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Samacheer Kalvi 12th Physics Chapter 9 Exercise Solutions

Tamil Nadu Class 12 Physics Chapter 9 Atomic and Nuclear Physics exercise solutions with Bohr model, energy levels, radioactivity, binding energy, nuclear reactions, and numericals.

Updated 25 May 202617 min read

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Step-by-step answers for all textbook questions in this chapter

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Chapter 9 — Exercise Solutions

Atomic and nuclear physics combines model-based theory with formula numericals. Write the physical meaning of symbols in derivations.

How to Use This Page

  • Start with the Key Concepts table to quickly revise definitions before attempting questions.
  • Work through MCQs and Fill in the Blanks first — they test recall and are quick to score.
  • Use the Worked Examples as templates for numerical problems; follow the same step-by-step structure in the exam.
  • Read Common Mistakes to Avoid before every mock test — most lost marks come from sign errors and unit confusion.
  • Tick off items on the Exam Checklist the night before your exam to ensure nothing is missed.

Key Concepts (Quick Recap)

TermDefinition
Bohr modelAtomic model where electrons revolve in stationary orbits without radiating energy; angular momentum is quantised in units of h/2pi.
Energy levelsDiscrete allowed energies of an electron in an atom; given by En = -13.6/n² eV for hydrogen.
Spectral seriesGroups of spectral lines arising from transitions to a common lower level: Lyman (to n=1), Balmer (to n=2), Paschen (to n=3), Brackett (to n=4), Pfund (to n=5).
RadioactivitySpontaneous disintegration of unstable nuclei by emitting alpha, beta, or gamma radiation.
Alpha decayNuclear decay emitting an alpha particle (He-4 nucleus); atomic number decreases by 2, mass number decreases by 4.
Beta decayNuclear decay emitting a beta particle (electron or positron); a neutron converts to a proton (beta-minus) or vice versa (beta-plus).
Gamma decayEmission of high-energy photons from an excited nucleus; no change in atomic or mass number, only energy release.
Half-lifeTime taken for half the radioactive nuclei in a sample to decay; T₁/2 = 0.693 / lambda.
Decay constantProbability of decay per unit time; lambda = 0.693 / T₁/2.
Mass defectDifference between the sum of individual nucleon masses and the actual nuclear mass; Delta m = Zmp + (A-Z)mn - M.
Binding energyEnergy equivalent of the mass defect; the energy required to completely separate a nucleus into its constituent nucleons.
Nuclear fissionSplitting of a heavy nucleus (e.g., U-235) into lighter fragments, releasing large energy.
Nuclear fusionCombining of light nuclei (e.g., hydrogen isotopes) into a heavier nucleus, releasing even more energy per nucleon than fission.
IsotopesAtoms of the same element with the same atomic number but different mass numbers (different neutron count).
Nuclear forceStrong short-range attractive force between nucleons; charge-independent, spin-dependent, and responsible for nuclear stability.

Formula Recap

ConceptFormula
Photon energyE = h nu = hc / lambda
Bohr energyEn = -13.6 / n² eV
Decay lawN = N0 e^(-lambda t)
Half-lifeT1/2 = 0.693 / lambda
Mass defectDelta m = Zmp + (A-Z)mn - M
Binding energyBE = Delta m c²

Numerical Pattern

1. Identify atomic transition, decay, or nuclear energy problem. 2. Write the formula. 3. Convert energy units if required. 4. Substitute values carefully. 5. State whether energy is emitted or absorbed.

I. Choose the Best Answer (MCQs)

1. The energy of an electron in the ground state of hydrogen atom is -13.6 eV. What is its energy in the n = 3 state?

(a) -1.51 eV   (b) -3.4 eV   (c) -4.53 eV   (d) -0.85 eV

Answer(a) -1.51 eV — E₃ = -13.6 / 3² = -13.6 / 9 = -1.51 eV.

