CBSE Class 12 Physics Nuclei MCQs Set 07

Answer: (b) 30

Explanation:
1. Neutron number N = A − Z
2. N = 56 − 26 = 30

Teacher's Note:
1. The lower number is Z (protons), the upper number is A (nucleons).
2. 82 is A + Z, a common slip.

Answer: (c) isotones

Explanation:
1. For ¹⁴₆C: N = 14 − 6 = 8. For ¹⁶₈O: N = 16 − 8 = 8.
2. Z and A are different but N is the same, so they are isotones.

Teacher's Note:
1. Isotopes share Z, isobars share A, isotones share N.
2. Always work out N before deciding.

Answer: (a) 3 : 4

Explanation:
1. R = R₀ A^1/3, so \( \frac{R_1}{R_2} = \left(\frac{27}{64}\right)^{1/3} \)
2. \( \frac{R_1}{R_2} = \frac{3}{4} \), i.e. 3 : 4

Teacher's Note:
1. 27 = 3³ and 64 = 4³, so the cube roots are easy.
2. 27 : 64 is the ratio of the volumes, not the radii.

Answer: (d) 6.0 fm

Explanation:
1. R = R₀ A^1/3 = 1.2 × (125)^1/3
2. R = 1.2 × 5 = 6.0 fm

Teacher's Note:
1. 150 fm comes from multiplying by A instead of A^1/3.
2. 1 fm = 10⁻¹⁵ m, so R = 6 × 10⁻¹⁵ m.

Answer: (b) A⁰ (independent of A)

Explanation:
1. Mass ∝ A and volume = \( \frac{4}{3}\pi R_0^3 A \) ∝ A.
2. Density = mass/volume, so A cancels: density is independent of A.

Teacher's Note:
1. All nuclei have roughly the same density, about 2.3 × 10¹⁷ kg m⁻³.
2. This follows directly from R ∝ A^1/3.

Answer: (a) 9 × 10¹³ J

Explanation:
1. E = mc² = 10⁻³ × (3 × 10⁸)²
2. E = 9 × 10¹³ J

Teacher's Note:
1. Convert grams to kilograms first.
2. 9 × 10¹⁶ J is the energy equivalent of 1 kg.

Answer: (c) 28.3 MeV

Explanation:
1. ΔM = 2(1.007825) + 2(1.008665) − 4.002603 = 0.030377 u
2. E_b = 0.030377 × 931.5 ≈ 28.3 MeV

Teacher's Note:
1. 7.07 MeV is the binding energy per nucleon, not the total.
2. Using hydrogen-atom masses makes the electron masses cancel.

Answer: (d) 56

Explanation:
1. The E_bn curve rises steeply for light nuclei, peaks and then falls slowly.
2. The peak, about 8.75 MeV, is near A = 56 (iron), the most stable nuclei.

Teacher's Note:
1. Nuclei lighter than iron can gain stability by fusion, heavier ones by fission.
2. ⁴He is unusually stable for its size, but its E_bn (7.07 MeV) is not the maximum.

Answer: (c) It acts only between protons.

Explanation:
1. The nuclear force acts between any two nucleons: n–n, p–p and n–p.
2. It is charge independent, so the statement that it acts only between protons is not correct.

Teacher's Note:
1. Neutrons have no charge, yet they are held in the nucleus by the nuclear force.
2. The repulsive core below about 0.8 fm stops the nucleus collapsing.

Answer: (a) 1 × 10¹²

Explanation:
1. Energy per fission = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J
2. Fissions per second = \( \frac{32}{3.2 \times 10^{-11}} \) = 1 × 10¹²

Teacher's Note:
1. Power is energy per second, so divide power by energy per event.
2. Convert MeV to joules before dividing.

Answer: (b) the fragments have a larger binding energy per nucleon

Explanation:
1. A nucleus with A ≈ 240 has E_bn ≈ 7.6 MeV; the middle-mass fragments have E_bn ≈ 8.5 MeV.
2. The products are more tightly bound, so the difference in binding energy is released. Answer: the fragments have a larger binding energy per nucleon.

