Nuclei

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Summary

Chapter Thirteen: Nuclei

Summary

The nucleus is the center of an atom, containing most of its mass (over 99.9%) and is much smaller than the atom itself.
Atomic mass is measured in atomic mass units (u), with 1 u defined as 1/12th the mass of a ¹²C atom.
A nucleus consists of protons and neutrons, with protons determining the atomic number (Z) and the total number of nucleons (protons + neutrons) giving the mass number (A).
Isotopes are nuclides with the same atomic number but different neutron numbers, while isobars have the same mass number and isotones have the same neutron number.
The nuclear radius can be estimated using the formula R = R₀ A¹/³, where R₀ is approximately 1.2 fm, indicating nuclear density is nearly constant.
Neutrons and protons are held together in the nucleus by the strong nuclear force, which does not differentiate between them.
The binding energy of a nucleus is the energy required to separate it into its individual nucleons, with binding energy per nucleon (Eᵇₙ) being a useful measure of stability.
Nuclear reactions can convert mass into energy, as described by Einstein's equation E = mc², highlighting the relationship between mass and energy in nuclear processes.

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.

Detailed Notes

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

Density of nuclear matter is independent of nucleus size.
Different methods of measuring nuclear size yield slightly different results.
Mass-energy equivalence is central to nuclear physics.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Nuclear Reactions: Students often think that nuclear reactions are balanced like chemical equations. In nuclear reactions, while the number of protons and neutrons is conserved, the total number of atoms may not be.
Confusing Mass and Energy: Many students fail to grasp the concept of mass-energy interconversion in nuclear reactions, leading to incorrect conclusions about energy changes.
Ignoring Binding Energy: Students may overlook the significance of binding energy and its relation to mass defect, which is crucial for understanding nuclear stability.

Exam Tips

Understand Key Concepts: Make sure to clearly understand the definitions of atomic mass unit, half-life, and binding energy. These are often tested.
Practice Calculations: Work through problems involving Q-values and binding energy calculations to become comfortable with the formulas and their applications.
Review Diagrams: Familiarize yourself with diagrams related to nuclear reactions and binding energy curves, as they can be helpful in visualizing concepts.
Focus on Units: Pay attention to the units used in nuclear physics, such as MeV for energy and seconds for time, to avoid calculation errors.

Practice & Assessment

Multiple Choice Questions

Conservation of the number of atoms of each element

Conservation of the number of protons and neutrons

No mass-energy interconversion

Only chemical changes occur

Correct Answer: B

Solution:

In nuclear reactions, the number of protons and neutrons are conserved, although elements may be transmuted.

8 MeV

7 MeV

6 MeV

10 MeV

Correct Answer: A

Solution:

The binding energy per nucleon is calculated as total binding energy divided by the mass number:

960 \text{ MeV} / 120 = 8 \text{ MeV}

1.0 fm

1.2 fm

1.5 fm

2.0 fm

Correct Answer: B

Solution:

The constant

R_0

is 1.2 fm, which is used in the formula

R = R_0 A^{1/3}

to calculate the radius of a nucleus.

It decreases with increasing mass number for very heavy nuclei.

It remains constant for all nuclei.

It increases indefinitely with increasing mass number.

It is highest for very light nuclei.

Correct Answer: A

Solution:

The binding energy per nucleon decreases for very heavy nuclei due to the increased repulsive forces between protons. This is why very heavy nuclei are more likely to undergo fission.

Energy is released because the nucleons become more tightly bound.

Energy is absorbed because the nucleons become less tightly bound.

No energy change occurs because the binding energy per nucleon is the same.

The process is impossible due to conservation laws.

Correct Answer: A

Solution:

When a heavy nucleus with lower binding energy per nucleon splits into lighter nuclei with higher binding energy per nucleon, energy is released as the nucleons become more tightly bound.

R = 1.2 \times 64^{1/3}

R = 1.2 \times 64^{2/3}

R = 1.2 \times 64

R = 1.2 \times 64^{-1/3}

Correct Answer: A

Solution:

The radius of a nucleus is given by

R = R_0 A^{1/3}

, where

R_0 = 1.2

fm.