2. In alpha decay, the atomic number of the daughter nucleus changes by:

(a) +1   (b) -1   (c) -2   (d) No change

Answer(c) -2 — An alpha particle carries away 2 protons, so the atomic number decreases by 2.

3. The half-life of a radioactive substance is 10 years. Its decay constant (in year⁻¹) is approximately:

(a) 0.693   (b) 0.0693   (c) 6.93   (d) 0.00693

Answer(b) 0.0693 — lambda = 0.693 / T₁/2 = 0.693 / 10 = 0.0693 year⁻¹.

4. Which spectral series of hydrogen lies in the ultraviolet region?

(a) Balmer   (b) Paschen   (c) Lyman   (d) Brackett

Answer(c) Lyman — Transitions ending at n = 1 (Lyman series) produce ultraviolet photons.

5. The binding energy per nucleon is maximum for:

(a) Hydrogen   (b) Uranium   (c) Iron (Fe-56)   (d) Helium

Answer(c) Iron (Fe-56) — Fe-56 has the highest binding energy per nucleon (~8.8 MeV), making it the most stable nucleus.

6. In beta-minus decay, a neutron converts into:

(a) A proton and a positron   (b) A proton, an electron, and an antineutrino   (c) Two protons   (d) A proton and a neutrino

Answer(b) A proton, an electron, and an antineutrino — n → p + e⁻ + anti-nu.

7. The radius of the nth Bohr orbit is proportional to:

(a) n   (b) n²   (c) 1/n²   (d) 1/n

Answer(b) n² — rn = 0.529 × n²/Z angstroms.

8. Nuclear forces are:

(a) Long-range and charge-dependent   (b) Short-range and charge-independent   (c) Long-range and charge-independent   (d) Short-range and charge-dependent

Answer(b) Short-range and charge-independent — Nuclear force acts within ~2 fm and does not depend on the charge of nucleons.

9. The number of atoms present in 1 amu of a substance is related to:

(a) Planck constant   (b) Avogadro number   (c) Boltzmann constant   (d) Rydberg constant

Answer(b) Avogadro number — 1 amu = 1/N_A grams, where N_A is Avogadro's number.

10. In nuclear fusion, two deuterium nuclei combine to form:

(a) He-4   (b) He-3   (c) H-3   (d) Li-6

Answer(b) He-3 — D + D → He-3 + n (or He-3 + proton in an alternate branch); the primary product is He-3.

II. Fill in the Blanks

1. The energy of the electron in the first Bohr orbit of hydrogen is ______ eV.  
Answer-13.6
2. The half-life of a radioactive element is related to its decay constant by T₁/2 = ______.  
Answer0.693 / lambda
3. In alpha decay, the mass number of the daughter nucleus decreases by ______.  
Answer4
4. The spectral series corresponding to transitions ending at n = 2 is called the ______ series.  
AnswerBalmer
5. Binding energy per nucleon of Fe-56 is approximately ______ MeV.  
Answer8.8
6. The radius of the nth Bohr orbit is proportional to ______.  
Answern²
7. In beta-minus decay, the atomic number of the daughter nucleus ______ by 1.  
Answerincreases
8. The force that holds nucleons together inside the nucleus is called the ______ force.  
Answernuclear (or strong)

III. True or False (with Corrections)

1. Statement: The energy of an electron in a Bohr orbit is always positive.
AnswerFalse. Correction: The energy of an electron in a Bohr orbit is negative, indicating it is bound to the nucleus.
2. Statement: Gamma decay changes both the atomic number and mass number of the nucleus.
AnswerFalse. Correction: Gamma decay only releases energy; it changes neither the atomic number nor the mass number.
3. Statement: Half-life is independent of the initial amount of the radioactive substance.
AnswerTrue. Half-life is a property of the nuclide and does not depend on how much material is present.
4. Statement: In beta-plus decay, a proton converts into a neutron, emitting a positron and a neutrino.
AnswerTrue. p → n + e⁺ + nu is the beta-plus decay process.
5. Statement: Nuclear fission releases more energy per nucleon than nuclear fusion.
AnswerFalse. Correction: Nuclear fusion releases more energy per nucleon than fission because lighter nuclei have a steeper rise in binding energy per nucleon.
6. Statement: The Lyman series of hydrogen lies in the visible region.
AnswerFalse. Correction: The Lyman series lies in the ultraviolet region; the Balmer series is in the visible region.
7. Statement: Mass defect is the difference between the mass of the nucleus and the sum of masses of its free nucleons.
AnswerTrue. The mass defect equals (Zmp + (A-Z)mn) - M, where M is the actual nuclear mass.