Teacher's Note:
1. The total number of nucleons is conserved in fission.
2. Higher binding energy means lower total mass of the products.

Answer: (d) the nuclei must overcome their Coulomb repulsion

Explanation:
1. Both nuclei are positively charged and repel each other.
2. Only nuclei with very large kinetic energy (very high temperature) can come within the range of the nuclear force. Answer: to overcome the Coulomb repulsion.

Teacher's Note:
1. This is why it is called thermonuclear fusion.
2. The core of the Sun is at about 1.5 × 10⁷ K.

Answer: (a) P

Explanation:
1. Force F = −dU/dr. At P (r < r₀) U falls as r increases, so F is positive, i.e. pushes the nucleons apart.
2. At Q, U is minimum and F = 0; at R, U rises with r, so F is attractive.
3. Answer: P

Teacher's Note:
1. The minimum of U is at about 0.8 fm, the equilibrium separation.
2. A negative slope of U means a repulsive force.

Answer: (b) 3.27 MeV

Explanation:
1. Mass of reactants = 2 × 2.014102 = 4.028204 u; mass of products = 3.016029 + 1.008665 = 4.024694 u
2. Δm = 0.003510 u, so Q = 0.003510 × 931.5 ≈ 3.27 MeV

Teacher's Note:
1. Q is positive because the products are lighter than the reactants.
2. 17.6 MeV is the D–T reaction, not D–D.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Volume ∝ R³ ∝ A and mass ∝ A, so density does not depend on A. The Assertion is correct.
2. The Reason is the radius formula and is correct.
3. Because volume and mass are both proportional to A, the Reason explains the Assertion.

Teacher's Note:
1. Nuclear density is about 2.3 × 10¹⁷ kg m⁻³.
2. Atomic densities vary a lot, nuclear densities do not.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The mass defect ΔM is always positive for a stable nucleus. The Assertion is correct.
2. Energy E_b = ΔM c² is released when the nucleus is formed. The Reason is correct.
3. The released energy carries away mass ΔM, which explains the Assertion.

Teacher's Note:
1. Mass defect and binding energy are linked by E = mc².
2. 1 u of mass defect corresponds to 931.5 MeV.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Fusing nuclei must overcome their Coulomb repulsion, which needs temperatures of about 10⁷ K. The Assertion is correct.
2. Hydrogen is indeed the most abundant element in the Sun, so the Reason is correct.
3. The abundance of hydrogen does not explain why high temperature is needed, so R is not the correct explanation.

Teacher's Note:
1. The correct explanation would be the Coulomb barrier between positive nuclei.
2. Choose (b) when R is true but on a different point.

Answer: (c) A is true, but R is false.

Explanation:
1. Fusion of light nuclei such as deuterium releases energy (about 3.27 MeV for D + D → ³He + n). The Assertion is correct.
2. E_bn of deuterium is only about 1.1 MeV, much less than that of the products. So the Reason is false.

Teacher's Note:
1. Energy is released whenever the products are more tightly bound.
2. Check the E_bn curve: it rises steeply at low A.

Answer: (d) A is false, but R is true.

Explanation:
1. m_n = 1.008665 u and m_p = 1.007276 u, so the neutron is slightly heavier. The Assertion is false.
2. Both are nucleons and both feel the nuclear force. The Reason is true.

Teacher's Note:
1. The masses are close but not equal.
2. The nuclear force is charge independent, which is why it treats both alike.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. E_bn falls from about 8.75 MeV near A = 56 to about 7.6 MeV for A = 238. The Assertion is correct.
2. The Coulomb repulsion acts between all pairs of protons and grows with Z, while the nuclear attraction saturates. The Reason is correct.
3. This growing repulsion is the cause of the fall, so R explains A.

Teacher's Note:
1. The nuclear force is short-ranged, but the Coulomb force is long-ranged.
2. This is why very heavy nuclei can release energy by fission.

