Detailed Notes On Atoms and Nuclei

Comprehensive Notes on Atoms

Class: XII | Subject: Physics

1. Introduction

Let's begin by discussing Robert Boyle, who studied how air expands and compresses. The ability of air to be compressed or expanded shows that it is made up of tiny particles with empty spaces between them. When air is compressed, these particles move closer together, reducing the empty space. Boyle's work marked a shift in the study of atoms, moving from simple reasoning to experimental observation. An atom is the smallest unit of an element that retains all the properties of that element.

In the 5th century BC, ancient Greek philosophers Leucippus and Democritus (460–370 BC) introduced the idea that matter is composed of indivisible particles called atoms. Later, in the 19th century, John Dalton developed his atomic theory:

• Atoms are indestructible and indivisible particles that make up matter.
• Atoms of the same element are identical in properties.
• Atoms can combine to form new substances (compounds).

Dalton’s idea that atoms are indestructible was disproved when J. J. Thomson discovered electrons in 1897. His experiments demonstrated that atoms contain smaller, negatively charged particles. This chapter explores atomic structure in depth.

Key Fact: The nucleus is about 100,000 times smaller than the atom itself and contains 99.9% of its mass.

2. Thomson’s Atomic Model (Plum Pudding Model)

J. J. Thomson conducted experiments with vacuum tubes where he observed that the cathode emitted rays consisting of negatively charged particles called electrons. He Proposed his model Plum Pudding Model of the atom in 1903. According to this model an atom is a sphere having a uniform positive charge in which electrons are embedded.

• The atom is a positively charged sphere in which negatively charged electrons are embedded.
• The total positive charge balances the total negative charge, making the atom neutral.
• Positive charge is spread uniformly throughout the atom.

This model explained the neutrality of atoms but was later replaced by more accurate models.

3. Geiger-Marsden Experiment

To study the internal structure of the atom, Ernest Rutherford proposed an experiment where alpha particles were directed at a thin gold foil. This experiment was conducted by his colleagues Geiger and Marsden between 1908 and 1913.

Alpha particles from a source were collimated, (focused into a narrow beam) and were made to fall on a gold foil.

It was found that most alpha particles passed straight through the foil while a few were deflected (scattered) through various scattering angles.

Only about 0.14% of the incident alpha particles were scattered through angles larger than \(0.1^\circ\).

About one alpha particle in 8000 was deflected through angle larger than \(90^\circ\) and a fewer still were deflected through angles as large as \(180^\circ\).

The Alpha particle itself is about 7350 times heavier than the electron. So neither an electron, nor a similar positively charged particle could cause a large scale deflection of an aplha particle. It means that alpha particle must have encountered a very heavy particle in its path with the same charge, later on that heavy partilce named as Nucleus.

Observations:

• Most alpha particles passed through the foil without deflection.
• A small fraction was deflected at small angles.
• Very few were deflected at large angles, with some bouncing back.

Conclusions:

• Most of the atom is empty space, allowing particles to pass through undeflected.
• A small, dense, positively charged nucleus exists at the center of the atom.

4. Rutherford’s Atomic Model

Based on the experiment, Rutherford proposed the following:

• The atom consists of a central nucleus that contains most of the atom's mass and is positively charged.
• Electrons revolve around the nucleus in circular orbits.
• The space between the nucleus and electrons is mostly empty.

While this model explained atomic structure, it had limitations:

• According to Maxwell’s theory, an electron in circular motion emits electromagnetic radiation. This should cause the electron to lose energy and spiral into the nucleus, which does not happen.
• Atoms are observed to be stable, and they emit radiation only when excited.
• Radiation emitted by atoms has discrete frequencies, not a continuous spectrum.

4.1 Limitations of Rutherford’s Model

Although Rutherford’s model provided significant insights, it had the following limitations:

• According to Maxwell’s theory, an electron in a circular orbit experiences acceleration and should emit electromagnetic radiation continuously.
• Continuous energy loss would cause the electron to spiral into the nucleus, leading to the atom's collapse, which does not happen.
• Atoms are observed to be stable and emit radiation only when excited.
• Emitted radiation has discrete frequencies, contradicting the prediction of a continuous spectrum.

These limitations were later resolved by Niels Bohr’s Model.