IV. Match the Following

Column AColumn B
1. Lyman series(a) Neutron → Proton + Electron + Antineutrino
2. Alpha particle(b) Ultraviolet region
3. Beta-minus decay(c) Helium nucleus
4. Binding energy(d) 0.693 / lambda
5. Half-life(e) Delta m × c²
6. Paschen series(f) Infrared region
Answer1–b, 2–c, 3–a, 4–e, 5–d, 6–f

Short Answer Points

  • Bohr model explains line spectrum of hydrogen.
  • Radioactivity is spontaneous disintegration of unstable nuclei.
  • Alpha, beta, and gamma radiations differ in charge, mass, and penetration.
  • Binding energy per nucleon indicates nuclear stability.

V. Answer Briefly (2-mark)

1. What is the significance of the negative sign in the Bohr energy levels?

The negative sign indicates that the electron is bound to the nucleus and cannot leave the atom without receiving energy equal to or greater than the magnitude of the binding energy. When the energy becomes zero, the electron is free (ionised). The more negative the energy, the more tightly bound the electron is to the nucleus.

2. Distinguish between alpha and beta radiation.

Alpha radiation consists of helium nuclei (2 protons + 2 neutrons) with high ionising power and low penetration, stopped by paper. Beta radiation consists of fast-moving electrons (or positrons) with moderate ionising power and moderate penetration, stopped by aluminium foil. Alpha particles carry +2 charge and 4 amu mass, while beta particles carry -1 charge and negligible mass.

3. Define half-life and state its relation to the decay constant.

Half-life is the time required for half the radioactive nuclei in a sample to decay. It is related to the decay constant lambda by the formula T₁/2 = 0.693 / lambda. A larger decay constant means a shorter half-life, indicating faster decay. Half-life is characteristic of the nuclide and does not depend on the initial quantity.

4. Why is nuclear fusion difficult to achieve on Earth?

Nuclear fusion requires extremely high temperatures (about 10⁸ K) to overcome the Coulomb repulsion between positively charged nuclei so they can come within range of the nuclear force. Containing such a hot plasma and sustaining the reaction long enough for net energy gain is technologically challenging. Stars achieve this naturally due to their immense gravitational pressure.

5. State the law of radioactive decay.

The law of radioactive decay states that the rate of disintegration of a radioactive substance is directly proportional to the number of undecayed nuclei present at that instant: dN/dt = -lambda N. Solving this gives N = N₀ e^(-lambda t), where N₀ is the initial number of nuclei, lambda is the decay constant, and t is the time elapsed.

Worked Examples

Example 1 — Bohr energy level

Find the energy of the electron in the n = 2 orbit of hydrogen.

Eₙ = −13.6 / n² eV E₂ = −13.6 / 2² = −13.6 / 4 = −3.4 eV

Example 2 — Energy of emitted photon

Find the energy released when an electron jumps from n = 3 to n = 2 in hydrogen.

E₃ = −13.6/9 = −1.51 eV, E₂ = −13.6/4 = −3.4 eV ΔE = E₃ − E₂ = −1.51 − (−3.4) = 1.89 eV (photon emitted)

Example 3 — Radioactive decay (half-life)

A sample has 80 g of a radioactive element with half-life 5 days. How much remains after 15 days?