Higher for

A = 240

Lower for

A = 240

Same for both

Cannot be determined

Correct Answer: B

Solution:

A very heavy nucleus like

A = 240

has lower binding energy per nucleon compared to

A = 120

The potential energy is positive, indicating repulsion.

The potential energy is zero, indicating no interaction.

The potential energy is negative, indicating attraction.

The potential energy is infinite, indicating strong repulsion.

Correct Answer: C

Solution:

For a separation greater than 0.8 fm, the nuclear force is attractive, leading to negative potential energy.

0.72 MeV

1.44 MeV

2.88 MeV

4.32 MeV

Correct Answer: B

Solution:

The potential barrier is given by the Coulomb repulsion:

V = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{2r}

, where

r = 2 \times 2.0

fm = 4.0 fm. Substituting the values,

V = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{4.0 \times 10^{-15}}

Joules. Converting to MeV,

V \approx 1.44

MeV.

10^-12

10^-4

10^-8

10^-16

Correct Answer: A

Solution:

The volume of a nucleus is about 10^-12 times the volume of the atom, indicating that an atom is mostly empty space.

Preventing the plasma from cooling down.

Confining the plasma since no container can withstand the high temperature.

Ensuring the plasma does not undergo spontaneous fission.

Increasing the plasma density to maintain reaction rates.

Correct Answer: B

Solution:

The main challenge in a fusion reactor is confining the plasma, as no material container can withstand the high temperatures required for fusion.

Approximately 1 year

Approximately 10 years

Approximately 100 years

Approximately 1000 years

Correct Answer: C

Solution:

The energy from the fusion of 2.0 kg of deuterium can keep a 100W lamp glowing for approximately 100 years.

Finding a suitable nuclear fuel

Confine the plasma at high temperatures

Cooling the reactor efficiently

Preventing radioactive waste

Correct Answer: B

Solution:

The challenge in controlled fusion reactors is to confine the plasma, as no container can withstand the high temperatures required.

To increase the temperature of the plasma

To prevent the plasma from cooling down

To prevent the plasma from coming into contact with the reactor walls

To enhance the fusion reaction rate

Correct Answer: C

Solution:

Magnetic confinement is used in fusion reactors to prevent the plasma, which is at extremely high temperatures, from coming into contact with the reactor walls, as no material can withstand such high temperatures.

Achieving sufficient plasma density

Preventing plasma from touching the reactor walls

Maintaining a constant plasma temperature

Ensuring uniform plasma composition

Correct Answer: B

Solution:

In a controlled thermonuclear fusion reactor, the plasma must be confined at extremely high temperatures, and no material can withstand direct contact with the plasma. Therefore, one of the primary challenges is to prevent the plasma from touching the reactor walls, which would lead to energy loss and potential damage to the reactor.

180 MeV

200 MeV

150 MeV

100 MeV

Correct Answer: A

Solution:

The average energy released per fission of

^{239}_{94}Pu

is 180 MeV.

It represents the energy required to initiate the reaction.

It indicates the speed of the reaction.

It determines whether a reaction is exothermic or endothermic.

It measures the stability of the reactants.

Correct Answer: C

Solution:

The Q value of a nuclear reaction is defined as the difference in the total mass energy of the reactants and products. A positive Q value indicates an exothermic reaction, while a negative Q value indicates an endothermic reaction.

They have the same number of protons but different numbers of neutrons.

They have different numbers of protons and the same number of neutrons.

They have the same number of protons and neutrons.

They have different chemical properties.

Correct Answer: A

Solution:

Isotopes of an element have the same number of protons but differ in the number of neutrons.

The number of protons and neutrons are both separately conserved.

The total mass of the reactants is always equal to the total mass of the products.

The total charge and total 'baryon number' are strictly conserved.

The total energy, including rest mass energy, is not conserved.

Correct Answer: C

Solution:

In nuclear reactions, while the number of protons and neutrons are separately conserved, the total charge and total 'baryon number' are strictly conserved. Mass-energy equivalence allows for the conversion of mass into energy and vice versa.

