JEE Physics Nuclei MCQs Set 17

Answer: (c) 8 MeV

Explanation:
1. For most nuclei the binding energy per nucleon is between 7 and 9 MeV.
2. So it is about 8 MeV.

Teacher's Note:
1. The curve is nearly flat near 8 MeV for mass numbers 30 to 170.

Answer: (b) \( 10^{17} \) kg/m\( ^3 \)

Explanation:
1. Nuclear radius \( R=1.2\times10^{-15}A^{1/3} \) m, so the volume per nucleon is about \( \frac43\pi(1.2\times10^{-15})^3=7.2\times10^{-45} \) m\( ^3 \).
2. \( \rho=\frac{1.67\times10^{-27}}{7.2\times10^{-45}}\approx2.3\times10^{17} \) kg/m\( ^3 \).

Teacher's Note:
1. Nuclear density is independent of A.

Answer: (b) \( 1:2^{1/3} \)

Explanation:
1. Momentum conservation: \( m_1v_1=m_2v_2 \), so \( m_1:m_2=1:2 \) (the faster one is lighter).
2. Radius \( R\propto A^{1/3} \), so \( R_1:R_2=1:2^{1/3} \).

Teacher's Note:
1. Mass number is proportional to mass.

Answer: (d) a neutron

Explanation:
1. Charge: \( 2+Z=(Z+2)+q \) gives \( q=0 \). Mass number: \( 4+A=(A+3)+m \) gives \( m=1 \).
2. A particle of charge 0 and mass number 1 is a neutron.

Teacher's Note:
1. Balance charge and mass number separately.

Answer: (b) 23.6 MeV

Explanation:
1. Helium binding energy \( 4 imes7.0=28 \) MeV; two deuterons \( 2 imes2 imes1.1=4.4 \) MeV.
2. Energy released \( 28-4.4=23.6 \) MeV.

Teacher's Note:
1. Each deuteron has 2 nucleons.

Answer: (d) \( {}^{35}\text{Cl}+{}^{35}\text{Cl} \)

Explanation:
1. Fusion releases energy only when the product lies on the rising side of the binding energy curve, below mass number about 56.
2. Chlorine-35 pairs would form a nucleus of mass number 70, beyond the peak, so no net energy is released.

Teacher's Note:
1. Iron-56 region has the highest binding energy per nucleon.

Answer: (d) kinetic energy is high enough to overcome the Coulomb repulsion between nuclei

Explanation:
1. Nuclei are positively charged and repel each other.
2. They must be given large kinetic energy (temperature of about \( 10^7 \) K or more) to come within the range of the nuclear force.

Teacher's Note:
1. Fusion needs temperatures of millions of kelvin.

Answer: (d) mass-energy and momentum

Explanation:
1. Mass is converted to energy and vice versa, so mass alone and energy alone are not separately conserved.
2. Total mass-energy, linear momentum, charge and nucleon number are conserved.

Teacher's Note:
1. Mass-energy is the conserved quantity.

Answer: (a) a neutron

Explanation:
1. Charge: \( 1+8=9+q \) gives \( q=0 \). Mass number: \( 1+18=18+m \) gives \( m=1 \).
2. So a neutron is liberated.

Teacher's Note:
1. Check charge and mass number balance.

Answer: (a) 25 MeV

Explanation:
1. Fusion of light nuclei releases a few to about 25 MeV per reaction (e.g. 4 protons to He-4 gives 26.7 MeV, D-D about 24 MeV).
2. Fission releases about 200 MeV per nucleus.

Teacher's Note:
1. Fission yields about 200 MeV per event; fusion about 20 MeV.

Answer: (d) the nature of the element

Explanation:
1. Radioactive decay is a nuclear process with a fixed decay constant for each nuclide.
2. Half-life \( =\frac{0.693}{\lambda} \) is unaffected by amount, temperature or pressure.

Teacher's Note:
1. Only the nuclide decides its half-life.

Answer: (a) only one of alpha, beta or gamma at a time

Explanation:
1. A given decay event emits an alpha or a beta particle; gamma may follow when the daughter is excited.
2. A single nucleus does not emit the three kinds at the same time, so the correct choice is one at a time.

Teacher's Note:
1. Gamma rays accompany alpha or beta decay but arise in a later step.

Answer: (c) charged particles emitted by the nucleus

Explanation:
1. Beta rays are fast electrons (or positrons) created in the nucleus and emitted from it.
2. They are charged particles, not electromagnetic waves and not orbital electrons.

Teacher's Note:
1. Gamma rays are the electromagnetic ones.