Atomic Spectra

Atomic Spectra

When a metallic object is heated, it emits radiation of different wavelengths. This radiation, when passed through a prism, produces a continuous spectrum. However, the situation is different when we heat hydrogen gas inside a glass tube to high temperatures. The emitted radiation consists of only a few selected wavelengths, resulting in a line spectrum, as shown in Fig. for the visible range.

The hydrogen spectrum exhibits the following wavelengths: 410 nm, 434 nm, 486 nm, and 656 nm. Importantly, there are no radiation emissions between these specific wavelengths. These lines in the spectrum are known as emission lines.

Hydrogen also emits radiation at wavelengths in the ultraviolet (UV) and infrared (IR) regions, in addition to visible light. These spectral lines can be categorized into series named after scientists. Starting from shorter wavelengths and moving to longer wavelengths, these series are:

• Lyman series (UV region)
• Balmer series (Visible region)
• Paschen series (IR region)
• Brackett series (IR region)
• Pfund series (IR region)

In each series, the separation between successive lines decreases as we move towards shorter wavelengths. Ultimately, the lines reach a limiting value. Schematic diagrams of the first three series are shown in following fig., with the limiting wavelength values indicated by dotted lines.

The wavelengths of the emission lines are governed by the equation:

\( \frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right) \)
Here, \( \lambda \) is the wavelength of the emitted radiation, \( R \) is the Rydberg constant, and \( n \) and \( m \) are integers, with \( n \) representing the starting level and \( m \) representing the final level of the electron.

For the Lyman series, \( n = 1 \); for the Balmer series, \( n = 2 \); and so on. As \( m \) increases, the wavelength decreases. The difference in wavelengths of successive lines in each series decreases as \( m \) increases. Ultimately, the wavelengths approach a limiting value of \( \frac{1}{R} \) when \( m \to \infty \).

This equation can be used to calculate the wavelength of the lines in each series, and we observe that the lines get closer together as the value of \( m \) increases. This phenomenon is consistent across all series.

Derivation of the Wavelength Formula

The derivation of the formula for the wavelength of the spectral lines of hydrogen is based on the energy levels of electrons in an atom and their transitions between these levels.

Derivation

The energy of an electron in the \(n\)-th orbit (or energy level) of a hydrogen atom is given by Bohr's formula:

\( E_n = - \frac{13.6}{n^2} \, \text{eV} \)

Where: \( E_n \) is the energy of the electron in the \(n\)-th orbit. \( n \) is the principal quantum number (which takes positive integer values \(n = 1, 2, 3, \dots\)). \( 13.6 \, \text{eV} \) is a constant known as the ionization energy of hydrogen.

When an electron transitions from a higher orbit (with quantum number \( m \)) to a lower orbit (with quantum number \( n \)), the energy difference \( \Delta E \) is given by:

\( \Delta E = E_m - E_n \)

Substituting the expression for \(E_m\) and \(E_n\):

\( \Delta E = \left(- \frac{13.6}{m^2}\right) - \left(- \frac{13.6}{n^2}\right) \)

\( \Delta E = 13.6 \left(\frac{1}{n^2} - \frac{1}{m^2}\right) \, \text{eV} \)

Thus, the energy difference is the energy of the photon emitted during the transition.

The energy of the emitted photon is related to its frequency \( \nu \) by Planck's relation:

\( E = h\nu \)

Where: \( E \) is the energy of the photon. \( h \) is Planck’s constant (\(6.626 \times 10^{-34} \, \text{J·s}\)). \( \nu \) is the frequency of the emitted radiation.

The frequency \( \nu \) is related to the wavelength \( \lambda \) by the equation:

\( c = \lambda \nu \)

Where: \( c \) is the speed of light in a vacuum (\(3 \times 10^8 \, \text{m/s}\)). \( \lambda \) is the wavelength of the emitted radiation.