It decreases

It remains the same

It increases

It becomes zero

Correct Answer: C

Solution:

When a very heavy nucleus splits into two smaller nuclei, the binding energy per nucleon increases, releasing energy.

Isobars are nuclides with the same atomic number but different mass numbers.

Isobars are nuclides with the same neutron number but different atomic numbers.

Isobars are nuclides with the same mass number but different atomic numbers.

Isobars are nuclides with the same number of protons and neutrons.

Correct Answer: C

Solution:

Isobars are nuclides that have the same mass number (A) but different atomic numbers (Z).

104.66 MeV

110.22 MeV

115.50 MeV

120.75 MeV

Correct Answer: B

Solution:

The binding energy is calculated using the formula:

B.E. = (Zm_p + Nm_n - M)c^2

. For

^{14}N

B.E. = (7 \times 1.007825 + 7 \times 1.008665 - 14.00307) \times 931.5 \text{ MeV/u} = 110.22 \text{ MeV}

It is weaker than the gravitational force.

It is a long-range force.

It is much stronger than the Coulomb force.

It acts only between protons.

Correct Answer: C

Solution:

The nuclear force is much stronger than the Coulomb force and acts between nucleons.

It is weaker than the Coulomb force.

It is stronger than the Coulomb force and binds nucleons together.

It only acts on electrons.

It is responsible for radioactive decay.

Correct Answer: B

Solution:

The nuclear force is much stronger than the Coulomb force and is responsible for binding protons and neutrons within the nucleus.

R = R_0 A^{1/3}

R = R_0 A^{2/3}

R = R_0 A^{1/2}

R = R_0 A

Correct Answer: A

Solution:

The radius of a nucleus is given by the formula

R = R_0 A^{1/3}

, where

R_0

is a constant.

It will become a black hole.

It will expand into a red giant.

It will explode as a supernova.

It will remain unchanged.

Correct Answer: B

Solution:

After hydrogen burning stops, the sun will expand into a red giant.

Only the number of protons

Only the number of neutrons

Both the number of protons and neutrons

Neither protons nor neutrons

Correct Answer: C

Solution:

In a nuclear reaction, both the number of protons and the number of neutrons are conserved.

It decreases

It remains the same

It increases

It becomes zero

Correct Answer: C

Solution:

When a nucleus with

A = 240

breaks into two

A = 120

nuclei, the nucleons get more tightly bound, increasing the binding energy per nucleon.

Proportional to

A

Proportional to

A^{1/3}

Proportional to

A^{2/3}

Independent of

A

Correct Answer: B

Solution:

The radius of a nucleus with mass number

A

is given by

R = R_0 A^{1/3}

, indicating it is proportional to

A^{1/3}

It will become a black hole.

It will explode as a supernova.

It will expand into a red giant.

It will remain unchanged.

Correct Answer: C

Solution:

After exhausting its hydrogen fuel, the sun will expand into a red giant as it begins to cool and collapse under gravity.

It is the radius of a hydrogen atom

It is a constant used to calculate the radius of a nucleus

It is the volume of an atom

It is the density of nuclear matter

Correct Answer: B

Solution:

R₀ is a constant used to calculate the radius of a nucleus, where R = R₀A^{1/3}.

1.0 x 10^3 kg/m³

2.3 x 10^17 kg/m³

5.0 x 10^10 kg/m³

9.8 x 10^5 kg/m³

Correct Answer: B

Solution:

The density of nuclear matter is approximately 2.3 x 10^17 kg/m³, which is much larger than that of ordinary matter.

Kilogram

Gram

Atomic mass unit (u)

Milligram

Correct Answer: C

Solution:

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

1/10th of the mass of a hydrogen atom

1/12th of the mass of a carbon-12 atom

Equal to the mass of a proton

Equal to the mass of an electron

Correct Answer: B

Solution:

The atomic mass unit (u) is defined as 1/12th of the mass of a carbon-12 atom.

The number of protons and neutrons are both separately conserved.