Number of half-lives = 15 / 5 = 3 Remaining = 80 × (1/2)³ = 80 × 1/8 = 10 g

VI. Answer in Detail (5-mark)

1. Describe the Bohr atom model and its postulates.

Niels Bohr proposed a model of the hydrogen atom based on quantum ideas to explain the observed line spectrum. The postulates are: (i) Electrons revolve around the nucleus in certain discrete stationary orbits without radiating energy. (ii) The angular momentum of the electron in a stationary orbit is quantised and is an integral multiple of h/2pi, i.e., mvr = nh/2pi where n = 1, 2, 3, ... (iii) When an electron jumps from a higher energy orbit to a lower energy orbit, it emits a photon of energy equal to the difference: h nu = E₂ - E₁. Conversely, an electron absorbs a photon of the same energy to jump to a higher orbit. (iv) The radius of the nth orbit is rn = n²a₀/Z where a₀ = 0.529 angstroms is the Bohr radius. (v) The energy of the electron in the nth orbit is En = -13.6 Z²/n² eV, where Z is the atomic number. The model successfully explained the hydrogen spectrum but could not explain spectra of multi-electron atoms, the Zeeman effect, or the fine structure of spectral lines.

2. Explain nuclear forces and their properties.

Nuclear force is the strong attractive force that binds protons and neutrons together in the nucleus despite the Coulomb repulsion between protons. Its key properties are: (i) It is a short-range force, effective only up to about 2-3 fm (1 fm = 10⁻¹⁵ m). (ii) It is charge-independent — the force between two protons, two neutrons, or a proton and a neutron is approximately the same. (iii) It is spin-dependent — nucleons with parallel spins experience a stronger attraction than those with anti-parallel spins. (iv) It is saturated — each nucleon interacts only with its nearest neighbours, not with all nucleons in the nucleus. (v) It is non-central, meaning it depends on the relative orientation of the nucleons, not just the distance between them. (vi) At very short distances (less than 0.5 fm), the nuclear force becomes repulsive, preventing the nucleus from collapsing. The exchange of mesons (primarily pi-mesons) between nucleons is the theoretical mechanism proposed by Yukawa to explain this force.

3. State and derive the radioactive decay law.

The radioactive decay law states that the number of undecayed nuclei in a radioactive sample decreases exponentially with time. Derivation: Let N be the number of undecayed nuclei at time t. The rate of disintegration is proportional to N, so dN/dt = -lambda N, where lambda is the decay constant. Rearranging: dN/N = -lambda dt. Integrating both sides from t = 0 (where N = N₀) to t, we get ln(N/N₀) = -lambda t, hence N = N₀ e^(-lambda t). This shows that the number of undecayed nuclei falls exponentially. The time for the number to reduce to half is the half-life: T₁/2 = ln 2 / lambda = 0.693 / lambda. The mean life (average lifetime of a nucleus) is tau = 1/lambda = T₁/2 / 0.693. Activity A = lambda N gives the disintegrations per second, measured in becquerel. The decay law is statistical — it applies to large ensembles of atoms, not to individual nuclei.

4. Compare nuclear fission and nuclear fusion.

Nuclear fission is the splitting of a heavy nucleus (like U-235 or Pu-239) into two lighter nuclei, accompanied by the release of neutrons and a large amount of energy (~200 MeV per fission). It occurs when a heavy nucleus absorbs a slow neutron and becomes unstable. Nuclear fusion is the combining of two light nuclei (like deuterium and tritium) into a heavier nucleus, releasing even more energy per nucleon (~17.6 MeV for D-T fusion). Key differences: (i) Fission involves heavy nuclei (A > 200); fusion involves light nuclei (A < 30). (ii) Fission can occur at room temperature with slow neutrons; fusion requires extremely high temperatures (~10⁸ K) to overcome Coulomb repulsion. (iii) Fission produces radioactive waste (fission fragments); fusion produces mainly helium, which is not radioactive. (iv) Fission is controllable and used in nuclear power reactors; fusion is not yet commercially viable for power generation. (v) Both release energy because the binding energy per nucleon increases toward Fe-56 — fission from the right of the peak, fusion from the left.