Rearranging for \( \nu \):

\( \nu = \frac{c}{\lambda} \)

Now, equating the two expressions for the energy of the photon:

\( h\nu = 13.6 \left(\frac{1}{n^2} - \frac{1}{m^2}\right) \)

Substitute \( \nu = \frac{c}{\lambda} \) into the equation:

\( h \left(\frac{c}{\lambda}\right) = 13.6 \left(\frac{1}{n^2} - \frac{1}{m^2}\right) \)

Now, solving for \( \frac{1}{\lambda} \):

\( \frac{hc}{\lambda} = 13.6 \left(\frac{1}{n^2} - \frac{1}{m^2}\right) \)

Finally, the expression for \( \frac{1}{\lambda} \) becomes:

\( \frac{1}{\lambda} = R \left(\frac{1}{n^2} - \frac{1}{m^2}\right) \)

Where \( R \) is the Rydberg constant, which is defined as:

\( R = \frac{13.6}{hc} \)

Conclusion

Atomic spectra, especially the hydrogen line spectrum, provide crucial insights into the energy levels of electrons and their transitions. These spectral lines obey specific relations, which can be used to calculate the wavelengths of emitted radiation. Understanding these spectral lines is fundamental to quantum mechanics and atomic physics.

Bohr’s Atomic Model

6. Bohr’s Atomic Model

Niels Bohr modified Rutherford’s model by applying the principles of quantum physics, which were being developed during that time. He realized that Rutherford’s model is essentially correct but lacks stability for electron orbits. Additionally, electrons in these stable orbits should not emit electromagnetic radiation, as required by classical (Maxwell’s) electromagnetic theory.

Bohr made three important postulates that defined his atomic model:

• Postulate 1: Electrons revolve around the nucleus in circular orbits. This assumption is the same as in Rutherford’s model, where the centripetal force necessary for the circular motion is provided by the electrostatic force of attraction between the electron and the nucleus.
• Postulate 2: The radius of the orbit of an electron can only take certain fixed values such that the angular momentum of the electron in these orbits is an integral multiple of \( \frac{h}{2\pi} \), where \( h \) is Planck’s constant.
• Postulate 3: The electron does not emit electromagnetic radiation while it is revolving in these stable orbits. Only when the electron jumps from one orbit to another does it emit or absorb electromagnetic radiation of a specific frequency corresponding to the difference in energy between the orbits.

Radii of the Orbits

Using the first two postulates, we can study the entire dynamics of the electron’s circular motion, including its energy. Let:

• \( m_e \) be the mass of the electron.
• \( v_n \) be the velocity of the electron in the \(n\)-th stable orbit.
• \( r_n \) be the radius of the \(n\)-th orbit.

The angular momentum of the electron is \( m_e v_n r_n \), and according to the second postulate, this angular momentum is quantized and equal to an integer multiple of \( \frac{h}{2\pi} \). Therefore, we can write:

\( m_e v_n r_n = n \frac{h}{2\pi} \quad \text{(Equation 1)}\)

The centripetal force necessary for the circular motion of the electron is provided by the electrostatic force between the electron and the nucleus. If the atomic number of the atom is \( Z \), then the total positive charge on the nucleus is \( Ze \), where \( e \) is the charge of the electron. The electrostatic force is given by Coulomb's law:

\( \frac{Z e^2}{4\pi \epsilon_0 r_n^2} \text{(Equation 2)} \)

where \( \epsilon_0 \) is the permeability of free space. Using the centripetal force formula \( m_e v_n^2 / r_n \), we equate the centripetal force and the electrostatic force, giving the equation:

\( m_e \frac{v_n^2}{r_n} = \frac{Z e^2}{4\pi \epsilon_0 r_n^2} \quad \text{(Equation 3)}\)

We now eliminate \( v_n \) from Equation (2) and Equation (3). Substituting \( v_n \) from Equation (15.2) into Equation (15.3), we get:

\( m_e n \frac{h}{2\pi} \frac{v_n}{r_n^2} = \frac{Z e^2}{4\pi \epsilon_0 r_n^2} \)

Simplifying this expression, we obtain:

\( r_n = \frac{n^2 h^2 \epsilon_0}{\pi m_e Z e^2} \quad \text{(Equation 4)}\)

This equation shows that the radius of the orbit is proportional to \( n^2 \), i.e., the square of the principal quantum number. As \( n \) increases, the radius increases.

For the hydrogen atom, \( Z = 1 \). Substituting the constants into Equation (15.4), we get the radius for \( n = 1 \) (the smallest orbit), known as the Bohr radius, \( r_1 \), as:

\( r_1 = 0.053 \, \text{nm} \)

This is the Bohr radius, denoted by \( a_0 \), and it represents the radius of the smallest orbit of the electron in a hydrogen atom.