The number of atoms of each element is conserved.

Mass is not converted into energy.

Nuclear reactions do not involve energy changes.

Correct Answer: A

Solution:

In nuclear reactions, the number of protons and neutrons are both separately conserved, although the number of atoms of each element is not necessarily conserved.

Nuclear forces increase indefinitely with distance.

Nuclear forces are effective only up to a few femtometers.

Nuclear forces are weaker than gravitational forces.

Nuclear forces are independent of the number of nucleons.

Correct Answer: B

Solution:

The saturation property refers to the fact that nuclear forces are short-ranged and effective only up to a few femtometers.

Energy is absorbed because the binding energy per nucleon decreases.

Energy is released because the binding energy per nucleon increases.

No energy change occurs because the total mass number is conserved.

Energy is absorbed because the total binding energy remains constant.

Correct Answer: B

Solution:

When a very heavy nucleus splits into two lighter nuclei, the binding energy per nucleon increases, resulting in energy release.

4.8 fm

5.2 fm

6.4 fm

7.2 fm

Correct Answer: A

Solution:

Using the formula

R = R_0 A^{1/3}

, we find

R = 1.2 \times 10^{-15} \times 64^{1/3} = 4.8 \text{ fm}

The density increases with mass number

A

The density decreases with mass number

A

The density is independent of mass number

A

The density varies randomly with mass number

A

Correct Answer: C

Solution:

The volume of the nucleus is proportional to

R^3

, which is proportional to

A

. Since the mass is also proportional to

A

, the density, which is mass/volume, is constant and independent of

A

The number of protons and neutrons is conserved, but the number of atoms of each element is not.

The number of atoms of each element is conserved, similar to chemical reactions.

The total mass is conserved, but energy is not.

The number of electrons is conserved, but protons and neutrons are not.

Correct Answer: A

Solution:

In nuclear reactions, the number of protons and neutrons is conserved, but elements can transmute, so the number of atoms of each element is not necessarily conserved.

The reaction is exothermic and releases 5 MeV of energy.

The reaction is endothermic and requires 5 MeV of energy.

The reaction is exothermic and requires 5 MeV of energy.

The reaction is endothermic and releases 5 MeV of energy.

Correct Answer: B

Solution:

A negative Q-value indicates that the reaction is endothermic, meaning it requires an input of energy. In this case, 5 MeV of energy is needed.

216 MeV

2160 MeV

840 MeV

8400 MeV

Correct Answer: B

Solution:

The energy released in the fission process can be calculated by finding the difference in total binding energy before and after the reaction. The total binding energy of the

A = 240

nucleus is

240 \times 7.6

MeV = 1824 MeV. The total binding energy of two

A = 120

nuclei is

2 \times 120 \times 8.5

MeV = 2040 MeV. The energy released is

2040 - 1824 = 216

MeV.

4.52 \times 10^{26} \text{ MeV}

2.26 \times 10^{26} \text{ MeV}

9.04 \times 10^{26} \text{ MeV}

1.13 \times 10^{26} \text{ MeV}

Correct Answer: A

Solution:

The number of moles of

Pu

in 1 kg is

\frac{1000}{239}

. The number of atoms is

\frac{1000}{239} \times 6.023 \times 10^{23}

. The total energy released is

\frac{1000}{239} \times 6.023 \times 10^{23} \times 180 = 4.52 \times 10^{26} \text{ MeV}

Isotopes have the same number of protons but different numbers of neutrons.

Isotopes have different numbers of protons and neutrons.

Isotopes have the same mass number but different atomic numbers.

Isotopes have identical chemical properties and different electronic structures.

Correct Answer: A

Solution:

Isotopes of an element contain the same number of protons but differ in the number of neutrons.

Yes, because energy is released

No, because energy is absorbed

Yes, because the fragments are more stable

No, because the fragments are less stable

Correct Answer: B

Solution:

The total binding energy of the initial nucleus is

56 \times 8.8 = 492.8

MeV. The total binding energy of the resulting nuclei is

2 \times 28 \times 8.1 = 453.6

MeV. Since the binding energy decreases, energy is absorbed, making the fission energetically unfavorable.

