CBSE · Class 12 · Physics

Class 12 Nuclei Mock Test (CBSE)

5 practice questions from Nuclei, in the style of your board paper, each with a worked answer and the marks for every step. Then take a full chapter test on paper, snap the pages and get every step marked.

Topics in Nuclei

  • Atomic masses and composition of nucleus
  • Size of the nucleus
  • Mass-energy relation
  • Nuclear binding energy
  • Nuclear force
  • Nuclear energy
  • Nuclear fission
  • Nuclear fusion

Nuclei practice questions with answers

  1. Q1Multiple choice1 mark
    Which of the following quantities serves as a direct measure of the stability of a nucleus?
    1. (a)Total binding energy
    2. (b)Mass number
    3. (c)Binding energy per nucleon
    4. (d)Neutron number
    Show answer and step marks
    1. Correct option: (c). Binding energy per nucleon determines how tightly nucleons are bound together, making it a direct measure of nuclear stability.+1
    Answer: (c)
  2. Q2Multiple choice1 mark
    Which of the following characteristics is true for nuclear forces between nucleons?
    1. (a)They are long-range electrostatic forces.
    2. (b)They depend strongly on the electric charge of the nucleons.
    3. (c)They are charge-independent and strongly attractive at short distances.
    4. (d)They obey the inverse-square law.
    Show answer and step marks
    1. Correct option: (c). Nuclear forces are essentially charge-independent and are strongly attractive, acting over very short ranges without following an inverse-square law.+1
    Answer: (c)
  3. Q3Multiple choice1 mark
    The radius of a nucleus with mass number AA is given by R=R0A1/3R = R_0 A^{1/3}. If the mass number of a nucleus increases from 8 to 64, the ratio of the nuclear radii of the two nuclei is:
    1. (a)1:81:8
    2. (b)1:41:4
    3. (c)1:21:2
    4. (d)2:12:1
    Show answer and step marks
    1. Correct option: (c) 1:21:2. Since R∝A1/3R \propto A^{1/3}, the ratio is (8/64)1/3=(1/8)1/3=1/2(8/64)^{1/3} = (1/8)^{1/3} = 1/2.+1
    Answer: (c)
  4. Q4Short answer2 marks
    Calculate the energy equivalent of 1 u1\text{ u} of atomic mass in MeV using the mass-energy relation.
    Show answer and step marks
    1. Use E=mc2E = mc^2 with m=1 u=1.6605×10−27 kgm = 1\text{ u} = 1.6605 \times 10^{-27}\text{ kg} and c=2.9979×108 m/sc = 2.9979 \times 10^8\text{ m/s}: E=1.4924×10−10 JE = 1.4924 \times 10^{-10}\text{ J}.+1
    2. Convert to MeV using 1 MeV=1.6022×10−13 J1\text{ MeV} = 1.6022 \times 10^{-13}\text{ J}: E≈931.5 MeVE \approx 931.5\text{ MeV} (rounded constants 1.66×10−271.66 \times 10^{-27} kg, 3×1083 \times 10^8 m/s and 1.6×10−131.6 \times 10^{-13} J give about 934934 MeV, also accepted).+1
    Answer: ≈931.5 MeV\approx 931.5\text{ MeV} (about 934934 MeV with rounded constants)
  5. Q5Short answer3 marks
    Calculate the binding energy per nucleon (in MeV\text{MeV}) of a 816O{}^{16}_{8}\text{O} nucleus. Given: atomic mass of 816O=15.9949 u{}^{16}_{8}\text{O} = 15.9949\text{ u}, mass of a hydrogen atom mH=1.00783 um_H = 1.00783\text{ u}, mass of a neutron mn=1.00867 um_n = 1.00867\text{ u}, and 1 u=931.5 MeV1\text{ u} = 931.5\text{ MeV}.
    Show answer and step marks
    1. Identify the number of protons (Z=8Z = 8) and neutrons (N=16−8=8N = 16 - 8 = 8) in the oxygen nucleus.+0.5
    2. Calculate the mass defect using atomic masses (electron masses cancel): Δm=[ZmH+Nmn]−M=[8(1.00783)+8(1.00867)]−15.9949=16.13200−15.9949=0.1371 u\Delta m = [Zm_H + Nm_n] - M = [8(1.00783) + 8(1.00867)] - 15.9949 = 16.13200 - 15.9949 = 0.1371\text{ u}.+1
    3. Calculate the total binding energy: BE=0.1371×931.5=127.71 MeVBE = 0.1371 \times 931.5 = 127.71\text{ MeV}.+1
    4. Binding energy per nucleon =127.7116=7.98 MeV/nucleon= \frac{127.71}{16} = 7.98\text{ MeV/nucleon}.+0.5
    Answer: 7.98 MeV/nucleon7.98\text{ MeV/nucleon}

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