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CBSE · Class 12 · Physics

Atoms

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The radius of the 3rd Bohr orbit in a hydrogen atom is: (r1 = 0.53 Å for hydrogen)
  1. 14.31 Å
  2. 1.59 Å
  3. 9.54 Å
  4. 4.77 Å
2.
The energy of an Li2+ ion with its electron in the n = 5 level is:
  1. −4.896 eV
  2. −8.16 eV
  3. 4.896 eV
  4. −0.544 eV
3.
The energy of the photon emitted when a hydrogen electron falls from n = 3 to n = 2 is:
  1. 1.51 eV
  2. 2.27 eV
  3. 3.4 eV
  4. 1.89 eV
4.
The energy needed to remove the electron from a hydrogen atom in the n = 4 level is:
  1. 1.7 eV
  2. 13.6 eV
  3. 3.4 eV
  4. 0.85 eV
5.
In Rutherford's experiment, most α-particles passed straight through the foil, showing that:
  1. the nucleus is negative
  2. the atom is solid
  3. most of the atom is empty space
  4. electrons are heavy
6.
Hydrogen atoms are excited to the n = 6 level. The largest number of different spectral lines they can emit on returning to the ground state is:
  1. 15
  2. 36
  3. 5
  4. 6
7.
The line emitted when a hydrogen electron falls from n = 6 to n = 5 belongs to the:
  1. Balmer series (visible)
  2. Pfund series (infrared)
  3. Paschen series (infrared)
  4. Lyman series (ultraviolet)
8.
Assertion (A): The energy of a hydrogen atom in the n = 6 level is −0.38 eV.
Reason (R): In the Bohr model, the energy of level n is −13.6/n eV.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
For the electron in the n = 3 orbit of a hydrogen atom, find its kinetic energy, potential energy and total energy.
10.
Find the energy and wavelength of the photon emitted when a hydrogen electron falls from n = 7 to n = 3. (hc = 1240 eV nm)
11.
An α-particle of kinetic energy 2 MeV heads straight at a aluminium nucleus (Z = 13). Find its distance of closest approach. (ke2 = 1.44 MeV fm)
12.
Find the speed of the electron in the n = 5 orbit of an He+ ion, and the radius of that orbit. (for hydrogen v1 = 2.18 × 106 m/s, r1 = 0.53 Å)
13.
Find the ratio of the radii, and of the energies, of the n = 5 and n = 4 orbits of hydrogen.
14.
How much energy must a hydrogen atom absorb to lift its electron from n = 1 to n = 5? What is the angular momentum of the electron in the upper orbit?
15.
Find the longest and shortest wavelengths of the Pfund series (lines ending on n = 5) of He+ ion. (hc = 1240 eV nm)
16.
Hydrogen atoms in the ground state are lifted to n = 9. How many different lines can they emit? Which transition gives the longest wavelength?
Section C — Word problems4 × 4 = 16 marks
17.
A hydrogen atom with its electron in the n = 2 level absorbs a photon of 3.02 eV. To which level does the electron rise? List the photon energies it can emit as it falls back to n = 2.
18.
Find the energy and wavelength of the photon emitted in the n = 7 → n = 4 transition of hydrogen, name its series, and find the longest wavelength in that series. (hc = 1240 eV nm)
19.
α-particles of 6 MeV and 3 MeV are aimed head-on at gold nuclei (Z = 79). Find the distance of closest approach for each, and explain what this says about the size of the nucleus. (ke2 = 1.44 MeV fm)
20.
For the n = 2 orbit of a hydrogen atom, find the radius, the speed of the electron, its total energy, and the energy needed to remove the electron from there. (for hydrogen r1 = 0.53 Å, v1 = 2.18 × 106 m/s, E1 = −13.6 eV)
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Physics

Answer key

Atoms · AW-12P-ATM-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 4.77 Årn = n2r1/Z = 32 × 0.53/1
2(a) −4.896 eVEn = −13.6Z2/n2 eV with Z = 3
3(d) 1.89 eV13.6(1/22 − 1/32)
4(d) 0.85 eV13.6 × 12/42
5(c) most of the atom is empty space—
6(a) 15n(n − 1)/2
7(b) Pfund series (infrared)lines ending on n = 5 make the Pfund series
8(c) A is true, but R is false.A is true: En = −13.6/62 = −0.38 eV. R is false.
9KE = 1.51 eV; PE = −3.02 eV; total = −1.51 eVKE = −E, PE = 2E, E = −13.6Z2/n2
101.23 eV; 1,005.22 nmΔE = 13.6(1/32 − 1/72); λ = 1240/ΔE
1118.72 fmK = k(2e)(Ze)/d ⇒ d = 1.44 × 2 × 13/2
128.72 × 105 m/s; 6.625 Åvn = Zv1/n; rn = n2r1/Z
13radii 25 : 16; energies 16 : 25r ∝ n2, E ∝ 1/n2
1413.06 eV; 5h/(2π)ΔE = 13.6(1/12 − 1/52); L = nh/2π
15longest 1,864.97 nm; shortest 569.85 nmlongest: n = 6 → 5; shortest: n = ∞ → 5; energies × Z2 = 4
1636 lines; n = 9 → 8the smallest energy gap gives the longest wavelength
17n = 6; 3→2: 1.89 eV, 4→2: 2.55 eV, 5→2: 2.856 eV, 6→2: 3.02 eV, 4→3: 0.66 eV, 5→3: 0.97 eV, 6→3: 1.13 eV, 5→4: 0.306 eV, 6→4: 0.47 eV, 6→5: 0.17 eV−13.6/22 + 3.02 = −13.6/n2
180.57 eV; 2,166.13 nm; Brackett; longest 4,052.29 nmλ = 1240/ΔE
1937.92 fm and 75.84 fm; the nucleus must be smaller than these distances, so it is only a few tens of femtometres across at mostd = 2Zke2/K
202.12 Å; 1.09 × 106 m/s; −3.4 eV; 3.4 eVrn = n2r1/Z; vn = Zv1/n; En = Z2E1/n2