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Class 12 · Physics · Chapter 12

Atoms: Class 12 PYQs 2016–2026 with solutions

All 38 CBSE board questions asked from Atoms in 17 board papers, with the year, set and marks of each. 19 are solved free below. The Class 12 Physics PYQ Book: Solved Board Papers 2016-2026 (CBSE) has every solution, all chapters and 17 full board papers.

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Atoms: board weightage

Board weightage 2016–2026

Year	2016	2017	2018	2019	2020	2022	2023	2024	2025	2026
Questions	2	1	1	5	4	4	6	5	5	5
Marks	4	2	3	11	6	10	10	9	6	7

Most asked topics

Distance of closest approach2

Bohr energy levels (A–R)1

Spacing of Bohr orbits1

Ratio of emitted wavelengths1

Bohr's second postulate; levels getting closer1

Energy of an orbiting electron (A–R)1

Time period of revolution ∝ n³1

Hydrogen lines on zinc1

38 questions from 17 papers. Repeated questions are shown once with every year they were asked.

Jump to: 1 mark 2 marks 3 marks

1 mark: MCQ and Assertion–Reason

Q1 · CBSE 2026 · 55/1/1 · Q81 mark

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)

2 d

(C)

\frac{d}{4}

(D)

4 d

Solution

Answer: (D) 4d

At closest approach all kinetic energy becomes potential energy: $\frac{1}{2} m v^{2} = \frac{k ( 2 e ) ( Z e )}{d}$ , so $d \propto \frac{1}{v ^{2}}$ .

Halving v makes $v^{2}$ one-fourth, so $d^{'} = 4 d$ .

Q2 · CBSE 2026 · 55/1/1 · Q161 mark

Assertion (A): In Bohr model of hydrogen atom, the energy levels are discrete and quantised.
Reason (R): In a hydrogen atom, the electrostatic force on the electron provides the necessary centripetal force to it to revolve around the nucleus.

(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Solution

Answer: (B) Both A and R are true, but R is not the correct explanation of A.

The energy levels are discrete because of Bohr's quantum condition $m v r = \frac{nh}{2 π}$ . The electrostatic force acting as centripetal force is true, but on its own it allows any orbit; it does not explain quantisation.

Q3 · CBSE 2026 · 55/2/1 · Q41 mark

In Bohr model of hydrogen atom, for large values of n, the distance between the consecutive orbits is proportional to

(A)

n

(B)

n

(C)

n^{2}

(D)

n^{3}

Solution

Answer: (B) n

$r_{n} \propto n^{2}$ , so $r_{n + 1} - r_{n} \propto (n + 1)^{2} - n^{2} = 2 n + 1 \approx 2 n$ for large n. The spacing is proportional to n.

Q4 · CBSE 2026 · 55/2/1 · Q121 mark

In Bohr model of hydrogen atom, the electron makes a transition from n = 5 to n = 1 state. As a result, a photon of wavelength $λ$ is emitted in the process. The wavelength of the photon emitted when an electron makes a transition from energy level n = 5 to n = 2 will be

(A)

\frac{8}{7} λ

(B)

\frac{24}{7} λ

(C)

\frac{16}{7} λ

(D)

\frac{32}{7} λ

Solution

Answer: (D) $\frac{32}{7} λ$

$\frac{1}{λ} = R (\frac{1}{1 ^{2}} - \frac{1}{5 ^{2}}) = \frac{24 R}{25}$ and $\frac{1}{λ ^{'}} = R (\frac{1}{2 ^{2}} - \frac{1}{5 ^{2}}) = \frac{21 R}{100}$ .

$\frac{λ ^{'}}{λ} = \frac{24/25}{21/100} = \frac{96}{21} = \frac{32}{7}$

Q5 · CBSE 2025 · 55/1/1 · Q161 mark

Assertion (A): The potential energy of an electron revolving in any stationary orbit in a hydrogen atom is positive.
Reason (R): The total energy of a charged particle is always positive.

(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Solution

Answer: (D) Both A and R are false.

The electron and the proton attract, so the potential energy $- \frac{k e ^{2}}{r}$ is negative. The total energy of a bound electron, $- \frac{k e ^{2}}{2 r}$ , is also negative.

