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Class 12 Science (PCM)

Physics Part 2

Nuclei

Nuclei

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Atomic Mass Unit and Isotopes
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Atomic Structure and Nucleus
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Average Atomic Mass from Isotopic Abundance
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Balancing Nuclear Reactions
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Chain Reaction in Fission
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Controlled Thermonuclear Fusion
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Decay Constant, Half-Life, Mean Life and Activity
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Energy Release in Fission from Binding Curve
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Isotopes, Isobars and Isotones
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Mass Defect
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Mass-Energy Equivalence
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Neutron-Induced Fission of Uranium-235
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Nuclear Binding Energy
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Nuclear Composition and Nuclide Notation
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Nuclear Density Calculation
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Nuclear Fission and Fusion
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Nuclear Forces
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Nuclear Fusion and Coulomb Barrier
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Nuclear Radius Formula and Scaling
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Nuclear Size and Density
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Proton-Proton Cycle in Stars
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Proton, Neutron and Nucleon
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Q-Value of Nuclear Reactions
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Radioactive Decay Products and Nuclear Transformation
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Radioactivity
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Saturation Property of Nuclear Force
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Stability from Binding Energy per Nucleon
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Binding Energy per Nucleon Curve
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Discovery of Neutron
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Low

Learning Objectives

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Learning Objectives

  • Understand the structure and composition of atomic nuclei.
  • Explain the concepts of nuclear fission and fusion.
  • Calculate the binding energy of various nuclei.
  • Analyze the stability of nuclei based on neutron-to-proton ratios.
  • Describe the processes and implications of radioactive decay.
  • Evaluate the significance of binding energy per nucleon in nuclear reactions.
  • Discuss the challenges and potential of controlled thermonuclear fusion.

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Revision Notes & Summary

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Chapter Thirteen: Nuclei

13.1 Introduction

  • The nucleus is the center of an atom, containing most of its mass (over 99.9%).
  • The nucleus is much smaller than the atom, with dimensions smaller by a factor of about 10⁴.
  • Experiments show that the volume of a nucleus is about 10⁻¹² times that of the atom.

13.2 Atomic Masses and Composition of Nucleus

  • Mass of a carbon atom (¹²C): 1.992647 x 10⁻²⁶ kg.
  • Atomic mass unit (u): 1 u = 1/12th mass of one atom of ¹²C = 1.660563 x 10⁻²⁷ kg.
  • Definitions:
    • Atomic number (Z): Number of protons in the nucleus.
    • Mass number (A): Total number of protons and neutrons (A = Z + N).
    • Isotopes: Nuclides with the same Z but different N.
    • Isobars: Nuclides with the same A.
    • Isotones: Nuclides with the same N.

13.3 Size of the Nucleus

  • Radius of a nucleus: R = R₀ A¹/³, where R₀ = 1.2 fm.
  • Density of nuclear matter is approximately 2.3 x 10¹⁷ kg/m³, independent of A.

13.4 Binding Energy

  • Binding energy (Eᵇₙ) is the energy required to separate a nucleus into its nucleons.
  • Binding energy per nucleon: Eₚₙ = Eᵇₙ / A.
  • The curve of binding energy per nucleon shows peaks at certain nuclides, indicating stability.

13.5 Nuclear Reactions

  • In nuclear reactions, both the number of protons and neutrons are conserved.
  • Mass-energy interconversion occurs, where the difference in binding energy appears as energy released or absorbed.

13.6 Controlled Thermonuclear Fusion

  • Fusion reactors aim to replicate the natural fusion process in stars, requiring temperatures around 10⁸ K.
  • The challenge lies in confining plasma at such high temperatures.

13.7 Exercises

  • Example problems include calculating binding energies and analyzing nuclear reactions.

Points to Ponder

  1. Density of nuclear matter is independent of nucleus size.
  2. Different methods of measuring nuclear size yield slightly different results.
  3. Mass-energy equivalence is central to nuclear physics.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

They have different chemical properties because they have different masses.

B.

They have similar chemical properties because they have identical electronic structures.

C.

They have different chemical properties because they have different numbers of neutrons.

D.

They have similar chemical properties because they have the same number of neutrons.
Correct Answer: B

Solution:

Isotopes have similar chemical properties because their electronic structures are identical, even though they have different masses.

Chapter Concept:

Isotopes, Isobars and Isotones

A.

R1/R2=(A1/A2)1/3R_1/R_2 = (A_1/A_2)^{1/3}

B.

R1/R2=(A2/A1)1/3R_1/R_2 = (A_2/A_1)^{1/3}

C.

R1/R2=(A1/A2)3R_1/R_2 = (A_1/A_2)^{3}

D.

R1/R2=(A2/A1)3R_1/R_2 = (A_2/A_1)^{3}
Correct Answer: A

Solution:

The nuclear radius varies as A1/3A^{1/3}, so the ratio of nuclear radii is R1/R2=(A1/A2)1/3R_1/R_2 = (A_1/A_2)^{1/3}.

Chapter Concept:

Nuclear Radius Formula and Scaling

A.

127.5 MeV

B.

120.0 MeV

C.

136.9 MeV

D.

140.0 MeV
Correct Answer: A

Solution:

The binding energy Eb=ΔM⋅c2=0.13691×931.5=127.5 MeVE_b = \Delta M \cdot c^2 = 0.13691 \times 931.5 = 127.5 \text{ MeV}.

Chapter Concept:

Nuclear Binding Energy

A.

The long-range nature of nuclear forces.

B.

The saturation property of nuclear force.

C.

The presence of electrons within the nucleus.

D.

