JEE Physics Nuclei MCQs Set 05

Answer: (a) M < (N·M_n + Z·M_p)

Explanation:
1. The nucleus is a bound system, and some mass is converted to binding energy
2. Hence the nuclear mass is less than the sum of the masses of its free nucleons (mass defect).

Teacher's Note:
1. Mass defect Δm = (N M_n + Z M_p) − M > 0.
2. Binding energy = Δm c².

Answer: (b) (a) and (b)

Explanation:
1. The fragments are more tightly bound, so their total binding energy exceeds that of the parent
2. The extra binding energy corresponds to a small mass defect converted to energy, so (a) and (b) are correct.

Teacher's Note:
1. Fission and fusion both increase the total binding energy.
2. The released energy equals the gain in binding energy.

Answer: (b) nuclear fission

Explanation:
1. An atom bomb is an uncontrolled chain reaction of fission of U-235 or Pu-239.

Teacher's Note:
1. A hydrogen bomb uses fusion, triggered by a fission bomb.
2. Super-critical mass gives K > 1 and an explosion.

Answer: (a) 3.125 × 10¹³

Explanation:
1. Energy per fission = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J
2. Number per second = 1000 / 3.2 × 10⁻¹¹ = 3.125 × 10¹³.

Teacher's Note:
1. Rate = power / energy per fission.
2. 1 MeV = 1.6 × 10⁻¹³ J.

Answer: (d) 931 MeV/c²

Explanation:
1. E = mc² = 1.66 × 10⁻²⁷ × (3 × 10⁸)² = 1.49 × 10⁻¹⁰ J = 931 MeV
2. So 1 u = 931 MeV/c².

Teacher's Note:
1. This conversion factor is used in every mass defect problem.
2. Remember 931.5 MeV/c² for more accuracy.

Answer: (a) 0.1%

Explanation:
1. The mass defect in fission is about 0.1% of the mass of the fissioned nucleus (about 200 MeV out of 235 × 931 MeV).

Teacher's Note:
1. Check: 200 / (235 × 931) ≈ 0.09%.
2. In fusion of hydrogen the figure is about 0.7%.

Answer: (a) light nuclei

Explanation:
1. Fusion of light nuclei increases the binding energy per nucleon and so releases energy
2. Middle elements are already near the maximum of the curve and heavy nuclei undergo fission instead.

Teacher's Note:
1. Fusion lies on the rising left-hand part of the binding energy curve.
2. Fission lies on the falling right-hand part.

Answer: (d) neutron

Explanation:
1. A free neutron decays into a proton, an electron and an anti-neutrino with a half-life of about 10 minutes
2. The other three are stable.

Teacher's Note:
1. A free proton cannot decay because it is lighter than a neutron.
2. Inside a nucleus, a neutron can be stable.

Answer: (a) gravitational force

Explanation:
1. The gravitational force is by far the weakest of the four fundamental forces.

Teacher's Note:
1. Order of strength: nuclear > electromagnetic > weak > gravitational.
2. Gravity matters only because masses are large.

Answer: (d) 10¹⁵

Explanation:
1. Atomic radius is about 10⁻¹⁰ m and nuclear radius about 10⁻¹⁵ m, a ratio of 10⁵
2. Volume scales as radius cubed, giving (10⁵)³ = 10¹⁵.

Teacher's Note:
1. Atoms are mostly empty space.
2. The volume ratio is the cube of the radius ratio.

Answer: (c) 28.4 MeV

Explanation:
1. Mass of 2 protons and 2 neutrons = 2(1.0073) + 2(1.0087) = 4.0320 u
2. Mass defect = 4.0320 − 4.0015 = 0.0305 u
3. Binding energy = 0.0305 × 931 = 28.4 MeV.

Teacher's Note:
1. Binding energy is expressed in MeV, not in u; u is a unit of mass.
2. Divide by 4 to get 7.1 MeV per nucleon.

Answer: (c) may be equal to its atomic number

Explanation:
1. A = Z + N, with N ≥ 0
2. For ordinary hydrogen (A = 1, Z = 1) there is no neutron, so A = Z there; for all others A > Z
3. So A may equal Z but is never less.

Teacher's Note:
1. A ≥ Z for every nucleus.
2. The equality holds only for the hydrogen nucleus (protium).

Answer: (c) an alpha particle

Explanation:
1. Mass number: 40 + A = 43 + 1, so A = 4
2. Charge: 20 + Z = 21 + 1, so Z = 2
3. A = 4, Z = 2 is an α-particle.

Teacher's Note:
1. Balance mass number and charge on both sides.
2. Such bombardment reactions are used to make new nuclides.

Answer: (a) only between light nuclei

Explanation:
1. Fusion releases energy only when the nuclei are light (mass number below about 56), so that the product has a higher binding energy per nucleon.

Teacher's Note:
1. Fusion of heavy nuclei would absorb energy.
2. The Coulomb barrier is lowest for hydrogen isotopes.