Q6 · CBSE 2025 · 55/2/1 · Q101 mark

Atomic spectral emission lines of hydrogen atom are incident on a zinc surface. The lines which can emit photoelectrons from the surface are members of

(A) Balmer series(B) Paschen series(C) Lyman series(D) Neither Balmer, nor Paschen nor Lyman series

Solution

Answer: (C) Lyman series

The work function of zinc is about 4.3 eV. Lyman lines (transitions to n = 1) have energies from 10.2 eV to 13.6 eV, which exceed it. Balmer photons (at most 3.4 eV) and Paschen photons (at most 1.5 eV) are too weak.

Q7 · CBSE 2025 · 55/2/1 · Q111 mark

The energy of an electron in a hydrogen atom in ground state is –13.6 eV. Its energy in an orbit corresponding to quantum number n is –0.544 eV. The value of n is

(A) 2(B) 3(C) 4(D) 5

Solution

Answer: (D) 5

$E_{n} = - \frac{13.6}{n ^{2}}$ : $n^{2} = \frac{13.6}{0.544} = 25$ , so $n = 5$ .

Q8 · CBSE 2025 · 55/2/1 · Q141 mark

Assertion (A): In Bohr model of hydrogen atom, the angular momentum of an electron in $n^{t h}$ orbit is proportional to the square root of its orbit radius $r_{n}$ .
Reason (R): According to Bohr model, electron can jump to its nearest orbits only.

(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Solution

Answer: (C) A is true, but R is false.

$L = \frac{nh}{2 π} \propto n$ and $r_{n} \propto n^{2}$ , so $L \propto r_{n}$ : A is true. The electron can jump between any two allowed orbits, not just neighbouring ones, so R is false.

Q9 · CBSE 2024 · 55/1/1 · Q91 mark

The transition of electron that gives rise to the formation of the second spectral line of the Balmer series in the spectrum of hydrogen atom corresponds to

(A)

n_{f} = 2

and

n_{i} = 3

(B)

n_{f} = 3

and

n_{i} = 4

(C)

n_{f} = 2

and

n_{i} = 4

(D)

n_{f} = 2

and

n_{i} = \infty

Full step-by-step solution is in the PYQ book (₹99).

Q10 · CBSE 2024 · 55/2/1 · Q91 mark

Energy levels A, B and C of an atom correspond to increasing values of energy i.e. $E_{A} < E_{B} < E_{C}$ . Let $λ_{1}$ , $λ_{2}$ and $λ_{3}$ be the wavelengths of radiation corresponding to the transitions C to B, B to A and C to A, respectively. The correct relation between $λ_{1}$ , $λ_{2}$ and $λ_{3}$ is

(A)

λ_{1}^{2} + λ_{2}^{2} = λ_{3}^{2}

(B)

\frac{1}{λ _{1}} + \frac{1}{λ _{2}} = \frac{1}{λ _{3}}

(C)

λ_{1} + λ_{2} + λ_{3} = 0

(D)

λ_{1} + λ_{2} = λ_{3}

Full step-by-step solution is in the PYQ book (₹99).

Q11 · CBSE 2024 · 55/2/1 · Q101 mark

An alpha particle approaches a gold nucleus in Geiger-Marsden experiment with kinetic energy K. It momentarily stops at a distance d from the nucleus and reverses its direction. Then d is proportional to

(A)

\frac{1}{K}

(B)

K

(C)

\frac{1}{K}

(D) K

Full step-by-step solution is in the PYQ book (₹99).

Q12 · CBSE 2023 · 55/1/1 · Q101 mark

The radius of the $n^{t h}$ orbit in Bohr model of hydrogen atom is proportional to

(a)

\frac{1}{n ^{2}}

(b)

\frac{1}{n}

(c)

n^{2}

(d) n

Full step-by-step solution is in the PYQ book (₹99).

Q13 · CBSE 2023 · 55/1/1 · Q111 mark

Hydrogen atom initially in the ground state, absorbs a photon which excites it to n = 5 level. The wavelength of the photon is

(a) 975 nm(b) 740 nm(c) 523 nm(d) 95 nm

Full step-by-step solution is in the PYQ book (₹99).

Q14 · CBSE 2023 · 55/2/1 · Q111 mark

A hydrogen atom makes a transition from n = 5 to n = 1 orbit. The wavelength of photon emitted is λ. The wavelength of photon emitted when it makes a transition from n = 5 to n = 2 orbit is

(a)

\frac{8}{7} λ

(b)

\frac{16}{7} λ

(c)

\frac{24}{7} λ

(d)

\frac{32}{7} λ

Full step-by-step solution is in the PYQ book (₹99).