Answer: (c) \( \frac1{\sqrt2} \)

Explanation:
1. \( \frac{N}{N_0}=\left(\frac12\right)^{t/T}=\left(\frac12\right)^{1/2} \).
2. This equals \( \frac1{\sqrt2} \).

Teacher's Note:
1. Use N = N0 (1/2)^(t/T).

Answer: (c) \( {}^{226}_{83}\text{X} \)

Explanation:
1. Three alpha emissions: \( A=238-12=226 \), \( Z=88-6=82 \).
2. One beta emission raises Z by 1: Z = 83, so the isotope is \( {}^{226}_{83}\text{X} \).

Teacher's Note:
1. Alpha: A - 4, Z - 2. Beta: Z + 1.

Answer: (d) 6 beta and 8 alpha

Explanation:
1. Alphas: \( (238-206)/4=8 \), lowering Z by 16 to 76.
2. Z must reach 82, so 6 beta particles are emitted.

Teacher's Note:
1. The uranium series ends in lead-206.

Answer: (b) 30 minutes

Explanation:
1. \( \frac1{16}=\left(\frac12\right)^4 \), so 2 hours is 4 half-lives.
2. Half-life = 120/4 = 30 minutes.

Teacher's Note:
1. 1/16 means four half-lives.

Answer: (a) Beta rays are the same as cathode rays

Explanation:
1. Both beta rays and cathode rays are streams of electrons, differing only in origin.
2. Gamma rays are photons, alpha particles are doubly ionised helium nuclei, and the neutron is slightly heavier than the proton.

Teacher's Note:
1. Check each statement against its definition.

Answer: (b) carbon

Explanation:
1. Two alpha particles remove 4 protons and 8 nucleons.
2. The remaining nucleus has Z = 6 and A = 14, i.e. carbon-14.

Teacher's Note:
1. Z = 6 identifies carbon.

Answer: (a) A and C are isotopes

Explanation:
1. After the alpha emission, Z drops by 2, and the two beta (electron) emissions raise Z by 2.
2. So C has the same Z as A but mass number smaller by 4: A and C are isotopes.

Teacher's Note:
1. Isotopes share Z; isobars share A.

Answer: (c) \( {}^A_Z X\to{}^A_Z X+f \)

Explanation:
1. Gamma decay leaves Z and A unchanged; an excited nucleus drops to a lower state and emits a photon f.
2. The other processes change Z or A.

Teacher's Note:
1. Gamma emission changes only the energy state.

Answer: (c) radiotracer technique

Explanation:
1. A radioactive isotope used to follow a process in the body is a radiotracer.
2. The radiotracer technique is the diagnostic tool described.

Teacher's Note:
1. PET and nuclear imaging rely on tracers.

Answer: (a) \( 10^9 \) K

Explanation:
1. Thermal energy must be comparable with the barrier: \( kT\sim7.7\times10^{-14} \) J, giving \( T\sim\frac{7.7\times10^{-14}}{1.38\times10^{-23}}=5.6\times10^{9} \) K.
2. Of the options, \( 10^9 \) K is the order of magnitude (the average-energy estimate with 3kT/2 gives a few times \( 10^9 \) K).

Teacher's Note:
1. Equate kT (or 3kT/2) to the Coulomb barrier.

Answer: (c) 6 fermi

Explanation:
1. \( R\propto A^{1/3} \): \( \frac{R_{Te}}{3.6}=\left(\frac{125}{27}\right)^{1/3}=\frac53 \).
2. So \( R_{Te}=6 \) fermi.

Teacher's Note:
1. 27 and 125 are perfect cubes, giving 3 and 5.

Answer: (d) after any time

Explanation:
1. Radioactive decay is a random process; the half-life is a statistical measure for a large number of nuclei.
2. The next decay can happen at any time.

Teacher's Note:
1. Decay times of individual nuclei are unpredictable.

Answer: (b) \( \frac{\log_e2}{\lambda}\text{ and }\frac1\lambda \)

Explanation:
1. Half-life \( T_{1/2}=\frac{\ln2}{\lambda} \) and mean life \( \tau=\frac1\lambda \).

Teacher's Note:
1. Mean life = 1.44 x half-life.

Answer: (b) \( {}^3_2\text{He}+{}^3_2\text{He}\to{}^4_2\text{He}+{}^1_1\text{H}+{}^1_1\text{H} \)

Explanation:
1. The Sun's energy comes from fusion of hydrogen to helium (the proton-proton chain).
2. The He-3 + He-3 reaction is the last step of that chain.

Teacher's Note:
1. Stars burn light nuclei by fusion.