 1 0^{- 34} J s$

Answer: About $3.2 \times 1 0^{- 34} J s$

Radius, speed and energy in the nth orbit (expressions)

NCERT section 12.4 and 12.4.1

Combining $\frac{m v ^{2}}{r} = \frac{e ^{2}}{4 π ε _{0} r ^{2}}$ with $m v r = \frac{nh}{2 π}$ gives the following expressions for hydrogen (the syllabus asks for the expressions; the derivation of the radius is not required):

Speed: $v_{n} = \frac{e ^{2}}{2 ε _{0} nh} = \frac{1}{n} \cdot \frac{e ^{2}}{2 ε _{0} h} \approx \frac{2.18 \times 1 0 ^{6}}{n} m s^{- 1}$
Radius: $r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}} = n^{2} a_{0}$ , where the Bohr radius $a_{0} = \frac{h ^{2} ε _{0}}{π m e ^{2}} \approx 5.3 \times 1 0^{- 11} m$
Energy: $E_{n} = - \frac{m e ^{4}}{8 n ^{2} ε _{0}^{2} h ^{2}} = - \frac{13.6}{n ^{2}} e V$

So $r_{n} \propto n^{2}$ , $v_{n} \propto \frac{1}{n}$ and $E_{n} \propto - \frac{1}{n ^{2}}$ . In any orbit, kinetic energy $= - E_{n}$ and potential energy $= 2 E_{n}$ .

Energy levels: n = 1 (ground state) −13.6 eV; n = 2, −3.4 eV; n = 3, −1.51 eV; n = 4, −0.85 eV; n → ∞, 0. The energy needed to remove the electron from the ground state, 13.6 eV, is the ionisation energy of hydrogen. The levels get closer together as n increases.

Worked example

For the electron in the n = 3 orbit of hydrogen, find (a) the radius, (b) the speed, (c) the total, kinetic and potential energies. [3 marks]

Try it first, then show the solutionSolution

(a) $r_{3} = 9 a_{0} = 9 \times 0.53 \overset{˚}{A} = 4.77 \overset{˚}{A} \approx 4.8 \times 1 0^{- 10} m$
(b) $v_{3} = \frac{2.18 \times 1 0 ^{6}}{3} \approx 7.3 \times 1 0^{5} m s^{- 1}$
(c) $E_{3} = - \frac{13.6}{9} \approx - 1.51 e V$ ; K = +1.51 eV; U = −3.02 eV

Answer: (a) about 4.8 Å (b) about $7.3 \times 1 0^{5} m s^{- 1}$ (c) E = −1.51 eV, K = 1.51 eV, U = −3.02 eV

Hydrogen line spectra (qualitative)

NCERT section 12.5

At room temperature almost all hydrogen atoms are in the ground state. When energy is supplied (by an electric discharge or heating), electrons jump to higher levels. They then fall back, in one or more steps, emitting a photon at each jump with $h ν = E_{i} - E_{f}$ . Since only certain energy differences exist, only certain wavelengths appear: an emission line spectrum.

The lines are grouped by the level the electron falls to (qualitative treatment):

Jumps ending on n = 1 give the Lyman series (ultraviolet).
Jumps ending on n = 2 give the Balmer series (the visible lines, with some in the near ultraviolet).
Jumps ending on n = 3 give the Paschen series (infrared); further series (Brackett, Pfund) end on n = 4 and 5, also in the infrared.

White light passed through cool hydrogen loses exactly the same wavelengths, because atoms in the ground state absorb photons that lift them to higher levels: an absorption spectrum of dark lines at the same positions. From level n, the largest number of different lines an electron can produce on its way down is $\frac{n ( n - 1 )}{2}$ .

Worked example

An electron in hydrogen falls from n = 3 to n = 2. Find the energy and wavelength of the photon emitted. In which series and region of the spectrum is this line? ( $h c \approx 1240 e V nm$ ) [2 marks]

Try it first, then show the solutionSolution

$Δ E = E_{3} - E_{2} = - 1.51 - (- 3.40) = 1.89 e V$
$λ = \frac{1240}{1.89} \approx 656 nm$
It ends on n = 2, so it is in the Balmer series, in the visible region (red).

Answer: 1.89 eV; about 656 nm; Balmer series, visible (red)

Physics in real life: Atoms

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Line spectra

Suppose a hydrogen discharge tube glows with a reddish-pink colour, and you look at it through a spectroscope that spreads out its light.

Explain why you see a few separate coloured lines rather than a continuous rainbow, and find the wavelength of the bright red line, which comes from the n = 3 to n = 2 jump. ( $h c \approx 1240 e V nm$ )

Try it first, then show the solutionSolution

Electrons in hydrogen can only have the energies $- 13.6/ n^{2}$ eV, so only certain energy differences, and hence certain wavelengths, can be emitted.
$Δ E = - 1.51 - (- 3.40) = 1.89 e V$
$λ = \frac{1240}{1.89} \approx 656 nm$ , in the red part of the visible spectrum

Answer: Only fixed energy jumps are possible, giving separate lines; the red line is at about 656 nm.

