Atoms and Nuclei — Previous Year Questions (Class 12 Physics)
The study of atomic structure and nuclear physics reveals the nature of matter at the smallest scales. Understanding radioactivity and nuclear reactions is
TL;DR: The study of atomic structure and nuclear physics reveals the nature of matter at the smallest scales. Understanding radioactivity and nuclear reactio…
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The study of atomic structure and nuclear physics reveals the nature of matter at the smallest scales. Understanding radioactivity and nuclear reactions is
Atoms and Nuclei — Previous Year Questions with Solutions
Q (2023, 3 marks): State the postulates of Bohr's model of the atom. How does it explain the stability of atoms?
Answer: Bohr's postulates:
1. Electrons move in discrete circular orbits around the nucleus without radiating energy.
2. Only certain orbits are allowed, characterized by quantized angular momentum: L = mvr = n(h/2π), where n = 1, 2, 3...
3. Energy is emitted or absorbed only when an electron transitions between allowed orbits.
4. The centripetal force is provided by Coulomb's force between electron and nucleus.
Stability explanation:
1. Electrons in allowed orbits do not radiate energy (despite being accelerated).
2. The atom is stable in its ground state (lowest energy orbit).
3. Energy is only required to transition to a higher orbit (excitation).
4. This explains why atoms don't collapse and why spectral lines are discrete.
Q (2022, 3 marks): Derive the expression for the radius of the nth Bohr orbit in hydrogen. Calculate the radius of the first Bohr orbit.
Answer: For hydrogen, centripetal force = Coulomb force:
mv^2/r = ke^2/r^2
where k = 1/(4πε0) = 9 × 10^9 N·m^2/C^2
From Bohr's quantization: mvr = n(h/2π)
Solving these equations:
rn = n^2 × (h^2 × 4πε0) / (4π^2 × m × k × e^2)
rn = n^2 × a0
Where a0 = 0.53 Å (Bohr radius) = 0.53 × 10^-10 m
For n = 1 (first Bohr orbit):
r1 = a0 = 0.53 × 10^-10 m = 0.53 Å
Q (2023, 3 marks): Explain the concept of radioactivity. Distinguish between alpha, beta, and gamma decay.
Answer: Radioactivity: Spontaneous emission of radiation by unstable nuclei to achieve stability.
Alpha decay (α):
- Emission of alpha particles (He nucleus, ⁴He)
- AZX → (A-4)(Z-2)Y + ⁴₂He
- Example: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He
- Reduces both mass and atomic number
Beta decay (β):
- Emission of beta particles (electrons from neutron decay)
- AZX → A(Z+1)Y + e⁻ + anti-neutrino
- Example: ¹⁴₆C → ¹⁴₇N + e⁻ + anti-ν
- Increases atomic number by 1
Gamma decay (γ):
- Emission of high-energy photons (electromagnetic radiation)
- AZX → AZX + γ
- Does not change mass or atomic number
- Nucleus transitions to lower energy state
Q (2022, 2 marks): Define half-life. If a sample of radioactive material has a half-life of 10 years, what fraction remains after 30 years?
Answer: Half-life (t1/2): Time required for half of the original number of nuclei to decay.
Number of nuclei remaining: N(t) = N0 × (1/2)^(t/t1/2)
Given: t1/2 = 10 years, t = 30 years
Number of half-lives: n = 30/10 = 3
Fraction remaining: N/N0 = (1/2)^3 = 1/8
Therefore, 1/8 (or 12.5%) of the original sample remains after 30 years.
Q (2023, 3 marks): Explain the concept of mass defect and binding energy. Calculate the binding energy of deuterium (1 neutron + 1 proton).
Answer: Mass defect (Δm): Difference between the sum of masses of individual nucleons and actual nuclear mass.
Binding energy (BE): Energy released when nucleons combine to form a nucleus.
BE = Δm × c^2
For deuterium (²₁H):
Mass of proton = 1.007276 u
Mass of neutron = 1.008665 u
Mass of electron = 0.000549 u
Atomic mass of deuterium = 2.014102 u
Nuclear mass of deuterium = 2.014102 - 0.000549 = 2.013553 u
Mass defect: Δm = (1.007276 + 1.008665) - 2.013553 = 0.002388 u
Binding energy: BE = 0.002388 × 931.5 MeV/u ≈ 2.22 MeV
(where 1 u = 931.5 MeV/c^2)
Q (2021, 2 marks): Explain the difference between fission and fusion reactions. Give examples of each.
Answer: Fission:
- Heavy nucleus splits into lighter nuclei, releasing energy.
- Requires a neutron to initiate (chain reaction possible).
- Example: ²³⁵₉₂U + n → ⁹¹₃₆Kr + ¹⁴²₅₆Ba + 3n + 200 MeV
- Used in nuclear reactors and bombs.
- Produces radioactive waste.
Fusion:
- Light nuclei combine to form a heavier nucleus, releasing energy.
- Requires extremely high temperature and pressure.
- Example: ²₁H + ³₁H → ⁴₂He + n + 17.6 MeV (deuterium-tritium fusion)
- Powers the sun and stars.
- Cleaner energy source with no radioactive byproducts.
- Difficult to control on Earth.
Frequently Asked Questions
What is the nuclear force and how is it different from electromagnetic force?
The nuclear force (strong force) is the force that holds nucleons together in the nucleus. Unlike electromagnetic force which acts over large distances and can be attractive or repulsive, the nuclear force is: (1) extremely strong but acts only over very short distances (~10^-15 m), (2) always attractive, (3) independent of charge (acts on protons and neutrons equally), (4) responsible for nuclear binding.
Why do heavy nuclei undergo radioactive decay?
Heavy nuclei are unstable because the nuclear force (which is short-range) cannot hold an increasingly large number of nucleons together. The Coulomb repulsion between protons increases with nuclear size, eventually exceeding the binding strength. Radioactive decay reduces the nucleus size or Z, lowering its total energy and achieving a more stable configuration.
More Class 12 Physics PYQs
- Current Electricity
- Ray Optics
- Electrostatics
- Moving Charges and Magnetism
- Electromagnetic Induction
- Alternating Current
🤖 Stuck on any of these? Ask Syllab's free AI Tutor to explain step by step →