PhysicsISC Class 12 · Chapter 12

Atoms ISC Class 12: Notes, Bohr Model Derivations and Numericals

Updated 14 October 2026 · Fact-checked

Atoms is Chapter 12 of ISC Class 12 Physics. It covers the Geiger-Marsden alpha-particle scattering experiment, Rutherford's nuclear model and the distance of closest approach, Bohr's postulates for the hydrogen atom, expressions for orbit radius, speed and energies, the energy-level diagram and the Lyman, Balmer, Paschen, Brackett and Pfund series.

What Atoms is about

How are atoms built? This chapter traces the answer from Rutherford to Bohr. Together with Nuclei it carries 6 of the 70 theory marks; Bohr-model derivations have been set as 3-mark parts of long questions in recent papers.

Alpha-particle scattering (Geiger-Marsden): a thin gold foil bombarded with alpha particles let most pass straight through, deflected a few through large angles and turned back about 1 in 8000. Rutherford concluded that the positive charge and almost all the mass of an atom are concentrated in a tiny nucleus, with electrons around it and the atom mostly empty. The size of the nucleus is estimated from the distance of closest approach $r_{0}$ , found by equating the alpha particle's kinetic energy to the electrostatic potential energy $\frac{1}{4 π ε _{0}} \frac{2 e \times Z e}{r _{0}}$ , giving $r_{0}$ of about $1 0^{- 14}$ to $1 0^{- 15}$ m. The mathematical theory of scattering is excluded.

Bohr's model (Z = 1): electrons move in certain stable orbits without radiating; their angular momentum is quantised, $m v r = \frac{nh}{2 π}$ ; radiation is emitted or absorbed only in a jump between orbits, $h f = E_{2} - E_{1}$ . From these you derive the radius $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ (∝ n²), speed $v_{n} = \frac{e ^{2}}{2 ε _{0} nh}$ (∝ 1/n), and energies: kinetic $\frac{13.6}{n ^{2}}$ eV, potential $- \frac{27.2}{n ^{2}}$ eV and total $E_{n} = - \frac{13.6}{n ^{2}}$ eV. The energy-level diagram explains the hydrogen line spectrum: $\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ , with the Lyman (UV), Balmer (visible), Paschen, Brackett and Pfund (infrared) series. The syllabus asks for wavelengths in nm, not Å.

