Atomic and Nuclear physics — Practice Quiz
Tamil Nadu Board · Class 12 · Physics
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Quick Quiz: Atomic and Nuclear physics
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In J.J. Thomson's experiment, the electric field E = 3 × 10⁴ V/m and the magnetic field B = 2 × 10⁻³ T are applied simultaneously, causing no deflection of the electron beam. When only the electric field is applied, the electron beam deflects. What is the specific charge (e/m) of the electron if the accelerating potential V = 1.5 × 10³ V?
In Rutherford's alpha scattering experiment, an alpha particle of kinetic energy 5 MeV is directed towards a gold nucleus (Z = 79). What is the distance of closest approach? (1 MeV = 1.6 × 10⁻¹³ J, e = 1.6 × 10⁻¹⁹ C, k = 9 × 10⁹ N m² C⁻²)
According to Bohr's model, the total energy of an electron in the nth orbit of hydrogen atom is En = –13.6/n² eV. Which of the following correctly explains why the total energy is NEGATIVE?
For a hydrogen-like atom, the ground state energy is –54.4 eV. An electron in this atom jumps from n = 4 to n = 2. What is the wavelength of the emitted photon? (h = 6.6 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)
Sample Questions
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?
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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
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)
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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.
A radioactive sample initially contains N₀ nuclei. After 3 half-lives, what fraction of the original nuclei have DECAYED (not remaining)?
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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.
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)
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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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