Memory hook: Fixed steps on the energy ladder, fixed colours in the light.

Ionisation energy

Imagine sending ultraviolet light of wavelength 80 nm into a container of hydrogen gas in its ground state.

Suppose a photon of wavelength 80 nm is absorbed by a ground-state hydrogen atom. Will the atom be ionised? If so, with what kinetic energy does the electron leave? ( $h c \approx 1240 e V nm$ )

Try it first, then show the solutionSolution

Photon energy $= \frac{1240}{80} = 15.5 e V$
Ionisation needs 13.6 eV, which is less than 15.5 eV, so the atom is ionised.
Kinetic energy of the freed electron $= 15.5 - 13.6 = 1.9 e V$

Answer: Yes; the electron leaves with about 1.9 eV.

Memory hook: 13.6 eV is the price of freedom for hydrogen's electron.

Size of the nucleus

Suppose a hydrogen atom were enlarged so that its nucleus (a proton about $1 0^{- 15} m$ across) became a marble 1 cm wide.

Taking the atom's size as about $1 0^{- 10} m$ , how big would the whole atom be on this scale?

Try it first, then show the solutionSolution

Ratio of sizes $= \frac{1 0 ^{- 10}}{1 0 ^{- 15}} = 1 0^{5}$
Atom size $= 1 0^{5} \times 1 c m = 1 0^{5} c m = 1 k m$
A marble at the centre of a region a kilometre across: the atom is almost entirely empty space.

Answer: About 1 km across.

Memory hook: A marble in a stadium-sized space: that is how empty an atom is.

Topics in this chapter

Atoms formula sheet

Distance of closest approach: $d = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{K}$; $\frac{e ^{2}}{4 π ε _{0}} = 1.44 M e V f m$ .
Bohr's quantum condition: $m v r = \frac{nh}{2 π}$
Bohr's frequency condition: $h ν = E_{i} - E_{f}$
Radius of the nth orbit (expression only): $r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}} = n^{2} a_{0}, a_{0} \approx 0.53 \overset{˚}{A}$
Speed in the nth orbit: $v_{n} = \frac{e ^{2}}{2 ε _{0} nh} \approx \frac{2.18 \times 1 0 ^{6}}{n} m s^{- 1}$
Energy in the nth orbit: $E_{n} = - \frac{m e ^{4}}{8 n ^{2} ε _{0}^{2} h ^{2}} = - \frac{13.6}{n ^{2}} e V$
Kinetic and potential energy: $K = - E_{n}, U = 2 E_{n}$
Photon wavelength from a transition: $λ = \frac{h c}{E _{i} - E _{f}}$; $h c \approx 1240 e V nm$ .
Maximum number of lines from level n: $N = \frac{n ( n - 1 )}{2}$

Atoms: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Alpha-particle scattering set-up

Describing the Geiger-Marsden experiment (3-5 marks).

Draw

Draw a radioactive source inside a lead block with a narrow opening.
Draw lead collimators (slits) producing a narrow beam of alpha particles.
Draw a thin gold foil in the path of the beam.
Draw a circular rotatable detector (zinc sulphide screen with a microscope) around the foil, showing particles scattered at various angles θ.

Label

Source (in lead block)
Collimator
Gold foil
Scattered alpha particles
θ
ZnS screen and microscope

Watch out: Show most arrows going straight through and one or two bouncing back.

Energy-level diagram of hydrogen

Transition and spectrum questions (2-3 marks).

Draw

Draw horizontal lines for n = 1 (−13.6 eV), n = 2 (−3.4 eV), n = 3 (−1.51 eV), n = 4 (−0.85 eV) and n = ∞ (0 eV), getting closer together upwards.
Draw downward arrows ending on n = 1 (Lyman), on n = 2 (Balmer) and on n = 3 (Paschen).

Label

n values
Energies in eV
Lyman (UV)
Balmer (visible)
Paschen (IR)
Ionisation (n = ∞)

Watch out: Space the levels roughly to scale (crowding near 0) so the diagram shows why higher series are in the infrared.

How the board asks Atoms

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Scattering experiment	Observations and conclusions; graph of number of scattered particles against angle; impact parameter and scattering angle; distance of closest approach.	Link each observation to a conclusion; small b gives large deflection; for d set $K = k (2 e) (Z e) / d$ .
Limitations of Rutherford's model	Two limitations, and how Bohr's postulates remove them.	Stability (accelerating charge radiates) and line spectra (continuous versus discrete); answer each with the matching Bohr postulate.
Bohr expressions	Ratios of radii, speeds, energies or periods for different n; kinetic and potential energies; angular momentum.	Use $r \propto n^{2}$ , $v \propto 1/ n$ , $E \propto - 1/ n^{2}$ , $T \propto n^{3}$ , $L \propto n$ .
Transitions and spectra	Wavelength or energy of the photon in a transition; series and region; number of lines; which transition gives the longest or shortest wavelength.	ΔE from the level energies, λ = 1240/ΔE nm; series named by the final level; longest λ means smallest ΔE.
Excitation by electrons or photons	An electron beam or photon of given energy hits hydrogen in the ground state: up to which level is it excited and how many lines are emitted?	Find the highest n with $E_{n} - E_{1}$ not exceeding the given energy; count lines with n(n − 1)/2.