Total mass

Total energy

Total charge and baryon number

Total volume

Correct Answer: C

Solution:

In nuclear reactions, the total charge and baryon number are conserved, even if the number of protons and neutrons may not be.

Atoms with the same number of protons but different number of neutrons.

Atoms with the same number of neutrons but different number of protons.

Atoms with the same atomic mass but different atomic numbers.

Atoms with different chemical properties.

Correct Answer: A

Solution:

Isotopes are defined as atoms with the same number of protons but different numbers of neutrons.

4.8 fm

5.6 fm

6.4 fm

7.2 fm

Correct Answer: A

Solution:

The radius

R

is calculated as

R = 1.2 \times 64^{1/3} = 1.2 \times 4 = 4.8

fm.

180 MeV

1.6 x 10^13 MeV

3.6 x 10^13 MeV

2.4 x 10^13 MeV

Correct Answer: B

Solution:

The energy released is calculated by multiplying the number of atoms in 1 kg of 239 94 Pu by the energy released per fission, which is 180 MeV.

By the gravitational force between them

By the Coulomb repulsion when they just touch each other

By the nuclear force when they are far apart

By the magnetic force between them

Correct Answer: B

Solution:

The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other.

Number of atoms and energy

Number of protons and neutrons

Mass and volume

Charge and temperature

Correct Answer: B

Solution:

In nuclear reactions, the number of protons and neutrons are conserved, although the number of atoms may change.

Nuclear force is weaker than gravitational force.

Nuclear force is stronger than Coulomb force.

Nuclear force acts over long distances.

Nuclear force is repulsive at all distances.

Correct Answer: B

Solution:

Nuclear force is much stronger than the Coulomb force, which allows it to bind protons and neutrons within the nucleus.

The mass of a hydrogen atom

The mass of a neutron

1/12th of the mass of a carbon-12 atom

The mass of an electron

Correct Answer: C

Solution:

The atomic mass unit (u) is defined as 1/12th of the mass of a carbon-12 atom.

1.5 x 10⁷ seconds

3.0 x 10⁷ seconds

4.5 x 10⁷ seconds

6.0 x 10⁷ seconds

Correct Answer: B

Solution:

The energy released per fusion reaction is 3.27 MeV, which is 3.27 x 1.6 x 10⁻¹³ J. The number of reactions from 2.0 kg of deuterium can be calculated using Avogadro's number and the molar mass of deuterium. The total energy can then be calculated and divided by the power of the lamp to find the time.

3.6 fm

4.8 fm

5.4 fm

2.7 fm

Correct Answer: A

Solution:

Using the formula

R = R_0 A^{1/3}

, the radius is

R = 1.2 \times 10^{-15} \times 27^{1/3} \approx 3.6

fm.

It is a short-range force, effective only up to a few femtometers.

It is charge-independent and acts equally between all nucleons.

It is responsible for the stability of atomic nuclei.

It decreases linearly with increasing distance between nucleons.

Correct Answer: D

Solution:

The nuclear force decreases rapidly and not linearly with increasing distance between nucleons, becoming negligible beyond a few femtometers.

They have different chemical properties.

They have the same number of neutrons.

They have the same number of protons but different numbers of neutrons.

They are always radioactive.

Correct Answer: C

Solution:

Isotopes of an element have the same number of protons but differ in the number of neutrons.

Nuclear forces have a longer range than Coulomb forces.

Nuclear forces are attractive, while Coulomb forces are repulsive.

Nuclear forces are much stronger than Coulomb forces.

Nuclear forces act only on protons.

Correct Answer: C

Solution:

Nuclear forces are much stronger than Coulomb forces, allowing them to overcome the repulsive forces between protons and hold the nucleus together.

Kilogram

Gram

Atomic mass unit (u)

Pound

Correct Answer: C

Solution:

The atomic mass unit (u) is used for expressing atomic or nuclear masses and is defined as 1/12th of the mass of a ¹²C atom.

