CBSE Class 10 Physics Nuclear Physics MCQs Set 38

Answer: (a) β-rays

Explanation:
1. Specific charge is charge divided by mass. The β-particle (electron) has the smallest mass for a unit charge, so its specific charge, about \(1.76\times10^{11}\) C kg\(^{-1}\), is the largest.
2. Protons and α-particles are thousands of times heavier, and neutrons have no charge.
3. So the correct option is (a) β-rays.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (c) isotopes

Explanation:
1. Both have atomic number 7 (the same number of protons) but different mass numbers 14 and 15.
2. Atoms with the same atomic number and different mass numbers are isotopes.
3. So the correct option is (c) isotopes.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (b) P → B; Q → D; R → A; S → C

Explanation:
1. Radioactive sodium-24 is used to detect blood clots, so P matches B. Heavy water is used as a moderator in reactors, so Q matches D.
2. Iodine-131 is used to check the thyroid gland, so R matches A, and radio therapy uses hard X-rays or gamma rays, so S matches C.
3. So the correct option is (b) P → B; Q → D; R → A; S → C.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (b) B A D C

Explanation:
1. First note the data and find the numbers of protons and neutrons (B). Then find the total mass of the separate nucleons (A).
2. The mass defect is the total mass of the nucleons minus the actual mass (D), and finally the binding energy is 931.5 × Δm (C). The order is B A D C.
3. So the correct option is (b) B A D C.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (a) 58.4 MeV

Explanation:
1. \(\ce{^{9}_{4}Be}\) has 4 protons and 5 neutrons. Their total mass is \(4\times1.0078 + 5\times1.0087 = 9.0747\) u.
2. Mass defect = 9.0747 - 9.012 = 0.0627 u. Binding energy = 0.0627 × 931.5 = 58.4 MeV.
3. So the correct option is (a) 58.4 MeV.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (a) \(7.4\times10^{-44}\) m\(^3\)

Explanation:
1. Nuclear radius is proportional to \(A^{1/3}\), so when A becomes 8 times, the radius becomes 2 times: \(r = 2.6\times10^{-15}\) m.
2. Volume \(=\frac{4}{3}\pi r^3 = 4.19\times(2.6\times10^{-15})^3 = 7.4\times10^{-44}\) m\(^3\).
3. So the correct option is (a) \(7.4\times10^{-44}\) m\(^3\).

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (c) 60 days

Explanation:
1. If 7/8 has disintegrated, 1/8 is left. \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), so 3 half-lives have passed.
2. Time = 3 × 20 = 60 days.
3. So the correct option is (c) 60 days.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (c) neutrons

Explanation:
1. Specific charge is charge per unit mass. A neutron has no charge, so its specific charge is zero, the minimum.
2. The other three carry charge and have non-zero specific charge.
3. So the correct option is (c) neutrons.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (b) isobars

Explanation:
1. Both nuclei have the same mass number 31 but different atomic numbers 14 and 15.
2. Nuclei with the same mass number and different atomic numbers are isobars.
3. So the correct option is (b) isobars.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (c) A → S; B → P; C → Q; D → R

Explanation:
1. Binding energy per nucleon measures the stability of a nucleus (A to S). A reproduction factor of 1 means a controlled chain reaction (B to P).
2. Uranium dating is used to find the age of rocks (C to Q), and artificial radioactivity can emit positrons (D to R).
3. So the correct option is (c) A → S; B → P; C → Q; D → R.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (d) D B A C E

Explanation:
1. The model starts with the heavy nucleus as a charged liquid drop (D). A neutron is then absorbed (B), which disturbs the internal forces and starts oscillations (A).
2. The oscillations form a neck with two parts (C), and then the repulsive forces split the nucleus (E). The order is D B A C E.
3. So the correct option is (d) D B A C E.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (d) \(5.04\times10^{-29}\)

Explanation:
1. Helium-4 has 2 protons and 2 neutrons: \(2\times1.0078 + 2\times1.0087 = 4.0330\) u.
2. Mass defect = 4.0330 - 4.0026 = 0.0304 u = \(0.0304\times1.66\times10^{-27} = 5.04\times10^{-29}\) kg.
3. So the correct option is (d) \(5.04\times10^{-29}\).