Answer: (b) 0.1370 u

Explanation:
1. ΔM = 8 m_H + 8 m_n − M = 8(1.007825) + 8(1.008665) − 15.994915
2. ΔM = 8.062600 + 8.069320 − 15.994915 = 0.1370 u

Teacher's Note:
1. Oxygen-16 has 8 protons and 8 neutrons.
2. 0.0304 u is the mass defect of helium-4.

Answer: (d) 127.6 MeV

Explanation:
1. E_b = ΔM × 931.5 = 0.137005 × 931.5
2. E_b ≈ 127.6 MeV

Teacher's Note:
1. Multiply the mass defect in u by 931.5 to get MeV directly.
2. This is the energy needed to pull the nucleus completely apart.

Answer: (a) 7.98 MeV

Explanation:
1. E_bn = \( \frac{E_b}{A} = \frac{127.6}{16} \)
2. E_bn ≈ 7.98 MeV

Teacher's Note:
1. Divide by A (16), not by Z (8).
2. 8.79 MeV is the peak value near iron.

Answer: (c) 3.0 fm

Explanation:
1. R = R₀ A^1/3 = 1.2 × 16^1/3 = 1.2 × 2.52
2. R ≈ 3.0 fm

Teacher's Note:
1. 16^1/3 is a little more than 2.5.
2. 19.2 fm comes from multiplying by A instead of its cube root.

Answer: (b) ¹⁶O is more stable because its binding energy per nucleon is larger

Explanation:
1. Stability is judged by the binding energy per nucleon, not by the total.
2. E_bn(¹⁶O) ≈ 7.98 MeV > E_bn(⁴He) = 7.07 MeV, so ¹⁶O is more stable.

Teacher's Note:
1. A larger total binding energy alone does not mean greater stability.
2. Fusing helium into heavier nuclei like oxygen releases energy in older stars.

Answer: (c) 17.6 MeV

Explanation:
1. Mass of reactants = 2.014102 + 3.016049 = 5.030151 u; mass of products = 4.002603 + 1.008665 = 5.011268 u
2. Δm = 0.018883 u, so Q = 0.018883 × 931.5 ≈ 17.6 MeV

Teacher's Note:
1. Subtract products from reactants; a positive answer means energy is released.
2. 26.7 MeV is the overall energy of hydrogen burning in the Sun.

Answer: (b) 3.5 MeV

Explanation:
1. Number of nucleons in the reactants = 2 + 3 = 5
2. Energy per nucleon = \( \frac{17.6}{5} \) ≈ 3.5 MeV

Teacher's Note:
1. Fission of U-235 gives only about 200/236 ≈ 0.85 MeV per nucleon.
2. So fusion releases more energy per unit mass of fuel.

Answer: (a) have enough kinetic energy to overcome their Coulomb repulsion

Explanation:
1. Deuterium and tritium nuclei are both positively charged and repel each other.
2. Only at very high temperature do they move fast enough to come within the range of the nuclear force. Answer: to overcome their Coulomb repulsion.

Teacher's Note:
1. No material container can hold a plasma this hot, so magnetic fields are used.
2. This is called controlled thermonuclear fusion.

Answer: (d) 3.5 × 10¹⁷

Explanation:
1. Energy per reaction = 17.6 × 1.6 × 10⁻¹³ ≈ 2.82 × 10⁻¹² J
2. Reactions per second = \( \frac{10^6}{2.82 \times 10^{-12}} \) ≈ 3.5 × 10¹⁷

Teacher's Note:
1. 1 MW = 10⁶ J s⁻¹.
2. 2.8 × 10⁻¹² is the energy per reaction in joules, not the number of reactions.

Answer: (c) total number of nucleons and total charge

Explanation:
1. Left side: A = 2 + 3 = 5, Z = 1 + 1 = 2. Right side: A = 4 + 1 = 5, Z = 2 + 0 = 2.
2. Mass, kinetic energy and binding energy all change. Answer: total number of nucleons and total charge.

Teacher's Note:
1. Mass decreases and the lost mass appears as kinetic energy.
2. Balancing A and Z is the first check on any nuclear equation.