Weightage: Atoms is part of the unit Atoms and Nuclei, which carries 6 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Alpha-particle scattering and Rutherford's nuclear model: In the Geiger-Marsden experiment a narrow beam of alpha particles (He nuclei, charge +2e) from a radioactive source struck a very thin gold foil, and a movable zinc-sulphide screen with a microscope counted scattered particles at different angles. Observations: most passed undeviated; a few were deflected through large angles; about 1 in 8000 bounced back (more than 90°). Conclusions: (1) the atom is mostly empty space; (2) all the positive charge and almost all the mass are concentrated in a tiny central nucleus, of radius about $1 0^{- 15}$ to $1 0^{- 14}$ m against $1 0^{- 10}$ m for the atom; (3) electrons revolve round the nucleus, held by electrostatic attraction. The atom has Z electrons and a nucleus of charge +Ze, with mass number A = Z + N. Limitations: an accelerating (circling) electron should radiate continuously and spiral into the nucleus, so the model cannot explain the stability of atoms or the line spectrum.; Worked example
Why did Rutherford conclude that the nucleus is very small and massive? Which observation shows each point?
Try it first, then show the solutionSolution
Most alpha particles passed undeviated: the atom is mostly empty, so the region of strong repulsion is tiny.
Very few were turned back through large angles: a heavy, highly charged centre must have repelled them head-on; a light or spread-out charge could not reverse a fast alpha particle.
Answer: Small: most particles pass straight through. Massive and charged: a few are deflected through very large angles.
Distance of closest approach: An alpha particle fired straight at a nucleus slows down as it approaches, stops momentarily at the distance of closest approach $r_{0}$ , and returns. At that point all its kinetic energy has become electrostatic potential energy: $K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{r _{0}}$ , so $r_{0} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{K}$ . The nucleus cannot be larger than $r_{0}$ , so this gives an upper limit to nuclear size, of the order of $1 0^{- 14}$ m for gold with 5 MeV alpha particles. $r_{0} \propto \frac{Z}{K}$ : doubling the alpha particle's energy halves $r_{0}$ .; Worked example
A 5 MeV alpha particle approaches a gold nucleus (Z = 79) head-on. Find the distance of closest approach.
Try it first, then show the solutionSolution
$K = 5 \times 1 0^{6} \times 1.6 \times 1 0^{- 19} = 8 \times 1 0^{- 13} J$
$r_{0} = \frac{9 \times 1 0 ^{9} \times 2 \times 79 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{8 \times 1 0 ^{- 13}}$
$r_{0} = 4.55 \times 1 0^{- 14} m$
Answer: About $4.6 \times 1 0^{- 14}$ m (about 45 fm)
Bohr's postulates: Bohr kept Rutherford's nucleus but added quantum rules for hydrogen (Z = 1). Postulate 1 (stationary orbits): the electron revolves only in certain circular orbits, in which it does not radiate energy; the centripetal force is the Coulomb attraction, $\frac{m v ^{2}}{r} = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}}$ . Postulate 2 (quantisation of angular momentum): in these orbits the angular momentum of the electron is an integral multiple of $\frac{h}{2 π}$ : $m v r = \frac{nh}{2 π}$ , n = 1, 2, 3, … (state it in words in answers). Postulate 3 (frequency condition): radiation is emitted or absorbed only when the electron jumps between orbits; the photon energy equals the energy difference: $h f = E_{n_{2}} - E_{n_{1}}$ .; Worked example
Find the angular momentum of an electron in the third orbit of hydrogen. ( $h = 6.63 \times 1 0^{- 34}$ J s)
Try it first, then show the solutionSolution
$L = \frac{nh}{2 π} = \frac{3 \times 6.63 \times 1 0 ^{- 34}}{2 π}$
$L = 3.17 \times 1 0^{- 34} k g m^{2} s^{- 1}$
Answer: $3.17 \times 1 0^{- 34}$ J s
Radius and speed of the nth orbit: From postulate 1: $m v^{2} r = \frac{e ^{2}}{4 π ε _{0}}$ … (i). From postulate 2: $v = \frac{nh}{2 π m r}$ … (ii). Substituting (ii) into (i): $m \frac{n ^{2} h ^{2}}{4 π ^{2} m ^{2} r ^{2}} r = \frac{e ^{2}}{4 π ε _{0}}$ , so $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ . Thus $r_{n} \propto n^{2}$ ; the first (Bohr) radius is $r_{1} = 0.053 nm$ , and $r_{n} = 0.053 n^{2}$ nm. Putting r back into (ii): $v_{n} = \frac{e ^{2}}{2 ε _{0} nh}$ , so $v_{n} \propto \frac{1}{n}$ ; $v_{1} = 2.19 \times 1 0^{6} m s^{- 1}$ (about c/137).; Worked example
Find the radius of, and the electron's speed in, the third orbit of hydrogen.
Try it first, then show the solutionSolution
$r_{3} = 0.053 \times 3^{2} = 0.477 nm$
$v_{3} = \frac{2.19 \times 1 0 ^{6}}{3} = 7.3 \times 1 0^{5} m s^{- 1}$
Answer: 0.477 nm; $7.3 \times 1 0^{5} m s^{- 1}$
Energy of the electron: kinetic, potential and total: Kinetic energy: from $m v^{2} = \frac{e ^{2}}{4 π ε _{0} r}$ , $K = \frac{1}{2} m v^{2} = \frac{e ^{2}}{8 π ε _{0} r}$ . Potential energy: $U = - \frac{e ^{2}}{4 π ε _{0} r}$ = −2K. Total: $E = K + U = - \frac{e ^{2}}{8 π ε _{0} r}$ . Substituting $r_{n}$ : $E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} n ^{2} h ^{2}} = - \frac{13.6}{n ^{2}} e V$ . So K = +13.6/n² eV and U = −27.2/n² eV. The negative total energy means the electron is bound: 13.6/n² eV must be supplied to free it. Ionisation energy of hydrogen (from n = 1) is 13.6 eV; the ground state is n = 1, and higher n are excited states with energies −3.4, −1.51, −0.85 eV, crowding together towards 0 (the ionisation limit).; Worked example
For the electron in the ground state of hydrogen, find the kinetic, potential and total energies. How much energy is needed to raise it to n = 2?
Try it first, then show the solutionSolution
$E_{1} = - 13.6 e V$ ; $K = + 13.6 e V$ ; $U = - 27.2 e V$
$E_{2} = - \frac{13.6}{4} = - 3.4 e V$
Energy needed $= - 3.4 - (- 13.6) = 10.2 e V$
Answer: K = 13.6 eV, U = −27.2 eV, E = −13.6 eV; 10.2 eV
Hydrogen spectrum and spectral series: When the electron falls from $n_{2}$ to $n_{1}$ it emits a photon of energy $E_{n_{2}} - E_{n_{1}}$ , so $\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ , with the Rydberg constant $R = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{3} c} = 1.097 \times 1 0^{7} m^{- 1}$ , in agreement with the observed lines. Series: Lyman ( $n_{1} = 1$ , ultraviolet), Balmer ( $n_{1} = 2$ , visible), Paschen ( $n_{1} = 3$ ), Brackett ( $n_{1} = 4$ ), Pfund ( $n_{1} = 5$ ) (all infrared). The first line of a series ( $n_{2} = n_{1} + 1$ ) has the longest wavelength; the series limit ( $n_{2} = \infty$ ) the shortest. An emission spectrum comes from atoms falling to lower levels (bright lines); an absorption spectrum comes from cool atoms absorbing photons that lift them to higher levels (dark lines at the same wavelengths). From level n, the number of possible emission lines is $\frac{n ( n - 1 )}{2}$ .; Worked example
Find the wavelength of the first line of the Balmer series (Hα) and the Balmer series limit. ( $R = 1.097 \times 1 0^{7} m^{- 1}$ )
Try it first, then show the solutionSolution
Hα (3 → 2): $\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$ , $λ = \frac{36}{5 \times 1.097 \times 1 0 ^{7}} = 656.3 nm$ (red)
Limit (∞ → 2): $λ = \frac{4}{R} = 364.6 nm$
Answer: 656.3 nm; 364.6 nm