4.8 fm

5.2 fm

5.6 fm

6.0 fm

Correct Answer: A

Solution:

Using the formula

R = R_0 A^{1/3}

, we can find the ratio of the radii:

\frac{R_{64}}{R_{27}} = \left(\frac{64}{27}\right)^{1/3}

. Solving this gives

R_{64} = 3.6 \times \left(\frac{64}{27}\right)^{1/3} = 4.8

fm.

1.22

1.34

1.45

1.67

Correct Answer: A

Solution:

The nuclear radius is proportional to

A^{1/3}

. Therefore, the ratio is

\left(\frac{197}{107}\right)^{1/3} \approx 1.22

Gravitational force

Coulomb force

Nuclear force

Magnetic force

Correct Answer: C

Solution:

The nuclear force is much stronger than the Coulomb force and is responsible for binding protons and neutrons in the nucleus.

They are weaker than gravitational forces.

They are stronger than Coulomb forces.

They are long-range forces.

They act only on electrons.

Correct Answer: B

Solution:

Nuclear forces are much stronger than Coulomb forces, allowing them to overcome the repulsion between protons.

True or False

Correct Answer: True

Solution:

According to the excerpt, the average energy released per fission of

^{239}_{94}Pu

is indeed 180 MeV.

Correct Answer: False

Solution:

While the number of protons and neutrons are generally conserved in nuclear reactions, this is not strictly true at very high energies. What is conserved is the total charge and total 'baryon number'.

Correct Answer: False

Solution:

In nuclear reactions, the number of protons and neutrons are separately conserved, according to the excerpt.

Correct Answer: True

Solution:

The nuclear force is significantly stronger than the Coulomb force, allowing it to overcome the repulsion between protons within a nucleus.

Correct Answer: False

Solution:

In nuclear reactions, the number of protons and the number of neutrons are separately conserved.

Correct Answer: True

Solution:

Both

^{239}_{94}Pu

and

^{235}U

have similar fission properties, which means they release energy in a similar manner during fission.

Correct Answer: False

Solution:

The atomic mass of chlorine is approximately 35.46 u, which is not an integral multiple of the mass of a hydrogen atom.

Correct Answer: True

Solution:

Experiments on scattering of α-particles demonstrated that the radius of a nucleus was smaller than the radius of an atom by a factor of about 10⁴.

Correct Answer: True

Solution:

Experiments on scattering of

\alpha

-particles demonstrated that the radius of a nucleus was smaller than the radius of an atom by a factor of about

10^4

, indicating that an atom is mostly empty space.

Correct Answer: True

Solution:

Both

^{239}_{94}\text{Pu}

and

^{235}\text{U}

undergo fission reactions that release a significant amount of energy, approximately 180 MeV per fission event.

Correct Answer: False

Solution:

The atomic mass unit (u) is defined as 1/12th of the mass of a carbon (¹²C) atom, not a hydrogen atom.

Correct Answer: False

Solution:

For very heavy nuclei, the binding energy per nucleon is lower compared to nuclei with intermediate mass numbers, but it generally increases for light nuclei as they fuse into heavier nuclei.

Correct Answer: False

Solution:

The mass of a proton is slightly less than the mass of a hydrogen atom because the mass of a hydrogen atom includes the mass of an electron.

Correct Answer: False

Solution:

Isotopes of a given element contain the same number of protons but differ in their number of neutrons.

Correct Answer: False

Solution:

While the number of protons and neutrons are often conserved in nuclear reactions, this is not strictly true at very high energies where other conservation laws, like baryon number, apply.

Correct Answer: False

Solution:

The nuclear force is much stronger than both the Coulomb force and the gravitational force. It is strong enough to overcome the repulsion between protons and bind nucleons within the nucleus.

Correct Answer: True

Solution:

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

Correct Answer: True

Solution:

The excerpt explains that once hydrogen burning stops, the sun will expand and become a red giant.

Correct Answer: True

Solution:

The excerpt mentions that there is enough hydrogen in the sun to keep it going for another 5 billion years.