Q15 · CBSE 2020 · 55/1/1 · Q91 mark

Which of the following statements is not correct according to Rutherford model?

(A) Most of the space inside an atom is empty.(B) The electrons revolve around the nucleus under the influence of coulomb force acting on them.(C) Most part of the mass of the atom and its positive charge are concentrated at its centre.(D) The stability of atom was established by the model.

Full step-by-step solution is in the PYQ book (₹99).

Q16 · CBSE 2020 · 55/1/1 · Q14(a)1 mark

Fill in the blank: According to Bohr's atomic model, the circumference of the electron orbit is always an ________ multiple of de Broglie wavelength.

Full step-by-step solution is in the PYQ book (₹99).

2 marks: very short answer

Q17 · CBSE 2025 · 55/1/1 · Q202 marks

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

Solution

$T = \frac{2 π r}{v}$ …(1)

Bohr's condition $m v r = \frac{nh}{2 π}$ gives $v = \frac{nh}{2 π m r}$ …(2)

From (1) and (2): $T = \frac{2 π r \cdot 2 π m r}{nh} = \frac{4 π ^{2} m r ^{2}}{nh}$ .

In Bohr's model $r_{n} = \frac{n ^{2} h ^{2}}{4 π ^{2} mk e ^{2}}$ , so $r^{2} \propto n^{4}$ :

$T \propto \frac{n ^{4}}{n} = n^{3}$

(Alternatively: $r \propto n^{2}$ and $v \propto \frac{1}{n}$ , so $T = \frac{2 π r}{v} \propto n^{3}$ .)

Q18 · CBSE 2023 · 55/1/1 · Q232 marks

The ground state energy of hydrogen atom is –13.6 eV. What is the potential energy and kinetic energy of an electron in the third excited state?

Solution

Third excited state means n = 4. Total energy $E_{4} = - \frac{13.6}{16} = - 0.85$ eV.

In Bohr's model, KE = –E and PE = 2E:

Kinetic energy $= + 0.85$ eV; potential energy $= 2 (- 0.85) = - 1.70 eV$ .

Q19 · CBSE 2023 · 55/2/1 · Q23(a)2 marks

What is meant by ionisation energy? Write its value for hydrogen atom.

Solution

Ionisation energy is the minimum energy needed to remove an electron from the ground state of an atom to infinity (i.e. to make it free).

For hydrogen: $E_{\infty} - E_{1} = 0 - (- 13.6 eV) = 13.6 eV$ .

Q20 · CBSE 2022 · 55/1/1 · Term 2 · Q2(a)2 marks

Draw the graph showing the variation of the number (N) of scattered alpha particles with scattering angle (θ) in Geiger – Marsden experiment. Infer two conclusions from the graph.

Solution

Conclusions:

1. Most α-particles pass through the foil undeflected or with very small deflections, so most of the atom is empty space.

2. A very small number are deflected through large angles (some by more than 90°, even bouncing back). So the positive charge and almost all the mass of the atom are concentrated in a tiny central region, the nucleus.

Q21 · CBSE 2022 · 55/2/1 · Term 2 · Q2(a)2 marks

(i) Define the terms: 'impact parameter' and 'distance of closest approach' for an α-particle in Geiger-Marsden scattering experiment.
(ii) What will be the value of the impact parameter for scattering angle (I) θ = 0° and (II) θ = 180°?

Solution

(i) Impact parameter (b): the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus.

Distance of closest approach ( $r_{0}$ ): the minimum distance from the nucleus that an α-particle moving head-on reaches, where its whole kinetic energy becomes potential energy and it momentarily stops.

(ii) (I) θ = 0° (no deflection): b is very large (b → ∞).

(II) θ = 180° (head-on collision, bounces back): b = 0.

Q22 · CBSE 2020 · 55/1/1 · Q26(a)2 marks

Using Bohr's atomic model, derive the expression for the radius of nth orbit of the revolving electron in a hydrogen atom.