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.

Spectral lines of hydrogen

Suppose you look at a glowing hydrogen discharge tube in a school laboratory: it shines a pinkish red.

The strongest red line comes from the jump n = 3 → n = 2. Find the photon energy and wavelength.

Try it first, then show the solutionSolution

$Δ E = - 1.51 - (- 3.4) = 1.89 e V$
$λ = \frac{1240}{1.89} = 656 nm$ (red)

Answer: 1.89 eV; about 656 nm, the red Hα line

Memory hook: Three to two glows red: 656 nm.

Absorption spectrum

Imagine astronomers studying a star whose spectrum shows a dark line at 656 nm against the bright continuous background.

What does the dark line tell them about the star's outer gas?

Try it first, then show the solutionSolution

A dark line means photons of this energy (1.89 eV) were absorbed.
1.89 eV matches the jump from n = 2 to n = 3 in hydrogen, so hydrogen atoms in the n = 2 state are present in the cooler gas in front of the star.

Answer: The outer gas contains hydrogen (some in the n = 2 level), which absorbs 656 nm light.

Memory hook: Bright line: atoms giving; dark line: atoms taking.

Ionisation energy

Suppose ultraviolet light from a hot star falls on hydrogen gas in space.

What is the longest wavelength that can ionise hydrogen atoms in their ground state?

Try it first, then show the solutionSolution

Ionisation energy = 13.6 eV
$λ_{ma x} = \frac{1240}{13.6} = 91.2 nm$

Answer: 91.2 nm (far ultraviolet); longer wavelengths cannot ionise ground-state hydrogen.

Memory hook: 13.6 eV to set the electron free.

Topics in this chapter

Atoms formula sheet

Distance of closest approach: $r_{0} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{K}$; K = kinetic energy of the alpha particle (J).
Bohr's quantisation condition: $m v r = \frac{nh}{2 π}$; n = 1, 2, 3, …
Frequency condition: $h f = E_{n_{2}} - E_{n_{1}}$; Emission or absorption.
Radius of nth orbit: $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}} = 0.053 n^{2} nm$; $r \propto n^{2}$ .
Speed in nth orbit: $v_{n} = \frac{e ^{2}}{2 ε _{0} nh} = \frac{2.19 \times 1 0 ^{6}}{n} m s^{- 1}$; $v \propto 1/ n$ .
Energies: $K_{n} = \frac{13.6}{n ^{2}} e V, U_{n} = - \frac{27.2}{n ^{2}} e V, E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} n ^{2} h ^{2}} = - \frac{13.6}{n ^{2}} e V$; U = −2K; E = −K.
Rydberg formula: $\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}), R = 1.097 \times 1 0^{7} m^{- 1}$; Use nm for answers.
Number of spectral lines: $N = \frac{n ( n - 1 )}{2}$; From level n down to the ground state.

Atoms: diagrams to practise

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

Geiger-Marsden scattering set-up and alpha paths

Describing the experiment and its observations.

Draw

Lead block with a radioactive source and a narrow channel producing an alpha beam.
Thin gold foil in the beam, inside an evacuated chamber; a ZnS screen with microscope on a circular track around the foil.
Inset: a nucleus with paths of alpha particles: far ones straight, nearer ones bent, one head-on returning.