Tricks to remember Atoms

Short memory aids. Say each one aloud once, then use it on a question the same day.

r goes up as n², v goes down as 1/n, E as −1/n²
Higher orbit: bigger (n²), slower (1/n), less tightly bound (energy closer to zero).
Use it when: Ratio questions.
13.6 ÷ n²
Hydrogen levels: −13.6, −3.4, −1.51, −0.85 eV for n = 1, 2, 3, 4.
Use it when: Any transition energy.
Lyman Uses Ultraviolet, Balmer Visible, Paschen IR
LU-BV-PI: Lyman (to n = 1) UV; Balmer (to n = 2) visible; Paschen (to n = 3) infrared.
Use it when: Naming series and regions.
K = −E, U = 2E
In any Bohr orbit the kinetic energy is the size of the total energy, and the potential energy is twice the total energy.
Use it when: Energy-split questions.
Closest approach: all K becomes U
At the turning point the alpha particle stops: $K = k (2 Z e^{2}) / d$ .
Use it when: Size-of-nucleus numericals.

Atoms: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Energy of the second orbit · 1 mark · MCQBoard level

The energy of an electron in the second orbit of hydrogen is:
(a) −13.6 eV
(b) −6.8 eV
(c) −3.4 eV
(d) −1.51 eV

Try it first, then show the solutionSolution

$E_{2} = - \frac{13.6}{2 ^{2}} = - 3.4 e V$

Answer: (c) −3.4 eV

Example 2: Large deflections · 1 mark · Assertion-reasonBoard level

Assertion (A): In the alpha-particle scattering experiment, only a very small fraction of alpha particles is deflected through large angles.
Reason (R): The nucleus occupies a very small fraction of the volume of the atom.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

Large deflections need a very close approach to the nucleus. Because the nucleus is so tiny, only a few particles come that close. R explains A.

Answer: (a)

Example 3: Ratios between orbits · 2 marks · Very short answerHarder board level

For the hydrogen atom, find the ratio of (a) the radii, (b) the speeds, (c) the energies of the electron in the first and third orbits. [2 marks]

Try it first, then show the solutionSolution

(a) $r \propto n^{2}$ : $r_{1} : r_{3} = 1 : 9$
(b) $v \propto \frac{1}{n}$ : $v_{1} : v_{3} = 3 : 1$
(c) $E \propto - \frac{1}{n ^{2}}$ : $E_{1} : E_{3} = 9 : 1$ (−13.6 eV and −1.51 eV)

Answer: 1 : 9; 3 : 1; 9 : 1

Example 4: Transition wavelengths · 3 marks · Short answerHarder board level

An electron in a hydrogen atom falls from n = 4 to n = 2. (a) Find the energy and wavelength of the emitted photon. (b) Name the series and region. (c) Of all lines that can come from n = 4, which has the longest wavelength? ( $h c \approx 1240 e V nm$ ) [3 marks]

Try it first, then show the solutionSolution

(a) $Δ E = - 0.85 - (- 3.4) = 2.55 e V$ ; $λ = \frac{1240}{2.55} \approx 486 nm$
(b) Ends on n = 2: Balmer series, visible (blue-green)
(c) Longest wavelength = smallest energy gap: 4 → 3, with $Δ E = - 0.85 - (- 1.51) = 0.66 e V$ , $λ \approx 1880 nm$ (infrared, Paschen)

Answer: (a) 2.55 eV, about 486 nm (b) Balmer, visible (c) 4 → 3 (about 1880 nm)

Example 5: Bohr's model: full answer · 5 marks · Long answerHarder board level

(a) State Bohr's postulates. (b) Write expressions for the radius, speed and energy of the electron in the nth orbit of hydrogen, and show how each depends on n. (c) Calculate the energy needed to move an electron from n = 1 to n = 3, and the wavelength of the photon emitted when it returns directly to n = 1. (d) Name the series of this line. [5 marks]

Try it first, then show the solutionSolution

(a) Stationary orbits without radiation; $m v r = nh /2 π$ ; $h ν = E_{i} - E_{f}$ .
(b) $r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}} \propto n^{2}$ ; $v_{n} = \frac{e ^{2}}{2 ε _{0} nh} \propto \frac{1}{n}$ ; $E_{n} = - \frac{m e ^{4}}{8 n ^{2} ε _{0}^{2} h ^{2}} = - \frac{13.6}{n ^{2}} e V \propto - \frac{1}{n ^{2}}$ .
(c) $E_{3} - E_{1} = - 1.51 - (- 13.6) = 12.09 e V$ .
Emitted photon: $λ = \frac{1240}{12.09} \approx 102.6 nm$ .
(d) Ends on n = 1: Lyman series (ultraviolet).