Answer: (c) 10¹⁷ kg/m³

Explanation:
1. Density ≈ 1.66 × 10⁻²⁷ A / [(4/3)π(1.2 × 10⁻¹⁵)³ A] ≈ 2.3 × 10¹⁷ kg/m³.

Teacher's Note:
1. This is the same for all nuclei.
2. A teaspoon of nuclear matter would weigh about a billion tonnes.

Answer: (b) m₃ < (m₁ + m₂)

Explanation:
1. The product is more tightly bound, so some mass is converted into released energy
2. Hence m₃ is less than m₁ + m₂.

Teacher's Note:
1. Mass defect in fusion = (m₁ + m₂) − m₃ > 0.
2. Energy released = Δm × 931 MeV.

Answer: (a) Z protons and A − Z neutrons

Explanation:
1. Z is the atomic number (number of protons) and A is the mass number (protons + neutrons), so neutrons = A − Z.

Teacher's Note:
1. The number of electrons in the neutral atom equals Z.
2. Isotopes have the same Z but different A.

Answer: (b) M(N, Z) = N·M_n + Z·M_p − B/c²

Explanation:
1. Binding energy B = Δm c² where Δm = (N M_n + Z M_p) − M
2. So M = N M_n + Z M_p − B/c².

Teacher's Note:
1. Dimension check: B/c² has the dimensions of mass, but Bc² does not.
2. The nucleus is lighter than its separate nucleons.

Answer: (b) 1.66 × 10⁻²⁷ kg

Explanation:
1. 931 MeV = 931 × 1.6 × 10⁻¹³ = 1.49 × 10⁻¹⁰ J
2. m = E/c² = 1.49 × 10⁻¹⁰ / 9 × 10¹⁶ = 1.66 × 10⁻²⁷ kg (one atomic mass unit).

Teacher's Note:
1. 931 MeV corresponds to exactly 1 u.
2. Use m = E/c² for any energy-to-mass conversion.

Answer: (d) 9 × 10¹⁰ J

Explanation:
1. m = 1 mg = 10⁻⁶ kg
2. E = mc² = 10⁻⁶ × 9 × 10¹⁶ = 9 × 10¹⁰ J.

Teacher's Note:
1. Convert milligram to kilogram first (1 mg = 10⁻⁶ kg).
2. This is enough energy to run a 1 MW power station for about a day.

Answer: (a) nuclear force > electric force

Explanation:
1. Nuclear force is strongest at nuclear distances (about 100 times the Coulomb repulsion), which is why the nucleus is stable
2. Gravity is much weaker than both.

Teacher's Note:
1. Heavy nuclei are unstable because the Coulomb force is long-range while the nuclear force is short-range.
2. The nuclear force is attractive between 0.8 fm and about 3 fm.

Answer: (a) graph (a)

Explanation:
1. B_N rises steeply for light nuclei, reaches a maximum of about 8.8 MeV at A = 56 (iron) and then falls slowly for heavy nuclei
2. Only graph (a) has this shape with the peak at A = 56.

Teacher's Note:
1. The other graphs have peaks or minima at the wrong place or the wrong shape.
2. The maximum near iron explains why iron is the most stable nucleus.

Answer: (a) \({}^{74}_{34}\mathrm{Se}\) and \({}^{71}_{31}\mathrm{Ga}\)

Explanation:
1. Isotones have the same number of neutrons
2. Se-74 has 74 − 34 = 40 neutrons and Ga-71 has 71 − 31 = 40 neutrons
3. The other pairs: isotopes, isobars, and 20 and 16 neutrons respectively.

Teacher's Note:
1. Isotopes: same Z. Isobars: same A. Isotones: same N.
2. Always compute N = A − Z for each nucleus.

Answer: (c) \({}^{239}_{93}\mathrm{Np}\) → \({}^{239}_{94}\mathrm{Pu}\) + β⁻ + anti-neutrino

Explanation:
1. Check A and Z on both sides
2. First: A 14 → 14 but Z 7 → 8, no. Second: A 24 → 26, no. Third: A 239 → 239 and Z 93 → 94 − 1 = 93, yes. Fourth: Z 8 → 7 (6 + 1), no.

Teacher's Note:
1. A possible nuclear reaction conserves both mass number and charge.
2. The third is an ordinary β⁻ decay.

Answer: (c) c − a − b

Explanation:
1. Energy released = (total binding energy of products) − (total binding energy of reactants)
2. The neutron has no binding energy, so Q = c − (a + b).

Teacher's Note:
1. Binding energy released = gain in binding energy.
2. Free nucleons have zero binding energy.

Answer: (d) less than 1

Explanation:
1. Some mass is converted into released energy, so the products weigh less than the parent.

Teacher's Note:
1. The mass defect is only about 0.1%, so the ratio is just under 1.
2. Mass and energy together are conserved.