Label

source in lead block
alpha beam
gold foil
ZnS screen and microscope
scattering angle θ
nucleus

Watch out: Show the head-on path stopping at $r_{0}$ and returning; label the impact parameter qualitatively if you wish.

Energy-level diagram of hydrogen with series

'Draw the energy level diagram and show the transitions for the Lyman and Balmer series'; reading transitions from a given diagram.

Draw

Horizontal lines at −13.6 (n = 1), −3.4 (n = 2), −1.51 (n = 3), −0.85 (n = 4) and −0.54 eV (n = 5), crowding towards 0 eV (n = ∞).
Downward arrows ending on n = 1 (Lyman), on n = 2 (Balmer), on n = 3 (Paschen).
Label each group and its region: UV, visible, IR.

Label

n values
energies in eV
0 eV ionisation limit
Lyman, Balmer, Paschen
UV, visible, IR

Watch out: Arrows pointing down = emission; up = absorption. Read the question's diagram for the direction before naming the spectrum.

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 of alpha scattering; why the nucleus is small; distance of closest approach.	Pair each observation with its conclusion; for $r_{0}$ , convert MeV to J and use $K = k (2 e) (Z e) / r_{0}$ .
Bohr derivations (3 marks)	Derive $r_{n}$ and show $r_{n} \propto n^{2}$ ; derive the total energy; show U = −2K.	Write both postulates, eliminate v, put Z = 1, then state the proportionality.
Ratios for different orbits	If the electron moves from n = 1 to n = 3 or 4, how do v, r, E change?	v ∝ 1/n, r ∝ n², E ∝ −1/n²; substitute, never guess.
Spectral-line numericals	Wavelength of a given line, series limit, energy of a photon, number of lines.	Use the Rydberg formula or energy differences with 1240 eV nm; answer in nm.
Reading an energy-level diagram	Emission or absorption? Which series? Which transition gives the longest wavelength?	Look at arrow direction and the final level; the smallest energy gap gives the longest λ.
Meaning of statements	Bohr's angular-momentum postulate in words; meaning of negative total energy.	Give a worded statement; negative energy means the electron is bound.

Tricks to remember Atoms

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

r grows as n², v falls as 1/n, E as −1/n²
'Radius squares, speed shares, energy shrinks squared.'
Use it when: Ratio questions on orbits.
Lazy Boys Play Basketball Pretty
Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5).
Use it when: Naming series by lower level.
U = −2K, E = −K
Potential is twice kinetic and negative; total equals minus kinetic.
Use it when: Energy relations in Bohr's model.
13.6, 3.4, 1.51, 0.85
13.6 divided by 1, 4, 9, 16: memorise the first four levels.
Use it when: Quick photon-energy calculations.
Closest approach ∝ Z/K
Bigger nucleus pushes back earlier; faster alpha gets closer.
Use it when: Scaling questions.

Atoms: board-level solved questions

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

Example 1: Which series is visible? · 1 mark · MCQBoard level

The series of hydrogen spectral lines that lies in the visible region is the:
(a) Lyman series
(b) Balmer series
(c) Paschen series
(d) Pfund series

Try it first, then show the solutionSolution

Transitions to n = 2 give photons of 1.9-3.4 eV, i.e. 365-656 nm, mostly visible.

Answer: (b) Balmer

Example 2: Negative total energy · 1 mark · Assertion-reasonBoard level

Assertion: The total energy of the electron in a hydrogen atom is negative.
Reason: The electron is bound to the nucleus, and energy must be supplied to remove it to infinity.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

Zero energy is taken for the free electron at rest at infinity; a bound electron has less energy than that, so its total energy is negative.

Answer: (a)

Example 3: Moving to the third orbit · 2 marks · Short answerHarder board level

In the ground state of hydrogen the electron's speed is $v_{0}$ and its orbit radius $r_{0}$ . Find its speed and radius in the third orbit. [2]

Try it first, then show the solutionSolution

$v \propto \frac{1}{n}$ : $v_{3} = \frac{v _{0}}{3}$
$r \propto n^{2}$ : $r_{3} = 9 r_{0}$

Answer: $v_{0} /3$ ; $9 r_{0}$

Example 4: Derivation: radius of the nth orbit · 3 marks · Short answerHarder board level

Using Bohr's postulates, obtain the radius of the electron's orbit in a hydrogen atom when its principal quantum number is n, and hence show that $r_{n} \propto n^{2}$ . [3]

Try it first, then show the solutionSolution

Coulomb attraction provides centripetal force (Z = 1): $\frac{m v ^{2}}{r} = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}} \Rightarrow m v^{2} r = \frac{e ^{2}}{4 π ε _{0}}$ … (i)
Quantisation: $m v r = \frac{nh}{2 π} \Rightarrow v = \frac{nh}{2 π m r}$ … (ii)
Substituting (ii) in (i): $\frac{n ^{2} h ^{2}}{4 π ^{2} m r} = \frac{e ^{2}}{4 π ε _{0}} \Rightarrow r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ . Since $ε_{0}$ , h, m, e are constants, $r_{n} \propto n^{2}$ .

