Class 12 Physics Chapter 13: Nuclei worksheet with answer key
A 20-mark worksheet on Nuclei for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (a) 10⁻¹²
Volume ∝ r³, so the fraction is (10⁻⁴)³ = 10⁻¹²: the atom is almost empty, though the nucleus holds more than 99.9% of its mass.
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Answer: (b) 0.0304 u
Eb = ΔMc², so ΔM = 28.3 MeV ÷ 931.5 MeV/u = 0.0304 u. 0.0076 u is the mass defect per nucleon (÷ 4); 0.0152 u divides by 2; 0.304 u slips a decimal.
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Answer: (a) photons able to do this would need far more energy than the bombardment could supply
To knock protons out, photons would have needed far more energy than the α-bombardment could provide. Chadwick therefore proposed new neutral particles, neutrons, of mass very nearly that of the proton, which can transfer energy to protons efficiently in collisions.
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Answer: (d) 5626Fe
The binding energy per nucleon has its maximum, about 8.75 MeV, near A = 56; it is lower for light nuclei (⁴He about 7.1 MeV, ¹²C about 7.7 MeV) and for heavy ones (²³⁸U about 7.6 MeV).
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Answer: (d) 168O
Isotones have the same neutron number N = A − Z. For 146C, N = 8; for 168O, N = 8. 147N (N = 7) is an isobar of 146C, 136C (N = 7) an isotope, and 158O has N = 7.
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(i) Same A = 40, different Z: isobars. (ii) N = 13 − 6 = 7 and N = 14 − 7 = 7, different Z: isotones. (iii) Same Z = 17, different A: isotopes. (iv) Both have Z = 1 (deuterium and tritium), different A: isotopes.
- (i) isobars (0.5)
- (ii) isotones, with N = 7 for both (0.5)
- (iii) isotopes (0.5)
- (iv) isotopes (0.5)
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136C → 126C + 10n. Mass after − mass before = 12.000000 + 1.008665 − 13.003355 = 0.005310 u (the electron masses are the same on both sides). Energy needed = 0.005310 × 931.5 MeV = 4.95 MeV.
- mass difference 0.005310 u, final minus initial (1)
- energy needed = 4.95 MeV (1)
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1 g of uranium gives 10¹⁴ × 10⁻³ = 10¹¹ J. Mass of coal = 10¹¹/10⁷ = 10⁴ kg (10 tonnes). Chemical energies are of the order of eV per atom and nuclear energies of the order of MeV, so for the same quantity of matter nuclear sources give vastly more energy.
- 10¹¹ J from 1 g and 10⁴ kg of coal (1)
- nuclear (MeV) against chemical (eV) energies per reaction (1)
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(a) Protons and neutrons are conserved, so the energy to be supplied is the loss in total binding energy: 39.25 − (28.30 + 8.48) = 2.47 MeV. (b) ⁷Li: 39.25/7 = 5.61 MeV; ⁴He: 28.30/4 = 7.08 MeV. (c) The binding energy curve is not smooth for light nuclei: it shows a peak at ⁴He, which is unusually tightly bound, taken as evidence of shell structure in nuclei; ⁷Li does not sit on such a peak.
- 2.47 MeV from the difference in total binding energy (1)
- Ebn: ⁷Li ≈ 5.61 MeV, ⁴He ≈ 7.08 MeV (1)
- ⁴He sits on a peak of the curve (shell structure) (1)
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(a) Energy per fission = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J. Fissions per second = 5.0 × 10⁸/3.2 × 10⁻¹¹ = 1.56 × 10¹⁹ s⁻¹. (b) In one day (86 400 s): 1.56 × 10¹⁹ × 86 400 = 1.35 × 10²⁴ nuclei = 1.35 × 10²⁴/6.02 × 10²³ = 2.24 mol; mass = 2.24 × 235 g ≈ 527 g, about 0.53 kg.
- energy per fission in joules (0.5)
- fissions per second ≈ 1.6 × 10¹⁹ (1)
- nuclei per day ≈ 1.35 × 10²⁴ (0.5)
- mass ≈ 0.53 kg (1)
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23892U → 23490Th + 42He; 23490Th → 23491Pa + e⁻ + ν̄; 23491Pa → 23492U + e⁻ + ν̄. Each β⁻ decay raises Z by 1 and leaves A unchanged. The final nucleus has the same Z (92) and A = 234: it is an isotope of the original, 23492U.
- α step: A − 4, Z − 2 (1)
- two β⁻ steps: Z + 1 each, A unchanged (1)
- final 23492U, an isotope of the parent (1)
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