Answer: 12.09 eV; about 103 nm; Lyman series

Example 6: Closest approach and nuclear size · 3 marks · Short answerHarder board level

An alpha particle of kinetic energy 6.0 MeV is fired head-on at a gold nucleus (Z = 79). (a) Find the distance of closest approach. (b) What does this tell you about the size of the gold nucleus? (c) How would d change if the kinetic energy were doubled? ( $\frac{e ^{2}}{4 π ε _{0}} = 1.44 M e V f m$ ) [3 marks]

Try it first, then show the solutionSolution

(a) $d = \frac{2 Z}{K} \times \frac{e ^{2}}{4 π ε _{0}} = \frac{2 \times 79 \times 1.44}{6.0} f m = \frac{227.5}{6.0} f m \approx 37.9 f m$
(b) The alpha particle is turned back before touching the nucleus, so the nucleus must be smaller than about $3.8 \times 1 0^{- 14} m$ in radius.
(c) $d \propto \frac{1}{K}$ , so doubling K halves d to about 19.0 fm.

Answer: (a) about 37.9 fm (b) the nucleus is smaller than this (c) it halves to about 19 fm

Example 7: Excitation by an electron beam · 3 marks · Short answerHOTS / competency

A beam of electrons of energy 12.8 eV passes through hydrogen gas in its ground state. (a) Up to which level can the atoms be excited? (b) How many spectral lines can be emitted? (c) Which of them lie in the visible region? [3 marks]

Try it first, then show the solutionSolution

(a) $E_{4} - E_{1} = 13.6 - 0.85 = 12.75 e V \leq 12.8 e V$ , while $E_{5} - E_{1} = 13.06 e V > 12.8 e V$ : up to n = 4.
(b) $\frac{4 \times 3}{2} = 6$ lines.
(c) Visible lines are Balmer lines (ending on n = 2) from n = 3 and n = 4: 3 → 2 (about 656 nm) and 4 → 2 (about 486 nm).

Answer: (a) n = 4 (b) 6 lines (c) 3 → 2 and 4 → 2

Example 8: Orbital period and n · 2 marks · Very short answerHOTS / competency

Using $r_{n} \propto n^{2}$ and $v_{n} \propto \frac{1}{n}$ , show how the time period of revolution of the electron depends on n. If the period in the first orbit is T, what is it in the second orbit? [2 marks]

Try it first, then show the solutionSolution

$T_{n} = \frac{2 π r _{n}}{v _{n}} \propto \frac{n ^{2}}{1/ n} = n^{3}$
$T_{2} = 2^{3} T = 8 T$

Answer: $T_{n} \propto n^{3}$ ; 8T

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Inside a hydrogen lamp

Suppose a hydrogen discharge lamp contains atoms that have been excited to levels up to n = 4. The energies of the levels are $E_{n} = - 13.6/ n^{2}$ eV. Take $h c \approx 1240 e V nm$ .

(i) · 1 mark
How many different spectral lines can the lamp emit?
Answer
Answer: 6
$\frac{4 \times 3}{2} = 6$
(ii) · 1 mark
Which transition gives the shortest wavelength?
Answer
Answer: 4 → 1
Largest energy gap: $13.6 - 0.85 = 12.75 e V$
(iii) · 1 mark
What is the wavelength of the 2 → 1 line?
Answer
Answer: About 122 nm
$Δ E = 10.2 e V$ ; $1240/10.2 \approx 122 nm$
(iv) · 1 mark
In which region of the spectrum is the 2 → 1 line?
Answer
Answer: Ultraviolet (Lyman series)
Transitions ending on n = 1 are in the ultraviolet.

Probing the nucleus

Suppose students repeat the scattering experiment in a simulation, firing alpha particles of kinetic energy 5.0 MeV at gold nuclei (Z = 79) with different impact parameters. Use $\frac{e ^{2}}{4 π ε _{0}} = 1.44 M e V f m$ .

(i) · 1 mark
Which particles are scattered through the largest angles?
Answer
Answer: Those with the smallest impact parameter.
A nearly head-on approach meets the strongest repulsion.
(ii) · 1 mark
What happens to a particle with zero impact parameter?
Answer
Answer: It stops momentarily and rebounds straight back (180°).
Head-on collision.
(iii) · 1 mark
Find its distance of closest approach.
Answer
Answer: About 45.5 fm
$2 \times 79 \times 1.44/5.0$
(iv) · 1 mark
Why do most particles pass almost undeflected?
Answer
Answer: Most have large impact parameters because the nucleus is tiny and the atom is mostly empty.
Only a few pass close enough to a nucleus to feel a strong force.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

The ground state energy of hydrogen is −13.6 eV. Find the kinetic and potential energies of the electron in this state. [2 marks] (2 marks)