Answer: $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ , so $r_{n} \propto n^{2}$

Example 5: The Hα line two ways · 3 marks · Short answerHarder board level

Find the wavelength of the photon emitted when the electron in a hydrogen atom jumps from n = 3 to n = 2, (a) from the energy levels and (b) from the Rydberg formula. [3]

Try it first, then show the solutionSolution

(a) $E_{3} = - 1.51 e V$ , $E_{2} = - 3.4 e V$ ; $Δ E = 1.89 e V$ ; $λ = \frac{1240}{1.89} = 656 nm$ .
(b) $\frac{1}{λ} = 1.097 \times 1 0^{7} (\frac{1}{4} - \frac{1}{9}) = 1.524 \times 1 0^{6} m^{- 1}$ , $λ = 656.3 nm$ .
Both agree: the first Balmer line, in the red.

Answer: About 656 nm

Example 6: Bohr's postulates and the energy of the electron · 5 marks · Long answerHarder board level

(a) State Bohr's postulates for the hydrogen atom. [2]
(b) Derive an expression for the total energy of the electron in the nth orbit, and show that its potential energy is twice its kinetic energy in magnitude. [2]
(c) Find the ionisation energy of hydrogen in joules. [1]

Try it first, then show the solutionSolution

(a) The electron moves in certain circular orbits without radiating; its angular momentum in these orbits is an integral multiple of h/2π; it emits or absorbs a photon of energy $h f = E_{n_{2}} - E_{n_{1}}$ only when it jumps between orbits.
(b) From $m v^{2} = \frac{e ^{2}}{4 π ε _{0} r}$ : $K = \frac{e ^{2}}{8 π ε _{0} r}$ ; $U = - \frac{e ^{2}}{4 π ε _{0} r} = - 2 K$ .
$E = K + U = - \frac{e ^{2}}{8 π ε _{0} r}$ ; with $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ : $E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} n ^{2} h ^{2}} = - \frac{13.6}{n ^{2}} e V$ .
(c) Ionisation energy $= 0 - E_{1} = 13.6 e V = 13.6 \times 1.6 \times 1 0^{- 19} = 2.18 \times 1 0^{- 18} J$ .

Answer: (b) $E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} n ^{2} h ^{2}}$ (c) $2.18 \times 1 0^{- 18}$ J

Example 7: Lines from the n = 4 level · 3 marks · Short answerHOTS / competency

Hydrogen atoms are excited to n = 4. (a) How many different spectral lines can they emit? (b) Which transition gives the longest wavelength, and in which region is it? (c) Which lines are visible? [3]

Try it first, then show the solutionSolution

(a) $\frac{n ( n - 1 )}{2} = \frac{4 \times 3}{2} = 6$ lines.
(b) Smallest energy gap: 4 → 3 (0.66 eV), λ = 1875 nm, Paschen series, infrared.
(c) Only transitions ending on n = 2 are visible: 4 → 2 (486 nm) and 3 → 2 (656 nm).

Answer: (a) 6 (b) 4 → 3, 1875 nm, infrared (c) 4 → 2 and 3 → 2

Example 8: Scaling the closest approach · 2 marks · Short answerHOTS / competency

A 5 MeV alpha particle approaches a gold nucleus (Z = 79) to within $4.55 \times 1 0^{- 14}$ m. Find the closest approach (a) for a 10 MeV alpha particle on gold and (b) for a 5 MeV alpha particle on silver (Z = 47). [2]

Try it first, then show the solutionSolution

$r_{0} \propto \frac{Z}{K}$ . (a) Doubling K halves $r_{0}$ : $2.28 \times 1 0^{- 14}$ m.
(b) $r_{0} = 4.55 \times 1 0^{- 14} \times \frac{47}{79} = 2.71 \times 1 0^{- 14}$ m.

Answer: (a) $2.3 \times 1 0^{- 14}$ m (b) $2.7 \times 1 0^{- 14}$ m

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Reading the hydrogen energy-level diagram

Suppose a spectroscope is used to study light from hydrogen atoms that have all been excited to the n = 4 level. The levels are −13.6, −3.4, −1.51 and −0.85 eV for n = 1 to 4.