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (d) 27A

Explanation:
1. Nuclear radius is proportional to \(A^{1/3}\), so \(A \propto r^3\).
2. If the radius becomes 3 times, A becomes \(3^3 = 27\) times, that is 27A.
3. So the correct option is (d) 27A.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (b) \(\frac{3}{4}\)

Explanation:
1. 3200 years is 2 half-lives, so the mass left is \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\) of the original.
2. The fraction disintegrated is \(1 - \frac{1}{4} = \frac{3}{4}\).
3. So the correct option is (b) \(\frac{3}{4}\).

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (c) A → S; B → R; C → P; D → Q

Explanation:
1. 1 amu is 1/12 of the mass of a carbon-12 atom (A to S), and the specific charge of the electron is \(1.7\times10^{11}\) C kg\(^{-1}\) (B to R).
2. The thorium series is the 4n series (C to P) and the uranium series is the (4n + 2) series (D to Q).
3. So the correct option is (c) A → S; B → R; C → P; D → Q.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (d) Both A and R are false.

Explanation:
1. A high binding energy means a lot of energy is needed to separate the nucleons, so separating them is difficult. A is false.
2. A higher binding energy means a more stable nucleus, so R is also false.
3. So the correct option is (d) Both A and R are false..

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (a) α, β, α, γ, β

Explanation:
1. P to Q: mass number falls by 4 and atomic number by 2, so an α-particle is emitted. Q to R: atomic number rises by 1 with the same mass number, so a β-particle is emitted.
2. R to S: mass number falls by 4 and atomic number by 2, so α. S to T: no change, so γ. T to U: atomic number rises by 1, so β. The order is α, β, α, γ, β.
3. So the correct option is (a) α, β, α, γ, β.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (d) A moderator is used to control the reaction.

Explanation:
1. A reactor uses uranium fuel, a controlled fission reaction takes place, and the products are radioactive.
2. The moderator only slows the neutrons down; the reaction is controlled by control rods such as cadmium. So option (d) is wrong.
3. So the correct option is (d) A moderator is used to control the reaction..

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (b) Its charge is equal in magnitude and sign to that of a proton.

Explanation:
1. A positron is the antiparticle of the electron: it has the same mass as an electron but a positive charge equal in magnitude to that of the proton.
2. So its charge is equal to that of a proton.
3. So the correct option is (b) Its charge is equal in magnitude and sign to that of a proton..

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (b) \(19.2\times10^{12}\) J

Explanation:
1. 235 g of uranium-235 is 1 mole, which has \(6.02\times10^{23}\) atoms.
2. Energy = \(6.02\times10^{23}\times200\) MeV \(= 1.2\times10^{26}\) MeV \(= 1.2\times10^{26}\times1.6\times10^{-13}\) J \(= 19.2\times10^{12}\) J.
3. So the correct option is (b) \(19.2\times10^{12}\) J.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (d) They have different kinetic energy and different momenta.

Explanation:
1. An α-particle is much heavier than a β-particle. Kinetic energy is \(\frac{1}{2}mv^2\) and momentum is mv.
2. With the same v and different masses, both the kinetic energy and the momentum are different.
3. So the correct option is (d) They have different kinetic energy and different momenta..

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (c) 3.125

Explanation:
1. 10 days is 5 half-lives, so the amount left is \(100\times\left(\frac{1}{2}\right)^5\).
2. This is 100/32 = 3.125 g.
3. So the correct option is (c) 3.125.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (a) It controls the nuclear reaction.

Explanation:
1. Cadmium absorbs neutrons strongly. The control rods of cadmium are moved in or out to absorb more or fewer neutrons.
2. So they control the rate of the nuclear reaction.
3. So the correct option is (a) It controls the nuclear reaction..

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.

Answer: (b) A B C

Explanation:
1. For A, 2 g to 0.25 g is a factor of 8 = \(2^3\), so 3 half-lives in 6 days: half-life 2 days.
2. For B, 50 g to 12.5 g is a factor of 4 = \(2^2\), so 2 half-lives in 6 days: half-life 3 days. For C, 200 g to 12.5 g is a factor of 16 = \(2^4\), so 4 half-lives in 16 days: half-life 4 days.
3. The increasing order is A, B, C.
4. So the correct option is (b) A B C.

Teacher's Note:
1. Write the given data in standard form and decide which law applies: half-life (amount left = original / 2 to the power n), binding energy (931.5 MeV for every u of mass defect), or nuclear notation (A = mass number, Z = atomic number). For match and sequence questions, fix the pairs or steps you are sure about first.