Previous year questions: Atoms

Board pattern for this chapter

The Atoms chapter appears in board papers as one-mark MCQs on orbit quantities, two-mark calculations on spectral lines and energy levels, and three-mark questions on the scattering experiment or a set of orbit quantities. All ten questions below are rewritten from actual CBSE board papers between 2022 and 2024. The marks in each heading tell you how long the answer should be, and the year tells you which paper to open for the original.

Q1 (CBSE Board 2022, 2 marks)

Define distance of closest approach and impact parameter.

Answer: The distance of closest approach is measured at the moment a head-on alpha particle stops and turns back; there the entire kinetic energy has been converted into electrical potential energy, so the centre-to-centre gap at that moment is d. The impact parameter is the perpendicular distance from the nucleus centre to the initial line of motion of the alpha particle. A small impact parameter gives a large deflection, and a large one gives a small deflection.

Q2 (CBSE Board 2022, 2 marks)

For an alpha particle scattered by a nucleus, what is the value of the impact parameter when the scattering angle is (i) 0 degrees and (ii) 180 degrees?

Answer: (i) At 0 degrees the alpha particle passes far from the nucleus without deviation, so the impact parameter is very large, approaching infinity. (ii) At 180 degrees the particle makes a head-on approach and turns back, which happens when the impact parameter is zero.

Q3 (CBSE Board 2022, 3 marks)

The energy of a hydrogen atom in the first excited state is -3.4 eV. Taking the Bohr radius as 0.53 Å, find (a) the radius of this orbit, (b) the angular momentum of the electron in this orbit, and (c) the kinetic and potential energy of the electron.

Answer: The first excited state is n = 2. (a) r n = n squared a0, so r2 = 4 × 0.53 = 2.12 Å. (b) The angular momentum is n h/2π, so L = 2h/2π = h/π, about 2.1 × 10 to the -34 J s. (c) K = -E = 3.4 eV and U = 2E = -6.8 eV, so the electron has 3.4 eV of kinetic energy and -6.8 eV of potential energy, summing to the given -3.4 eV.

Q4 (CBSE Board 2023, 1 mark)

In the Bohr model of the hydrogen atom, the radius of the nth orbit is proportional to

(A) 1/n (B) 1/n squared (C) n (D) n squared

Answer: Option (D), n squared. The relation is r n = n squared a0, so doubling n makes the orbit four times larger and tripling it makes it nine times larger.

Q5 (CBSE Board 2023, 2 marks)

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

Answer: The third excited state is n = 4. Its energy is E4 = -13.6/16 = -0.85 eV. For a bound electron in this Coulomb field K = -E and U = 2E, so K = 0.85 eV and U = -1.7 eV. The sum, -0.85 eV, matches the total energy.

Q6 (CBSE Board 2023, 2 marks)

Calculate the wavelength of the second line of the Lyman series of hydrogen. Take R = 1.1 × 10 to the 7 per metre.

Answer: The Lyman series has n f = 1. The first line is the n = 2 to n = 1 transition and the second is n = 3 to n = 1. Using 1/λ = R(1/1 squared - 1/3 squared) = R(1 - 1/9) = 8R/9, λ = 9/(8R) = 9/(8 × 1.1 × 10 to the 7), which is about 1.02 × 10 to the -7 m or 102 nm. This lies in the ultraviolet region.

Q7 (CBSE Board 2023, 2 marks)

The ground state energy of hydrogen in the Bohr model is -13.6 eV. An electron makes a transition from the level at -1.51 eV to the level at -3.4 eV. Calculate the wavelength of the emitted spectral line.

Answer: The photon energy is the difference, E = -1.51 - (-3.4) = 1.89 eV. Using 1240/λ in nm = E in eV gives λ = 1240/1.89, which is about 656 nm. The transition is from n = 3 to n = 2, so this is the first line of the Balmer series, the red line of hydrogen.

Q8 (CBSE Board 2024, 1 mark)

The transition of an electron that produces the second spectral line of the Balmer series in the hydrogen spectrum 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 = 1 and n i = 4

Answer: Option (C), n f = 2 and n i = 4. Balmer lines end on n = 2; the first line is n = 3 to n = 2 and the second is n = 4 to n = 2, with wavelengths related by 1/λ = R(1/4 - 1/16) = 3R/16.

Q9 (CBSE Board 2024, 2 marks)

In the Bohr model of the hydrogen atom, find the ratio of the period of revolution of the electron in the orbit n = 2 to that in the orbit n = 1.

Answer: The period is T = 2π r n/v n. Since r n goes as n squared and v n goes as 1/n, T goes as n cubed. Therefore T2/T1 = 2 cubed/1 cubed = 8, so the period in the second orbit is eight times the period in the first orbit.

Q10 (CBSE Board 2024, 2 marks)

Energy levels A, B and C of an atom correspond to increasing energies, E A less than E B less than E C. Radiation of wavelengths λ1, λ2 and λ3 corresponds to the transitions C to B, B to A and C to A respectively. Obtain the relation between the three wavelengths.

Answer: The energy differences satisfy (E C - E A) = (E C - E B) + (E B - E A). Using E = hc/λ for each transition gives hc/λ3 = hc/λ1 + hc/λ2, so 1/λ3 = 1/λ1 + 1/λ2. Equivalently λ3 = λ1 λ2/(λ1 + λ2), so the direct transition has the shortest wavelength of the three.

What the board rewards

This chapter rewards careful formula use. For orbit numericals, write the three relations and mark which quantity the question wants before substituting. If the question gives a level energy, the radius and speed still come from n alone, so find n first. For spectral line questions, identify n f from the series name and n i from the ordinal of the line, then use 1/λ = R(1/n f squared - 1/n i squared) or 1240/λ in nm. Write the series letter first, because examiners look for it even before the number. For the scattering experiment, name the observation and the inference in the same sentence; a description of the apparatus alone earns little. In three-mark scattering answers, the marks usually sit in the inferences: most of the atom is empty, the nucleus is tiny, and the nucleus holds nearly all the mass and positive charge.

Two quick checks before the exam

First, write the energy ladder from memory: -13.6, -3.40, -1.51, -0.85 eV for n = 1 to 4, then 0 at infinity. Second, convert the gaps to wavelengths using 1240/energy and check that they land where the series should be: the n = 2 to 1 gap of 10.2 eV gives about 122 nm, in the ultraviolet; the n = 3 to 2 gap gives 656 nm, in the red. If those two numbers feel familiar, most spectral numericals become a matter of substitution. Practise the two routes side by side once, and then use whichever the data in the question makes shorter. Keep the sign convention straight as well: level energies are negative, photon energies are positive, and the two must not be mixed in one equation. A final check is the series name: any wavelength between roughly 90 and 400 nm belongs to Lyman, and anything near 656 nm is the first Balmer line.