(i) · 1 mark
Is the spectrum observed an emission or an absorption spectrum?
Answer
Answer: Emission.
The atoms fall from n = 4 to lower levels, giving out photons: bright lines.
(ii) · 1 mark
Which transition gives the shortest wavelength? Find it.
Answer
Answer: 4 → 1; about 97 nm
Largest gap: $- 0.85 - (- 13.6) = 12.75 e V$ ; $λ = 1240/12.75 = 97.3 nm$ (Lyman, UV).
(iii) · 2 marks
Find the wavelength of the 4 → 2 line and name its series.
Answer
Answer: 486 nm; Balmer
$Δ E = - 0.85 - (- 3.4) = 2.55 e V$ ; $λ = 1240/2.55 = 486 nm$ (blue-green).
(iv) · 1 mark
How much energy would ionise an atom already in the n = 4 level?
Answer
Answer: 0.85 eV
From −0.85 eV to 0 eV.

Probing the nucleus

Suppose a research group fires alpha particles of 5 MeV head-on at a thin gold foil (Z = 79), as in the Geiger-Marsden experiment.

(i) · 1 mark
Find the distance of closest approach.
Answer
Answer: $4.55 \times 1 0^{- 14}$ m
$r_{0} = \frac{9 \times 1 0 ^{9} \times 2 \times 79 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{8 \times 1 0 ^{- 13}}$
(ii) · 1 mark
What does this distance tell us about the size of the gold nucleus?
Answer
Answer: The nucleus is smaller than about $4.6 \times 1 0^{- 14}$ m.
The alpha particle turns back before touching the nucleus, so this is an upper limit to its radius.
(iii) · 1 mark
Most of the alpha particles pass straight through the foil. What does this show?
Answer
Answer: The atom is mostly empty space.
Only particles passing very close to a nucleus are deflected appreciably.
(iv) · 2 marks
Why could Rutherford's model not explain the stability of the atom?
Answer
Answer: A circling electron accelerates and should radiate energy continuously, spiralling into the nucleus.
Classical electromagnetism says accelerated charges radiate; the orbit would shrink and the emitted spectrum would be continuous, whereas atoms are stable and give line spectra. Bohr's stationary-orbit postulate fixed this.

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.

An electron in hydrogen moves from the ground state to n = 3. How do its speed and orbit radius change? (2 marks)

Step	Marks
$v \propto 1/ n$ : speed becomes $v_{0} /3$	1
$r \propto n^{2}$ : radius becomes $9 r_{0}$	1

Show that the radius of the nth Bohr orbit of hydrogen is proportional to n². (3 marks)

Step	Marks
$\frac{m v ^{2}}{r} = \frac{e ^{2}}{4 π ε _{0} r ^{2}}$ (Z = 1 used)	1
$m v r = nh /2 π$ used to eliminate v	1
$r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ , hence $r_{n} \propto n^{2}$ stated	1

Describe the Geiger-Marsden experiment, state two observations and their conclusions, and find the distance of closest approach of a 5 MeV alpha particle to a gold nucleus (Z = 79). (5 marks)

Step	Marks
Set-up: collimated alpha beam, thin gold foil, movable ZnS detector	1
Most alpha particles undeviated → atom mostly empty	1
Very few deflected through large angles → tiny, massive, positive nucleus	1
$K = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{r _{0}}$ with K in joules	1
$r_{0} = 4.55 \times 1 0^{- 14}$ m	1