Solution

Electrostatic attraction provides the centripetal force:

$\frac{m v ^{2}}{r} = \frac{1}{4 π ε _{0}} \cdot \frac{e ^{2}}{r ^{2}} \Rightarrow m v^{2} r = \frac{e ^{2}}{4 π ε _{0}} \dots (1)$

Quantisation of angular momentum: $m v r = \frac{nh}{2 π} \Rightarrow v = \frac{nh}{2 π m r}$ …(2)

Substituting (2) in (1):

$r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}}$

Q23 · CBSE 2020 · 55/2/1 · Q262 marks

In Geiger-Marsden scattering experiment, the trajectory of α-particles in Coulomb's field of a heavy nucleus is shown in the figure. (a) What do 'b' and 'θ' represent in the figure? (b) What will be the value of 'b' for (i) θ = 0°, and (ii) θ = 180°?

Solution

(a) b is the impact parameter: the perpendicular distance of the initial velocity of the α-particle from the centre of the nucleus. θ is the scattering angle: the angle through which the α-particle is deflected.

(b) (i) θ = 0°: b is very large (infinite). (ii) θ = 180°: b = 0 (head-on collision).

Q24 · CBSE 2019 · 55/1/1 · Q9(a)2 marks

State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong.

Full step-by-step solution is in the PYQ book (₹99).

Q25 · CBSE 2019 · 55/1/1 · Q9(b)2 marks

Calculate the orbital period of the electron in the first excited state of hydrogen atom.

Full step-by-step solution is in the PYQ book (₹99).

Q26 · CBSE 2019 · 55/2/1 · Q8(a)2 marks

A photon emitted during the de-excitation of electron from a state n to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2 eV, in a photo cell, with a stopping potential of 0.55 V. Obtain the value of the quantum number of the state n.

Full step-by-step solution is in the PYQ book (₹99).

Q27 · CBSE 2019 · 55/2/1 · Q8(b)2 marks

A hydrogen atom in the ground state is excited by an electron beam of 12.5 eV energy. Find out the maximum number of lines emitted by the atom from its excited state.

Full step-by-step solution is in the PYQ book (₹99).

Q28 · CBSE 2017 · 55/1 · Set 1 · Q92 marks

A 12.5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted.

Full step-by-step solution is in the PYQ book (₹99).

Q29 · CBSE 2016 · 55/1 · Set 1 · Q8(a)2 marks

When is $H_{α}$ line in the emission spectrum of hydrogen atom obtained? Calculate the frequency of the photon emitted during this transition.

Full step-by-step solution is in the PYQ book (₹99).

Q30 · CBSE 2016 · 55/1 · Set 1 · Q8(b)2 marks

Calculate the wavelength of radiation emitted when electron in a hydrogen atom jumps from n = ∞ to n = 1.

Full step-by-step solution is in the PYQ book (₹99).

3 marks: short answer

Q31 · CBSE 2026 · 55/2/1 · Q263 marks

(a) State Bohr's second postulate and mention its significance.
(b) Prove that, in Bohr model of hydrogen atom as principal quantum number n becomes large, the energy levels get closer and closer.

Solution

(a) The electron revolves only in those orbits for which its angular momentum is an integral multiple of $\frac{h}{2 π}$ : $m v r = \frac{nh}{2 π}$ , $n = 1, 2, 3, \dots$

Significance: angular momentum is quantised, so only certain stable (stationary) orbits are allowed, and the electron does not radiate energy in them.

(b) $E_{n} = - \frac{13.6}{n ^{2}}$ eV. For consecutive levels:

$Δ E = E_{n + 1} - E_{n} = 13.6 [\frac{1}{n ^{2}} - \frac{1}{( n + 1 ) ^{2}}] = 13.6 \frac{2 n + 1}{n ^{2} ( n + 1 ) ^{2}} eV$

For large n, $Δ E \approx \frac{13.6 \times 2 n}{n ^{4}} = \frac{27.2}{n ^{3}}$ eV, which decreases as n increases. So the energy levels get closer and closer.

Q32 · CBSE 2024 · 55/1/1 · Q273 marks

Write the drawbacks of Rutherford's atomic model. How did Bohr remove them? Show that different orbits in Bohr's atom are not equally spaced.

Solution

Drawbacks of Rutherford's model:

1. Stability: an electron revolving in an orbit is accelerated and, by classical theory, must radiate energy continuously, so it should spiral into the nucleus and the atom would collapse.