Common Mistakes to Avoid

  • ❌ Writing Bohr energy as +13.6/n² eV → ✅ Always include the negative sign: En = -13.6/n² eV, indicating a bound state.
  • ❌ Confusing half-life (T₁/2) with mean life (tau) → ✅ Remember T₁/2 = 0.693 × tau; mean life is longer than half-life.
  • ❌ Forgetting to convert amu to MeV (1 amu = 931.5 MeV) in binding energy numericals → ✅ Always multiply mass defect by 931.5 to get energy in MeV.
  • ❌ Mixing up alpha and beta decay changes in atomic/mass numbers → ✅ Alpha: Z-2, A-4; Beta-minus: Z+1, A unchanged; Beta-plus: Z-1, A unchanged.
  • ❌ Ignoring that Lyman series is UV, not visible → ✅ Lyman = UV (to n=1), Balmer = visible (to n=2), Paschen and beyond = IR.
  • ❌ Writing N = N₀ e^(+lambda t) instead of N = N₀ e^(-lambda t) → ✅ The decay law has a negative exponent because the number decreases over time.

Exam Checklist

  • Nuclear equations are balanced — atomic numbers and mass numbers are conserved on both sides.
  • Negative sign included in all Bohr energy level calculations.
  • Correct units used: eV for energy, fm for nuclear radius, amu or MeV/c² for mass.
  • 1 amu = 931.5 MeV conversion applied correctly in mass defect and binding energy problems.
  • Half-life formula (T₁/2 = 0.693/lambda) and decay law (N = N₀ e^(-lambda t)) memorised and applied.
  • Alpha, beta-minus, beta-plus, and gamma decay changes in Z and A are correctly written.
  • Spectral series names matched with their transition levels and spectral regions.
  • Binding energy per nucleon curve is sketched and interpreted — peak at Fe-56.
  • Distinguishing fission (heavy nucleus splits) from fusion (light nuclei combine).
  • All final numerical answers include proper units and a brief qualitative statement.

Additional Numerical Practice

Example 4 — Binding energy calculation

Calculate the binding energy of a helium-4 nucleus. Given: mass of He-4 = 4.002603 u, mass of proton = 1.007825 u, mass of neutron = 1.008665 u.

Mass defect: Delta m = (2 × 1.007825 + 2 × 1.008665) - 4.002603 = (2.015650 + 2.017330) - 4.002603 = 4.032980 - 4.002603 = 0.030377 u Binding energy = Delta m × 931.5 = 0.030377 × 931.5 ≈ 28.3 MeV Binding energy per nucleon = 28.3 / 4 ≈ 7.07 MeV

Example 5 — Activity calculation

A radioactive sample has a decay constant of 0.01 s⁻¹. If there are 10⁶ nuclei initially, find the activity at t = 0 and after 100 s.

Activity at t = 0: A₀ = lambda × N₀ = 0.01 × 10⁶ = 10⁴ disintegrations/s After t = 100 s: N = N₀ e^(-lambda t) = 10⁶ × e^(-0.01 × 100) = 10⁶ × e⁻¹ = 10⁶ × 0.368 = 3.68 × 10⁵ A = lambda × N = 0.01 × 3.68 × 10⁵ = 3.68 × 10³ disintegrations/s

Example 6 — Wavelength of spectral line

Find the wavelength of light emitted when an electron in hydrogen jumps from n = 3 to n = 2.

E₃ = -13.6 / 9 = -1.51 eV E₂ = -13.6 / 4 = -3.4 eV Delta E = E₃ - E₂ = -1.51 - (-3.4) = 1.89 eV lambda = hc / Delta E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (1.89 × 1.6 × 10⁻¹⁹) = 1.989 × 10⁻²⁵ / (3.024 × 10⁻¹⁹) = 6.57 × 10⁻⁷ m = 657 nm (visible, red — Balmer series)

Frequently Asked Questions

Bohr energy levels, photon energy transition, decay law, half-life, mass defect, and binding energy formulas are important.