Where students lose marks in Atoms

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

Mistake: Not putting Z = 1 and leaving the radius unsimplified
CISCE examiners' comments, 2024 and 2025: in deriving the radius of the nth orbit, candidates skipped steps, left the answer in terms of k, did not substitute Z = 1 for hydrogen, or did not link the result to $r_{n} \propto n^{2}$ .
Wrong working
$r = \frac{k Z e ^{2}}{m v ^{2}}$ (stops here).
Right working
Eliminate v with $m v r = nh /2 π$ , put Z = 1, write $r_{n} = \frac{ε _{0} n ^{2} h ^{2}}{π m e ^{2}}$ and state $r_{n} \propto n^{2}$ .
Do this: End every Bohr derivation with the proportionality the question asks for.
Usually costs: 1-2 marks.
Mistake: Wrong ratios for speed and radius
CISCE examiners' comments, 2024: asked for the electron's speed in the third orbit and the radius of the fourth orbit, candidates wrote 3v₀ instead of v₀/3 and r₀/4 instead of 16r₀, or used r ∝ n.
Wrong working
n = 4: $r = 4 r_{0}$ , $v = 4 v_{0}$ .
Right working
$r = 16 r_{0}$ , $v = v_{0} /4$ .
Do this: Write v ∝ 1/n and r ∝ n² and substitute.
Usually costs: 1 mark each.
Mistake: Stating the angular-momentum postulate as a bare formula
CISCE examiners' comments, 2024 and 2025: asked to state Bohr's angular-momentum postulate, candidates wrote expressions instead of a worded statement, quoted the first postulate, or could not explain the meaning of negative total energy.
Wrong working
mvr = nh/2π.
Right working
The electron can revolve only in those orbits in which its angular momentum is an integral multiple of h/2π (mvr = nh/2π).
Do this: Statement in words, then the formula.
Usually costs: 1 mark.
Mistake: Not recognising an absorption spectrum
CISCE examiners' comments, 2024: shown an energy-level diagram with upward transitions, most candidates did not distinguish absorption from emission and wrote 'hydrogen spectrum' or 'line spectrum', and some named the wrong series.
Wrong working
Line spectrum; Lyman series (for transitions from n = 2 upward).
Right working
Upward arrows = absorption spectrum; transitions starting from n = 2 belong to the Balmer series.
Do this: Check the arrow direction and the lower level before answering.
Usually costs: 1 mark each.
Mistake: Using Å or forgetting to convert eV
The syllabus specifies nm, not Å, for wavelengths; candidates also mix eV and J in $E = h c / λ$ .
Wrong working
λ = 6563 Å; $λ = \frac{h c}{1.89}$ with h in J s.
Right working
λ = 656 nm; use 1240 eV nm with energies in eV, or convert eV to J first.
Do this: Pick one unit system per calculation.
Usually costs: 1 mark.

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 ratio of the radii of the second and first Bohr orbits of hydrogen is:
(a) 2
(b) 4
(c) 1/2
(d) 1/4
Hint
r ∝ n².
Answer
(b) 4
Q2 · 1 mark · MCQBoard level
In the alpha-scattering experiment, most alpha particles passed straight through the foil. This shows that:
(a) the nucleus is negatively charged
(b) the atom is mostly empty space
(c) electrons are heavy
(d) the foil is thick
Hint
Nothing in the way.
Answer
(b)
Q3 · 1 mark · MCQBoard level
The Lyman series of hydrogen lies in the:
(a) infrared
(b) visible
(c) ultraviolet
(d) X-ray region
Hint
Transitions to n = 1 have large energies.
Answer
(c) ultraviolet
Q4 · 1 mark · Very short answerBoard level
State, in words, Bohr's postulate of quantisation of angular momentum.
Hint
Integral multiple.
Answer
An electron can revolve only in orbits in which its angular momentum is an integral multiple of h/2π.
Q5 · 2 marks · Short answerBoard level
Find the total energy of the electron in the n = 2 and n = 3 states of hydrogen, and its kinetic and potential energies in the ground state.
Hint
$E_{n} = - 13.6/ n^{2}$ eV.
Answer
$E_{2} = - 3.4$ eV, $E_{3} = - 1.51$ eV. Ground state: K = 13.6 eV, U = −27.2 eV.
Q6 · 2 marks · Short answerHarder board level
Find the wavelength of the first line of the Lyman series. ( $R = 1.097 \times 1 0^{7} m^{- 1}$ )
Hint
2 → 1.
Answer
$\frac{1}{λ} = R (1 - \frac{1}{4}) = \frac{3 R}{4}$ ; λ = 121.5 nm.
Q7 · 2 marks · Short answerHarder board level
Find the shortest wavelength (series limit) in the Balmer series.
Hint
∞ → 2.
Answer
λ = 4/R = 364.6 nm.
Q8 · 2 marks · Short answerHOTS / competency
Explain the meaning of the negative total energy of the electron, and why Rutherford's model could not explain atomic stability while Bohr's could.
Hint
Bound state; radiating charges.
Answer
Negative energy means the electron is bound; energy must be supplied to free it. In Rutherford's model the orbiting electron accelerates, radiates and spirals in; Bohr simply postulated stationary orbits in which it does not radiate.
Q9 · 3 marks · Short answerHarder board level
Find the distance of closest approach of a 5 MeV alpha particle to a gold nucleus (Z = 79). What does it show about nuclear size?
Hint
$r_{0} = \frac{2 Z k e ^{2}}{K}$ with K in J.
Answer
$r_{0} = 4.55 \times 1 0^{- 14}$ m; the nucleus must be smaller than this, about $1 0^{- 14}$ m or less, compared with $1 0^{- 10}$ m for the atom.
Q10 · 3 marks · Short answerHarder board level
Find the radius of the third orbit and the speed of the electron in it, using $r_{1} = 0.053$ nm and $v_{1} = 2.19 \times 1 0^{6} m s^{- 1}$ . Find the electron's angular momentum there.
Hint
r ∝ n², v ∝ 1/n, L = nh/2π.
Answer
$r_{3} = 0.477$ nm; $v_{3} = 7.3 \times 1 0^{5} m s^{- 1}$ ; $L = 3 h /2 π = 3.17 \times 1 0^{- 34}$ J s.
Q11 · 3 marks · Short answerHarder board level
Distinguish between emission and absorption spectra of hydrogen. Which lines would cool hydrogen gas (all atoms in n = 1) absorb from white light, and in which region?
Hint
Absorption starts from the occupied level.
Answer
Emission: bright lines from atoms falling to lower levels. Absorption: dark lines where atoms take in photons to rise to higher levels. Atoms in n = 1 absorb only Lyman-series wavelengths (1 → 2, 1 → 3, …), all in the ultraviolet, so no visible dark lines appear.
Q12 · 5 marks · Long answerHarder board level
(a) Describe the alpha-scattering experiment and state its conclusions. [3]
(b) A 10 MeV alpha particle approaches a gold nucleus head-on. Find its distance of closest approach. [2]
Hint
r₀ ∝ 1/K.
Answer
(a) Alpha beam on thin gold foil; ZnS detector at various angles; most undeviated, a few deflected through large angles, about 1 in 8000 reflected back; so the atom is mostly empty with a tiny, massive, positively charged nucleus. (b) $r_{0} = \frac{9 \times 1 0 ^{9} \times 2 \times 79 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{1.6 \times 1 0 ^{- 12}} = 2.28 \times 1 0^{- 14}$ m.
Q13 · 2 marks · Short answerHarder board level
Find the longest wavelength of radiation that can ionise a hydrogen atom in its ground state.
Hint
Ionisation energy 13.6 eV.
Answer
λ = 1240/13.6 = 91.2 nm