The high density of nuclear matter.
Correct Answer: B

Solution:

The saturation property of nuclear force means that each nucleon interacts mainly with nearby nucleons within the short range of nuclear force, leading to the approximate constancy of binding energy per nucleon for medium and heavy nuclei.

Chapter Concept:

Saturation Property of Nuclear Force

A.

Medium-mass nuclei have higher binding energy per nucleon, making them more tightly bound.

B.

Light nuclei are unstable due to a lack of neutrons.

C.

Heavy nuclei are unstable because they have too many protons.

D.

Medium-mass nuclei have more protons than neutrons, which enhances stability.
Correct Answer: A

Solution:

Medium-mass nuclei have higher binding energy per nucleon, which means they are more tightly bound and thus more stable compared to very light or very heavy nuclei.

Chapter Concept:

Stability from Binding Energy per Nucleon

A.

26

B.

30

C.

56

D.

82
Correct Answer: B

Solution:

The number of neutrons NN is given by N=A−ZN = A - Z. For this nuclide, N=56−26=30N = 56 - 26 = 30.

Chapter Concept:

Nuclear Composition and Nuclide Notation

A.

The reaction requires an external energy input to overcome the Coulomb barrier.

B.

The reaction releases energy because the final nucleus is more tightly bound than the initial nuclei.

C.

The reaction results in a net loss of energy because of the mass defect.

D.

The reaction does not involve any energy change as it is a simple rearrangement of particles.
Correct Answer: B

Solution:

In the fusion reaction, energy is released because the resulting deuteron is more tightly bound, which means it has a higher binding energy per nucleon compared to the initial protons.

Chapter Concept:

Controlled Thermonuclear Fusion

A.

The total number of nucleons is conserved.

B.

The total charge is not conserved.

C.

The reaction violates the conservation of mass-energy.

D.

The number of protons on both sides is different.
Correct Answer: A

Solution:

In nuclear reactions, both the total number of nucleons and the total charge are conserved. In the given reaction, the total number of nucleons on both sides is 18, and the charge is conserved as well.

Chapter Concept:

Balancing Nuclear Reactions

A.

216 MeV

B.

240 MeV

C.

120 MeV

D.

432 MeV
Correct Answer: A

Solution:

The gain in binding energy per nucleon is 8.5−7.6=0.98.5 - 7.6 = 0.9 MeV. Since there are 240 nucleons, the total energy released is 240×0.9=216240 \times 0.9 = 216 MeV.

Chapter Concept:

Mass-Energy Equivalence

A.

Nuclei with higher binding energy per nucleon are generally more stable.

B.

Very light and very heavy nuclei are more stable than medium-mass nuclei.

C.

The binding energy per nucleon is highest for very heavy nuclei.

D.

The stability of a nucleus is independent of its binding energy per nucleon.
Correct Answer: A

Solution:

Higher binding energy per nucleon means a more tightly bound and generally more stable nucleus. Medium-mass nuclei are more stable than very light or very heavy nuclei.

Chapter Concept:

Stability from Binding Energy per Nucleon

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True or False

Correct Answer: True

Solution:

The nucleus is the dense central part of an atom and contains protons and neutrons, which account for more than 99.9% of the atomic mass.

Chapter Concept :

Atomic Structure and Nucleus

Correct Answer: True

Solution:

The nuclear density is constant because the nuclear volume is proportional to the mass number AA, as given by the formula R=R0A1/3R = R_0 A^{1/3}, where R0=1.2 fmR_0 = 1.2 \text{ fm}. This results in a density that is independent of AA.

Chapter Concept :

Nuclear Size and Density

Correct Answer: True

Solution:

Nuclear force is indeed much stronger than the Coulomb force and is effective over a very short range of a few femtometres, as described in the text.

Chapter Concept :

Nuclear Forces

Correct Answer: False

Solution:

Isotopes have similar chemical properties because their electronic structures are identical, despite having different numbers of neutrons.

Chapter Concept :

Isotopes, Isobars and Isotones

Correct Answer: True

Solution:

The atomic mass unit (u) is indeed defined as 1/12th of the mass of a carbon-12 (12C^{12}C) atom, which provides a standardized way to express atomic masses.

Chapter Concept :

Atomic Mass Unit and Isotopes

Correct Answer: True

Solution:

Nuclear fusion results in the formation of a more tightly bound nucleus, as indicated by a higher binding energy per nucleon compared to the initial lighter nuclei. This difference in binding energy per nucleon results in the release of energy.

Chapter Concept :

Nuclear Fission and Fusion

Correct Answer: True

Solution:

Controlled thermonuclear fusion involves heating nuclear fuel to temperatures around 10810^8 K to achieve conditions similar to those in stars, where fusion naturally occurs.

Chapter Concept :

Controlled Thermonuclear Fusion

Correct Answer: True

Solution:

The atomic mass unit (u) is indeed defined as 1/12th of the mass of a carbon-12 atom, which is a standard definition used for expressing atomic masses.

Chapter Concept :

Atomic Mass Unit and Isotopes

Correct Answer: True

Solution:

Controlled thermonuclear fusion requires heating the nuclear fuel to extremely high temperatures, around 10810^8 K, to achieve the necessary conditions for fusion reactions to occur.

Chapter Concept :

Controlled Thermonuclear Fusion

Correct Answer: True

Solution:

Einstein's mass-energy equivalence relation E=mc2E = mc^2 shows that mass can be converted into energy. In nuclear calculations, 1 u is equivalent to 931.5 MeV/c², which is used to quantify the energy released or absorbed during nuclear reactions.

Chapter Concept :

Mass-Energy Equivalence

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