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Chapter 8 of 11
Important Questions

Atomic and Nuclear physics

Tamil Nadu Board · Class 12 · Physics

Most important questions from Atomic and Nuclear physics for Tamil Nadu Board Class 12 Physics board exam 2026. MCQs, short answer, and long answer questions with marks.

44 questions25 flashcards5 concepts

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44 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

The impact parameter in Rutherford's scattering experiment is related to scattering angle θ by b = K cot(θ/2). If the impact parameter is doubled, what happens to the scattering angle?

Show answer

The scattering angle decreases but not by exactly half, since the relationship is through cot(θ/2), not a linear one.

Step 1: The relation is b = K cot(θ/2), which means cot(θ/2) = b/K. Step 2: If b doubles to 2b, then cot(θ'/2) = 2b/K = 2 cot(θ/2). Step 3: Since cot is not a linear function, θ'/2 ≠ θ/2 / 2. The new angle θ' is smaller than θ, but not exactly half. For example, if θ = 90°, cot(45°) = 1. Doubling b gives cot(θ'/2) = 2, so θ'/2 = arccot(2) ≈ 26.6°, θ' ≈ 53.1°, which is NOT half of 90°. Step 4: This shows the non-linear inverse relationship. Option A is wrong (angle increases with smaller impact parameter). Option B oversimplifies the nonlinear relation. Option D is completely wrong — they are s

2multiple choice
1 marks

Calculate the binding energy per nucleon of ⁵⁶Fe nucleus. Given: atomic mass of ⁵⁶Fe = 55.9349 u, mass of proton = 1.00728 u, mass of neutron = 1.00867 u, mass of electron = 0.00055 u. (1 u = 931 MeV)

Show answer

8.8 MeV

Step 1: For ⁵⁶Fe, Z = 26, N = 56 – 26 = 30. Step 2: Total mass of constituents = Z×m_H + N×m_n = 26×1.00783 + 30×1.00867 (using hydrogen atom mass = proton + electron = 1.00728 + 0.00055 = 1.00783 u). = 26.2036 + 30.2601 = 56.4637 u. Step 3: Mass defect Δm = 56.4637 – 55.9349 = 0.5288 u. Step 4: Total BE = 0.5288 × 931 = 492.3 MeV. Step 5: BE per nucleon = 492.3/56 ≈ 8.8 MeV. This is the maximum value on the BE curve, confirming iron is the most stable nucleus. Option A (7.5 MeV) represents lighter elements, Option C is too low, Option D exceeds the maximum known value on the curve.

3multiple choice
1 marks

A radioactive sample initially contains N₀ nuclei. After 3 half-lives, what fraction of the original nuclei have DECAYED (not remaining)?

Show answer

7/8

Step 1: After n half-lives, the number of nuclei remaining is N = (1/2)ⁿ × N₀. Step 2: After 3 half-lives: N remaining = (1/2)³ × N₀ = N₀/8. Step 3: Number decayed = N₀ – N₀/8 = 7N₀/8. Step 4: Fraction decayed = 7/8. This is a very common conceptual trap — students often confuse 'remaining' with 'decayed'. Option A (1/8) is the fraction REMAINING, not decayed. Option C (3/8) has no physical basis. Option D (5/8) would mean only 2 half-lives worth of decay, not 3.

4multiple choice
1 marks

In Millikan's oil drop experiment, an oil drop of radius r = 1.5 × 10⁻⁶ m and density ρ = 900 kg/m³ is held stationary in an electric field E = 5 × 10⁴ V/m. Density of air σ = 1.2 kg/m³, g = 10 m/s². How many elementary charges does the drop carry? (e = 1.6 × 10⁻¹⁹ C)

Show answer

4

Step 1: For a stationary drop: qE = (4/3)πr³(ρ – σ)g. Step 2: q = (4/3)πr³(ρ – σ)g / E. Step 3: q = (4/3) × 3.14 × (1.5×10⁻⁶)³ × (900 – 1.2) × 10 / (5×10⁴). Step 4: Volume = (4/3)π(1.5×10⁻⁶)³ = 4.19 × (3.375×10⁻¹⁸) = 1.414×10⁻¹⁷ m³. Step 5: q = (1.414×10⁻¹⁷ × 898.8 × 10) / (5×10⁴) = (1.271×10⁻¹³) / (5×10⁴) = 2.54×10⁻¹⁸ / (5×10⁴)... Correcting: q = 1.414×10⁻¹⁷ × 8988 / 5×10⁴ = 1.271×10⁻¹³ / 5×10⁴ ≈ 6.4×10⁻¹⁹ C. Number of charges n = q/e = 6.4×10⁻¹⁹ / 1.6×10⁻¹⁹ = 4. Options A, B, D give wrong integer values arising from arithmetic errors in volume calculation or density difference.

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Frequently Asked Questions

What are the important topics in Atomic and Nuclear physics for Tamil Nadu Board Class 12 Physics?
Key topics in Atomic and Nuclear physics include Process of gas discharge at decreasing pressures showing emergence of cathode rays, Sequence of Thomson's experiment showing force balance and deflection measurement, Millikan experiment procedure showing charge quantization discovery. These are the concepts Tamil Nadu Board Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Atomic and Nuclear physics — Tamil Nadu Board Class 12 Physics?
Understand the core concepts first, then work through the 44 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
How many important questions are there in Atomic and Nuclear physics?
There are 44 practice questions available for Atomic and Nuclear physics. These cover multiple question types including MCQs, short answer, and long answer questions.

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