Step	Marks
K = −E = 13.6 eV	1
U = 2E = −27.2 eV	1

State Bohr's postulates. Using them, explain how hydrogen produces a line spectrum. [3 marks] (3 marks)

Step	Marks
Stationary orbits without radiation; $m v r = nh /2 π$	1
$h ν = E_{i} - E_{f}$ when an electron jumps between orbits	1
Only certain energy levels exist, so only certain energy differences and hence certain frequencies are emitted: a line spectrum	1

(a) Describe the alpha-particle scattering experiment and state its conclusions. (b) What is the distance of closest approach? Find it for a 5.0 MeV alpha particle and a gold nucleus (Z = 79). (c) Give two limitations of Rutherford's model. [5 marks] (5 marks)

Step	Marks
(a) Set-up: source, collimator, thin gold foil, rotatable ZnS screen and microscope; observations: most pass through, few deflected through large angles	1
Conclusions: atom mostly empty; positive charge and mass in a tiny nucleus	1
(b) At closest approach, K = U: $d = \frac{2 Z e ^{2}}{4 π ε _{0} K}$	1
$d = \frac{2 \times 79 \times 1.44}{5.0} f m \approx 45.5 f m$	1
(c) Cannot explain the stability of atoms (accelerating electron should radiate and spiral in); cannot explain line spectra	1

Where students lose marks in Atoms

The slips examiners see most often, and what to write instead.

Mistake: Giving a positive sign to the orbit energy
Forgetting that bound states have negative energy.
Do this: $E_{n} = - 13.6/ n^{2}$ eV; the minus sign shows the electron is bound.
Usually costs: 1 mark.
Mistake: Naming the series by the starting level
Confusing the start and end of a jump.
Wrong working
Jump 3 → 2 called Paschen (n = 3)
Right working
It ends on n = 2, so it is Balmer
Do this: The series is named by the final (lower) level.
Usually costs: 1 mark.
Mistake: Mixing up longest and shortest wavelength lines
Forgetting λ ∝ 1/ΔE.
Do this: Smallest energy gap gives the longest wavelength. In a series, the first line (from the next level up) is the longest; the series limit (from n = ∞) is the shortest.
Usually costs: 1 mark.
Mistake: Deriving the radius of the nth orbit at length
Older syllabi required it.
Do this: The 2026-27 syllabus asks only for the expressions of radius, speed and energy; quote them and use them.
Usually costs: Time.
Mistake: Writing spectral series formulas such as the Rydberg formula in answers
Old notes include them.
Do this: Spectral series formulae are not in the syllabus; compute wavelengths from $h ν = E_{i} - E_{f}$ with the Bohr energies.
Usually costs: Time, and confusion.

NCERT exercise guide: Atoms

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 12.1 and 12.2 (models of the atom)

What it tests: Comparing Thomson's and Rutherford's models, and predicting the scattering results for different targets.

How to approach it: Thomson's spread-out charge cannot produce large deflections; Rutherford's concentrated nucleus can. Heavier, more highly charged nuclei scatter more strongly.

Same kind as Exercise 12.2

If the gold foil in the scattering experiment is replaced by an equally thin foil of aluminium (Z = 13), how will the number of large-angle deflections change? Explain.

Show the solutionSolution

The repulsive force on an alpha particle at a given distance is proportional to the nuclear charge Ze.
Aluminium's nuclear charge (13e) is much smaller than gold's (79e), so fewer alpha particles are turned through large angles.

Answer: Fewer large-angle deflections, because the aluminium nucleus has a much smaller charge.

Exercises 12.3 to 12.5 (energy levels and transitions)

What it tests: Frequency of radiation from a given energy gap, kinetic and potential energies in the ground state, and the energy needed to raise an atom to a higher level.

How to approach it: $ν = Δ E / h$ ; K = −E, U = 2E; absorbed energy $= E_{n} - E_{1}$ .

Same kind as Exercise 12.3

Two energy levels in an atom are 2.0 eV apart. Find the frequency and wavelength of the radiation emitted when an electron falls from the upper to the lower level. ( $h = 6.63 \times 1 0^{- 34} J s$ )

Show the solutionSolution

$Δ E = 2.0 \times 1.6 \times 1 0^{- 19} = 3.2 \times 1 0^{- 19} J$
$ν = \frac{3.2 \times 1 0 ^{- 19}}{6.63 \times 1 0 ^{- 34}} \approx 4.83 \times 1 0^{14} H z$
$λ = \frac{c}{ν} = \frac{3 \times 1 0 ^{8}}{4.83 \times 1 0 ^{14}} \approx 6.2 \times 1 0^{- 7} m$ (about 620 nm)

Answer: $4.8 \times 1 0^{14} H z$ ; about 620 nm

Exercises 12.7 to 12.9 (radii and excitation)

What it tests: Radii of higher orbits from the Bohr radius, which lines appear when hydrogen is bombarded with electrons of given energy, and quantum numbers of large orbits.