Correct Answer: True

Solution:

Experiments on scattering of α-particles have shown that the radius of a nucleus is about 10,000 times smaller than that of an atom, indicating that an atom is mostly empty space.

Correct Answer: True

Solution:

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

Correct Answer: True

Solution:

The nuclear force is much stronger than both the Coulomb force and the gravitational force, which is much weaker than even the Coulomb force.

Correct Answer: False

Solution:

The nuclear force is much stronger than the Coulomb force, as it needs to overcome the repulsion between protons to bind them into the nucleus.

Correct Answer: True

Solution:

The atomic mass unit (u) is indeed defined as 1/12th of the mass of a

^{12}C

atom.

Correct Answer: False

Solution:

The nuclear force is much stronger than the Coulomb force, as it needs to overcome the repulsion between protons within the nucleus.

Correct Answer: False

Solution:

The nuclei of isotopes of a given element contain the same number of protons, but differ from each other in their number of neutrons.

Correct Answer: True

Solution:

Experiments on scattering of α-particles demonstrated that the radius of a nucleus was smaller than the radius of an atom by a factor of about 10⁴, implying the volume of a nucleus is about 10⁻¹² times the volume of the atom.

Correct Answer: True

Solution:

According to the excerpt, the average energy released per fission of

^{239}_{94}Pu

is indeed 180 MeV.

Correct Answer: False

Solution:

The atomic mass unit (u) is defined as 1/12th of the mass of a carbon (¹²C) atom, not a hydrogen atom.

Correct Answer: True

Solution:

The nuclear matter density is nearly constant and independent of the mass number

A

, as shown by the relation

R = R_0 A^{1/3}

Correct Answer: True

Solution:

It is estimated that after the hydrogen burning stops, the sun will begin to cool, collapse under gravity, and its outer envelope will expand, turning it into a red giant.

Correct Answer: True

Solution:

The relation

R = R_0 A^{1/3}

implies that the volume of the nucleus is proportional to

A

, leading to a constant nuclear density across different nuclei.

Correct Answer: True

Solution:

After the hydrogen in the sun is exhausted, it will expand and become a red giant as it begins to burn other elements.

Correct Answer: True

Solution:

The nuclear force is indeed much stronger than the Coulomb force, which allows it to overcome the repulsion between protons in the nucleus.

Correct Answer: True

Solution:

The volume of a nucleus is indeed about 10⁻¹² times the volume of an atom, as the nucleus is much smaller compared to the overall size of the atom.

Correct Answer: True

Solution:

The nuclear force is much stronger than both the Coulomb force and the gravitational force, allowing it to bind protons and neutrons within the nucleus.

Correct Answer: True

Solution:

Experiments on scattering of α-particles have demonstrated that the radius of a nucleus is smaller than the radius of an atom by a factor of about 10⁴.

Correct Answer: False

Solution:

A very heavy nucleus, say

A = 240

, has lower binding energy per nucleon compared to that of a nucleus with

A = 120

. Thus, if a nucleus

A = 240

breaks into two

A = 120

nuclei, nucleons get more tightly bound.

Correct Answer: False

Solution:

Isotopes of an element have identical chemical properties because they have identical electronic structures.

Correct Answer: False

Solution:

The motion of atomic electrons is determined by the Coulomb force, not the nuclear force.

Correct Answer: False

Solution:

Very heavy nuclei have a lower binding energy per nucleon compared to lighter nuclei. This is why energy is released when a heavy nucleus fissions into smaller nuclei.

Correct Answer: True

Solution:

It is estimated that there is enough hydrogen in the sun to keep it going for another 5 billion years.

Correct Answer: True

Solution:

The excerpt states that the density of nuclear matter is approximately 2.3 x

10^{17}

kg/m³, which is very large compared to ordinary matter.

Correct Answer: True

Solution:

The density of nuclear matter is indeed approximately 2.3 x

10^{17}

kg/m³, which is much larger than that of ordinary matter.