2. Spectrum: as the electron spirals in, its frequency of revolution changes continuously, so the atom should emit a continuous spectrum; real atoms emit line spectra.

Bohr's remedy: (i) electrons revolve only in certain stationary orbits in which they do not radiate; (ii) in these orbits angular momentum is quantised, $m v r = \frac{nh}{2 π}$ ; (iii) radiation is emitted or absorbed only when the electron jumps between orbits, with $h ν = E_{i} - E_{f}$ , giving discrete lines.

Spacing: $r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}} = n^{2} r_{1}$ . So $r_{1}, r_{2}, r_{3}, \dots = r_{1}, 4 r_{1}, 9 r_{1}, \dots$ and the gaps are $3 r_{1}, 5 r_{1}, 7 r_{1}, \dots$ , i.e. $r_{n + 1} - r_{n} = (2 n + 1) r_{1}$ . The gap grows with n, so the orbits are not equally spaced.

Q33 · CBSE 2024 · 55/2/1 · Q263 marks

Using Bohr's postulates, derive the expression for the radius of the $n^{t h}$ orbit of an electron in a hydrogen atom. Also find the numerical value of Bohr's radius $a_{0}$ .

Solution

The electrostatic attraction provides the centripetal force:

$\frac{m v ^{2}}{r} = \frac{1}{4 π ε _{0}} \cdot \frac{e ^{2}}{r ^{2}} \Rightarrow m v^{2} r = \frac{e ^{2}}{4 π ε _{0}} \dots (1)$

Bohr's quantisation condition: $m v r = \frac{nh}{2 π} \Rightarrow v = \frac{nh}{2 π m r}$ …(2)

Substituting (2) in (1): $m \cdot \frac{n ^{2} h ^{2}}{4 π ^{2} m ^{2} r ^{2}} \cdot r = \frac{e ^{2}}{4 π ε _{0}}$

$r_{n} = \frac{n ^{2} h ^{2} ε _{0}}{π m e ^{2}}$

For n = 1 (Bohr radius):

$a_{0} = \frac{h ^{2} ε _{0}}{π m e ^{2}} = \frac{( 6.63 \times 1 0 ^{- 34} ) ^{2} \times 8.854 \times 1 0 ^{- 12}}{3.14 \times 9.1 \times 1 0 ^{- 31} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}} \approx 5.3 \times 1 0^{- 11} m = 0.53 \overset{˚}{A}$

Q34 · CBSE 2023 · 55/1/1 · Q30(a)3 marks

Draw the energy level diagram for hydrogen atom. Mark the transitions corresponding to the series lying in the ultraviolet region, visible region and infrared region.

Solution

$E_{n} = - \frac{13.6}{n ^{2}}$ eV.

Ultraviolet: Lyman series, transitions from n = 2, 3, 4, … to n = 1.

Visible: Balmer series, transitions to n = 2.

Infrared: Paschen series, transitions to n = 3 (Brackett and Pfund series, ending on n = 4 and 5, are also infrared).

Q35 · CBSE 2022 · 55/1/1 · Term 2 · Q43 marks

Define the term – Distance of closest approach. How will it be affected, for an α - particle, if kinetic energy of the particle is doubled?

Full step-by-step solution is in the PYQ book (₹99).

Q36 · CBSE 2022 · 55/2/1 · Term 2 · Q83 marks

The energy of a hydrogen atom in the first excited state is –3.4 eV. Find: (a) the radius of this orbit. (Take Bohr radius = 0.53 Å) (b) the angular momentum of the electron in the orbit. (c) the kinetic and potential energy of the electron in the orbit.

Full step-by-step solution is in the PYQ book (₹99).

Q37 · CBSE 2019 · 55/2/1 · Q193 marks

Draw a plot of α-particle scattering by a thin foil of gold to show the variation of the number of the scattered particles with scattering angle. Describe briefly how the large angle scattering explains the existence of the nucleus inside the atom. Explain with the help of impact parameter picture, how Rutherford scattering serves a powerful way to determine an upper limit on the size of the nucleus.

Full step-by-step solution is in the PYQ book (₹99).

Q38 · CBSE 2018 · 55/1 · Set 1 · Q183 marks

(a) State Bohr's postulate to define stable orbits in hydrogen atom. How does de Broglie's hypothesis explain the stability of these orbits? (b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon.

Full step-by-step solution is in the PYQ book (₹99).

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