How to approach it: $r_{n} = n^{2} a_{0}$ ; find the highest level reachable with the given energy, then list all downward jumps.

Same kind as Exercise 12.8

Hydrogen in its ground state is bombarded with electrons of energy 12.2 eV. Up to which level can the atoms be excited, and how many spectral lines can then be emitted?

Show the solutionSolution

Energies needed from n = 1: to n = 2, 10.2 eV; to n = 3, 12.09 eV; to n = 4, 12.75 eV.
12.2 eV is enough for n = 3 but not n = 4.
Lines from n = 3: $\frac{3 \times 2}{2} = 3$ (3→2, 3→1, 2→1).

Answer: Up to n = 3; three lines

Atoms: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 4.4 marks per paper; 22 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Atoms

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 55/1/1, Q8 · 1 mark · MCQ

The ‘distance of closest approach’ of an alpha-particle is ‘d’ when it moves with a velocity $v$ head-on towards the target nucleus. If the velocity of alpha particle is halved, the new ‘distance of closest approach’ will be –

(A) $\frac{d}{2}$
(B)2d
(C) $\frac{d}{4}$
(D)4d

Try it first, then show our solutionSolution

At closest approach, $\frac{1}{2} m v^{2} = \frac{k ( 2 e ) ( Z e )}{d}$ , so $d \propto \frac{1}{v ^{2}}$ .
Halving $v$ makes the new distance $4 d$ .
1 mark

Answer: 4d, option (D)

Where the marks are

Uses energy conservation to get $d \propto 1/ v^{2}$ , giving 4d (1 mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/5/1, Q26(b) · 3 marks · Short answer

(i) Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.
(ii) If Bohr's quantization postulate (angular momentum = $\frac{nh}{2 π}$ ) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun ? Explain.

Try it first, then show our solutionSolution

(i) Graph: number of scattered α-particles N(θ) (log scale) on y-axis vs scattering angle θ (0° to 180°) on x-axis — very large at small angles and falling steeply as θ increases ( $N \propto 1/ sin^{4} (θ /2)$ ), with only a very few particles at angles close to 180°.
1 mark
Conclusions:
• Most α-particles pass undeflected, so most of the atom is empty space.
• A very few are deflected through large angles (some > 90°), so the positive charge and nearly all the mass of the atom are concentrated in a tiny central nucleus.
1 mark
(ii) Angular momentum of a planet is enormous: for the Earth, $L = m v r \approx 6 \times 1 0^{24} \times 3 \times 1 0^{4} \times 1.5 \times 1 0^{11} \approx 2.7 \times 1 0^{40}$ J s, so $n = \frac{2 π L}{h} \sim 1 0^{74}$ . For such huge n, the difference between successive allowed angular momenta and energies is negligibly small compared to the values themselves, so the allowed orbits appear continuous; quantization is unobservable and is not spoken of.
1 mark

Answer: (i) N(θ) falls sharply with θ; atom mostly empty, positive charge and mass concentrated in tiny nucleus; (ii) n ~ 10^74 for planets, so quantized levels are practically continuous

Where the marks are

Correct graph of N vs θ (1 mark)
Two conclusions: atom mostly empty; tiny massive positive nucleus (½ + ½ mark)
Explains huge quantum number (~10⁷⁴) makes quantization unnoticeable (1 mark)

See it in the full 2026 board paper, set 55/5/1

2025, set 55/1/1, Q20 · 2 marks · Very short answer

Prove that, in Bohr model of hydrogen atom, the time period of revolution of an electron in $n^{t h}$ orbit is proportional to $n^{3}$ .

Try it first, then show our solutionSolution

Time period of revolution in the $n^{t h}$ orbit: $T = \frac{2 π r _{n}}{v _{n}}$ ...(1).
½ mark
Bohr's quantisation condition: $m v_{n} r_{n} = \frac{nh}{2 π} \Rightarrow v_{n} = \frac{nh}{2 π m r _{n}}$ ...(2). Substituting in (1): $T = \frac{4 π ^{2} m r _{n}^{2}}{nh}$ .
½ mark
Radius of the $n^{t h}$ orbit: $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ , so $r_{n} \propto n^{2}$ .
½ mark
Hence $T = \frac{4 π ^{2} m}{nh} \cdot \frac{ε _{0}^{2} n ^{4} h ^{4}}{π ^{2} m ^{2} e ^{4}} = \frac{4 ε _{0}^{2} n ^{3} h ^{3}}{m e ^{4}}$ , so $T \propto n^{3}$ . (Shortcut: $r \propto n^{2}$ and $v \propto \frac{1}{n}$ give $T = \frac{2 π r}{v} \propto n^{3}$ .)
½ mark

Answer: T = 4ε₀²n³h³/(me⁴), so T ∝ n³

Where the marks are

Writes $T = 2 π r / v$ (½ mark)
Uses $m v r = nh /2 π$ to express v (½ mark)
Uses $r_{n} \propto n^{2}$ , i.e. the expression for the orbit radius (½ mark)
Combines the results to get $T \propto n^{3}$ (½ mark)