Nuclei

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Summary

Chapter Thirteen: Nuclei

Summary

Learning Objectives

Learning Objectives

Detailed Notes

Chapter Thirteen: Nuclei

13.1 Introduction

13.2 Atomic Masses and Composition of Nucleus

13.3 Size of the Nucleus

13.4 Binding Energy

13.5 Nuclear Reactions

13.6 Controlled Thermonuclear Fusion

13.7 Exercises

Points to Ponder

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.Which of the following is a characteristic of nuclear reactions?

Correct Answer: B

Q2.What is the binding energy per nucleon for a nucleus with a mass number A=120A = 120A=120 if its total binding energy is 960 MeV?

Correct Answer: A

Q3.If the radius of a nucleus is given by R=R0A1/3R = R_0 A^{1/3}R=R0​A1/3, what is the constant R0R_0R0​?

Correct Answer: B

Q4.The binding energy per nucleon for a nucleus is a measure of its stability. Which of the following statements is true regarding the binding energy per nucleon?

Correct Answer: A

Q5.Given that the binding energy per nucleon for a nucleus with mass number A=240A = 240A=240 is lower than that for a nucleus with A=120A = 120A=120, what can be inferred if a nucleus of A=240A = 240A=240 undergoes fission into two nuclei of A=120A = 120A=120?

Correct Answer: A

Q6.What is the approximate radius of a nucleus with mass number A=64A = 64A=64?

Correct Answer: A

Q7.What is the binding energy per nucleon for a nucleus with mass number A=240A = 240A=240 compared to A=120A = 120A=120?

Correct Answer: B

Q8.What is the potential energy of a pair of nucleons at a separation distance greater than 0.8 fm?

Correct Answer: C

Q9.Calculate the height of the potential barrier for a head-on collision of two deuterons, given that the radius of a deuteron is 2.0 fm.

Correct Answer: B

Q10.What is the approximate volume ratio of a nucleus to its atom?

Correct Answer: A

Q11.In a controlled thermonuclear fusion reactor, the plasma is heated to extremely high temperatures. What is the primary challenge in maintaining the plasma state?

Correct Answer: B

Q12.How long can a 100W electric lamp be kept glowing by the fusion of 2.0 kg of deuterium?

Correct Answer: C

Q13.What is the main challenge in developing controlled thermonuclear fusion reactors?

Correct Answer: B

Q14.In a controlled thermonuclear fusion reactor, the plasma is confined using magnetic fields. What is the primary reason for using magnetic confinement?

Correct Answer: C

Q15.In a controlled thermonuclear fusion reactor, maintaining the plasma state is crucial. Which of the following is a primary challenge associated with plasma confinement?

Correct Answer: B

Q16.What is the average energy released per fission of 94239Pu^{239}_{94}Pu94239​Pu?

Correct Answer: A

Q17.In the context of nuclear reactions, what is the significance of the 'Q value'?

Correct Answer: C

Q18.What is the primary difference between isotopes of a given element?

Correct Answer: A

Correct Answer: C

Q20.What happens to the binding energy per nucleon when a very heavy nucleus splits into two smaller nuclei?

Correct Answer: C

Q21.Which of the following statements best describes the concept of nuclear isobars?

Correct Answer: C

Q22.What is the binding energy of a nitrogen nucleus 14N^{14}N14N, given its atomic mass is 14.00307 u?

Correct Answer: B

Q23.Which of the following best describes the nuclear force?

Correct Answer: C

Q24.What is the role of nuclear force in a nucleus?

Correct Answer: B

Q25.What is the relationship between the radius of a nucleus and its mass number AAA?

Correct Answer: A

Q26.What happens to the sun after hydrogen burning stops?

Correct Answer: B

Q27.In a nuclear reaction, what is conserved?

Correct Answer: C

Q28.If a nucleus with mass number A=240A = 240A=240 breaks into two nuclei with A=120A = 120A=120, what happens to the binding energy per nucleon?

Correct Answer: C

Q29.What is the radius of a nucleus with mass number AAA given by the formula R=R0A1/3R = R_0 A^{1/3}R=R0​A1/3, where R0R_0R0​ is a constant?

Correct Answer: B