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.Main conclusion of the alpha-scattering experiment?Show answer
The atom has a tiny, dense, positively charged nucleus; the rest is mostly empty space.
2.Distance of closest approach?Show answer
$r_{0} = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{K}$ .
3.Bohr radius?Show answer
0.053 nm.
4.Energy of the nth level of hydrogen?Show answer
$E_{n} = - 13.6/ n^{2}$ eV.
5.Relation between kinetic and potential energy in a Bohr orbit?Show answer
U = −2K; E = −K.
6.Rydberg formula?Show answer
$1/ λ = R (1/ n_{1}^{2} - 1/ n_{2}^{2})$ .
7.Which series lie in the infrared?Show answer
Paschen, Brackett and Pfund.
8.Ionisation energy of hydrogen?Show answer
13.6 eV.

Best order to study it

Learn the Geiger-Marsden experiment: set-up, observations, conclusions.
Derive and practise the distance of closest approach.
Learn Bohr's three postulates in words.
Derive r_n, v_n and the energies; memorise 0.053 nm, 2.19 × 10⁶ m/s and −13.6/n² eV.
Learn the spectral series with the Rydberg formula and the energy-level diagram.
Practise transitions, series limits, number of lines and emission versus absorption.

Quick revision checklist

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

Can you describe alpha scattering and give two conclusions?
Can you calculate the distance of closest approach?
Can you state Bohr's postulates in words?
Can you derive r_n ∝ n² with Z = 1?
Can you derive E_n and show U = −2K?
Can you find speed and radius ratios between orbits?
Can you find any line's wavelength in nm by two methods?
Can you read an energy-level diagram for emission or absorption and name the series?

Atoms: frequently asked questions

Is the mathematical theory of alpha scattering required?

No. The syllabus excludes the mathematical theory of scattering; you need the experiment, observations, conclusions and the distance of closest approach from energy conservation.

Should wavelengths be given in Å or nm?

The ISC syllabus says to use nm and not Å for λ in this chapter.

Which derivations are needed for Bohr's model?

Expressions for orbital velocity, radius, kinetic energy, potential energy and total energy of the electron in hydrogen (Z = 1), plus the energy-level diagram and line wavelengths.

Why are the energy levels negative?

Energy is measured from the free electron at rest at infinity (zero). A bound electron has less energy than this, so its energy is negative; 13.6/n² eV must be supplied to free it.

Are radioactivity and X-rays in this unit for 2027?

No. Radioactivity (decay law, half-life, mean life) and X-rays were removed from the 2027 syllabus.

How many marks do Atoms and Nuclei carry?

Together, 6 of the 70 theory marks.

Based on: CISCE ISC 2027 Physics syllabus (revised, uploaded February 2026); CISCE ISC 2027 Physics specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Physics, 2025; CISCE Analysis of Pupil Performance, ISC Physics, 2024