See it in the full 2025 board paper, set 55/1/1

Every Atoms question, by year

For 2027

Sample paper 2026-27: Q11 (1 mark, solved), Q13 (1 mark, solved), Q21(A) (2 marks, solved), Q21(B) (2 marks, solved)

2026

Set 55/1/1: Q8 (1 mark, solved), Q16 (1 mark, solved)

Set 55/5/1: Q10 (1 mark, solved), Q26(b) (3 marks, solved)

Sample paper 2025-26: Q7 (1 mark, solved), Q13 (1 mark, solved), Q21(I) (2 marks, solved)

2025

Set 55/1/1: Q16 (1 mark, solved), Q20 (2 marks, solved)

Set 55/7/1: Q10 (1 mark, solved), Q27 (3 marks, solved)

Sample paper 2024-25: Q8 (1 mark, solved), Q14 (1 mark, solved)

2024

Set 55/1/1: Q9 (1 mark, solved), Q27 (3 marks, solved)

Set 55/5/1: Q9 (1 mark, solved), Q16 (1 mark, solved), Q20 (2 marks, solved), Q21 (2 marks, solved)

Sample paper 2023-24: Q4 (1 mark, solved), Q8 (1 mark, solved), Q12 (1 mark, solved), Q24 (3 marks, solved)

2023

Set 55/1/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q23 (2 marks, solved), Q30(a) (3 marks, solved)

Set 55/5/1: Q10 (1 mark, solved), Q15 (1 mark, solved), Q30(a) (3 marks, solved), Q30(b) (3 marks, solved)

Practice questions on Atoms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The radius of the first Bohr orbit is $a_{0}$ . The radius of the third orbit is:
(a) $3 a_{0}$
(b) $6 a_{0}$
(c) $9 a_{0}$
(d) $a_{0} /9$
Hint
$r \propto n^{2}$ .
Answer
(c) $9 a_{0}$
Q2 · 1 mark · MCQBoard level
The ionisation energy of hydrogen in its ground state is:
(a) 3.4 eV
(b) 10.2 eV
(c) 13.6 eV
(d) 27.2 eV
Hint
Energy to take the electron from n = 1 to n = ∞.
Answer
(c) 13.6 eV
Q3 · 1 mark · Very short answerBoard level
State Bohr's quantisation condition for angular momentum.
Hint
Multiple of h/2π.
Answer
The angular momentum of the electron in a stationary orbit is a whole-number multiple of $h /2 π$ : $m v r = nh /2 π$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The total energy of an electron in a hydrogen atom is negative.
Reason (R): The electron is bound to the nucleus.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Zero energy is a free electron at rest.
Answer
(a)
Q5 · 2 marks · Very short answerHarder board level
The electron in a hydrogen atom moves from n = 1 to n = 2. By what factor do (a) its speed, (b) its kinetic energy, (c) the radius of its orbit change?
Hint
$v \propto 1/ n$ , $K \propto 1/ n^{2}$ , $r \propto n^{2}$ .
Answer
(a) halves (b) becomes one quarter (c) becomes four times.
Q6 · 2 marks · Very short answerHarder board level
An electron in a hydrogen atom falls from n = 5 to n = 2. Find the energy and wavelength of the emitted photon, name its series and colour region, and compare its wavelength with that of the 3 → 2 line (656 nm). ( $h c \approx 1240 e V nm$ )
Hint
$E_{5} = - 13.6/25$ eV.
Answer
$E_{5} = - 0.544 e V$ , so $Δ E = - 0.544 - (- 3.40) \approx 2.86 e V$ and $λ = \frac{1240}{2.86} \approx 434 nm$ . It ends on n = 2: Balmer series, violet end of the visible spectrum. It is about $\frac{656}{434} \approx 1.5$ times shorter than the red 3 → 2 line, because the energy gap is larger.
Q7 · 3 marks · Short answerHarder board level
Explain, with reasons, two limitations of Rutherford's model and how Bohr's postulates overcome them.
Hint
Stability and spectra.
Answer
(1) An orbiting electron accelerates and, by classical theory, should radiate energy and spiral into the nucleus, so atoms would be unstable. Bohr: electrons in stationary orbits do not radiate. (2) A spiralling electron would emit a continuous spectrum, but atoms give line spectra. Bohr: radiation is emitted only in jumps between fixed levels, $h ν = E_{i} - E_{f}$ , so only certain frequencies appear.
Q8 · 3 marks · Short answerHarder board level
Find the shortest and the longest wavelengths in the Balmer series of hydrogen, using the Bohr energies. ( $h c \approx 1240 e V nm$ )
Hint
Balmer ends on n = 2; longest from n = 3, shortest from n = ∞.
Answer
Longest: 3 → 2, ΔE = 1.89 eV, λ ≈ 656 nm. Shortest (series limit): ∞ → 2, ΔE = 3.4 eV, λ ≈ 365 nm.
Q9 · 3 marks · Short answerHOTS / competency
A photon of energy 12.75 eV is absorbed by a hydrogen atom in its ground state. To which level is the electron raised? How many lines can be emitted as it returns, and which has the longest wavelength?
Hint
$E_{n} - E_{1} = 12.75$ eV.
Answer
$- 13.6/ n^{2} = - 13.6 + 12.75 = - 0.85 e V \Rightarrow n = 4$ . Six lines possible. Longest wavelength: 4 → 3 (smallest gap, 0.66 eV, about 1880 nm).
Q10 · 2 marks · Very short answerHOTS / competency
The kinetic energy of the alpha particles in a scattering experiment is doubled. How does the distance of closest approach to a given nucleus change? What happens if the target nucleus has twice the charge (same K)?
Hint
$d \propto Z / K$ .
Answer
Doubling K halves d. Doubling Z (same K) doubles d.
Q11 · 2 marks · Very short answerHarder board level
Find the speed of the electron in the first Bohr orbit of hydrogen, and the number of revolutions it makes per second. ( $v_{1} = 2.18 \times 1 0^{6} m s^{- 1}$ , $r_{1} = 0.53 \times 1 0^{- 10} m$ )
Hint
Frequency = v/2πr.
Answer
Speed $2.18 \times 1 0^{6} m s^{- 1}$ ; $f = \frac{2.18 \times 1 0 ^{6}}{2 π \times 0.53 \times 1 0 ^{- 10}} \approx 6.5 \times 1 0^{15}$ revolutions per second.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Give two conclusions of the alpha-scattering experiment.Show answer
Most of the atom is empty space; the positive charge and almost all the mass are concentrated in a tiny nucleus.
2.What is the impact parameter?Show answer
The perpendicular distance of the initial velocity line of the alpha particle from the centre of the nucleus.
3.State Bohr's quantisation condition.Show answer
$m v r = nh /2 π$ .
4.Write the energy of the electron in the nth orbit of hydrogen.Show answer
$E_{n} = - 13.6/ n^{2}$ eV.
5.How does the radius of the nth orbit depend on n?Show answer
$r_{n} \propto n^{2}$ ( $r_{n} = 0.53 n^{2}$ Å).
6.What is the ionisation energy of hydrogen?Show answer
13.6 eV.
7.Which series of hydrogen lies in the visible region?Show answer
The Balmer series (transitions ending on n = 2).
8.Energy of the photon for the transition n = 2 → n = 1?Show answer
$- 3.4 - (- 13.6) = 10.2 e V$ .

Best order to study it

Start with the alpha-scattering experiment: set-up, observations and conclusions.
Learn the impact parameter, the scattering graph and the distance of closest approach.
Study Rutherford's model, the energy of an orbiting electron and the model's two limitations.
Learn Bohr's three postulates and how they fix those limitations.
Learn the expressions for $r_{n}$ , $v_{n}$ and $E_{n}$ and practise ratio questions.
Finish with the energy-level diagram, transitions, the named series and line counting.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you describe the alpha-scattering set-up and its two key observations?
Can you explain how the impact parameter decides the scattering angle?
Can you calculate the distance of closest approach?
Can you show that the total energy of an orbiting electron is negative?
Can you state two limitations of Rutherford's model?
Can you state Bohr's three postulates?
Can you write $r_{n}$ , $v_{n}$ and $E_{n}$ and find their ratios for two orbits?
Can you draw the hydrogen energy-level diagram with the first four levels?
Can you find the wavelength of a photon from a given transition and name its series?
Can you find how many lines are emitted when atoms are excited to level n?

Atoms: frequently asked questions

Do I need to derive the radius of the Bohr orbit?

No. The 2026-27 syllabus lists the expressions for the radius of the nth orbit, the speed and the energy; the NCERT book keeps only the expression for the radius. Learn and use the expressions.

Are the Lyman, Balmer and Paschen formulas asked?

No. The spectral series formulae are not in the syllabus, and hydrogen line spectra are treated qualitatively. You should know which jumps give which series and region, and be able to find a line's wavelength from h\nu = E_i - E_f.

Is de Broglie's explanation of Bohr's second postulate in the syllabus?

No. NCERT section 12.6 is not named in the CBSE content for 2026-27.

Why is the energy of an electron in an atom negative?

We take the energy of a free electron at rest far from the nucleus as zero. A bound electron has less energy than that, so its energy is negative; supplying energy equal to its magnitude frees it.

Why did Rutherford use a gold foil?

Gold can be hammered into extremely thin foils, so each alpha particle is likely to meet at most one nucleus, and its large nuclear charge gives clear, strong scattering.

What kinds of questions come from this chapter?

In our question bank, mostly 1-mark MCQs and assertion-reason items, plus 2-3 mark questions on energy levels, transitions and Rutherford scattering.

Based on: CBSE Physics (042) syllabus 2026-27; NCERT Physics Part II, Textbook for Class XII, Chapter 12 (Reprint 2026-27)