Velocity of the Electron

Similarly, by eliminating \( r_n \) from Equation (15.2) and Equation (15.3), we can derive the velocity \( v_n \) of the electron in the \(n\)-th orbit. We obtain:

\( v_n = \frac{Z e^2}{2\epsilon_0 h m_e n} \quad \text{(Equation 5)}\)

From Equation (15.5), we see that the electron’s velocity in the orbit decreases as \( n \) increases, which corresponds to the electron being farther from the nucleus and moving more slowly in higher orbits.

Energy of the Electron

The total energy of an electron in the \(n\)-th orbit is the sum of its kinetic and potential energy. The kinetic energy \( K \) is given by:

\( K = \frac{1}{2} m_e v_n^2 \)

The potential energy \( U \) due to electrostatic attraction is given by Coulomb’s law:

\( U = - \frac{Z e^2}{4 \pi \epsilon_0 r_n} \)

The total energy \( E \) is then:

\( E = K + U = - \frac{Z e^2}{8 \pi \epsilon_0 r_n} \)

Substituting \( r_n \) from Equation (15.4), we get the total energy of the electron as:

\( E = - \frac{Z^2 e^4 m_e}{8 h^2 \epsilon_0^2 n^2} \)

Conclusion

Bohr’s model successfully explained the quantization of electron orbits in hydrogen and the observed spectral lines. The radius of the electron’s orbit is proportional to \( n^2 \), and the velocity of the electron in each orbit is inversely proportional to \( n \). The derivation of these equations provides insight into the energy levels and the stability of the hydrogen atom. Furthermore, the third postulate introduced the idea of discrete energy levels and the emission or absorption of radiation when electrons transition between these levels.

Electron Energy and Atomic Energy Levels

The negative energy of electrons indicates that the electron is bound to the nucleus. Energy must be supplied to free the electron, making the total energy zero.

The energy of an electron in an atom is given by:

\[ E = -\frac{13.6 Z^2}{n^2} \ \text{eV} \quad \text{---(15.8)} \]

Where \( Z \) is the atomic number and \( n \) is the principal quantum number.

Ground and Excited States

The first orbit ( n = 1) which has minimum energy, is called the ground state of the atom.
• Ground State (\(n = 1\)): Energy is minimum at \(-13.6 \ \text{eV}\) for hydrogen.
• Excited States (\(n > 1\)): Energy increases as \( n \) increases, e.g., \(-3.4 \ \text{eV}\) for \(n = 2\).

As \( n \to \infty \), the energy approaches zero, representing a free electron. Orbits with higher values of n and therefore, higher values of energy are called the excited states of the atom.

Excitation and Ionization Energies

The ionization energy of an atom is the minimum amount of energy required to be given to an electron in the ground state of that atom to set the electron free. It is the binding energy of hydrogen atom

The energy required to move an electron between states is:

• Excitation Energy: For \( n = 2 \):

\[ E_{\text{excitation}} = (-3.4) - (-13.6) = 10.2 \ \text{eV} \]

• Ionization Energy: Energy required to completely remove the electron:

\[ E_{\text{ionization}} = 13.6 \ \text{eV} \]

Bohr’s Third Postulate

An electron transitioning between orbits emits or absorbs energy as a photon:

\[ E_{\text{photon}} = E_m - E_n = h\nu \]

Expressed in terms of wavelength:

\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n^2} - \frac{1}{m^2} \right) \quad \text{---(15.11)} \]

Here \( R_H = 1.097 \times 10^7 \ \text{m}^{-1} \) is the Rydberg constant.

Limitations of Bohr’s Model

• Limited Applicability: Only explains hydrogen spectra.
• Spectral Line Intensity: Cannot explain variations in spectral line intensity.
• Lack of Theoretical Basis: Stability of orbits is assumed without explanation.

De Broglie’s Explanation

De Broglie proposed the wave-particle duality for electrons, where electron orbits are standing waves:

\[ 2\pi r_n = n\lambda \quad \text{(for \( n = 1, 2, 3,... \))} \]

Using \( \lambda = \frac{h}{p} \), the angular momentum becomes:

\[ L_n = m_e v_n r_n = \frac{nh}{2\pi} \]

This aligns with Bohr’s postulates, providing a theoretical foundation.

Atomic Nucleus - Detailed Notes

15.7. Atomic Nucleus

The atomic nucleus is the central part of an atom and contains most of its mass. It is made up of subatomic particles, which are smaller than an atom. Let’s study the structure and properties of the atomic nucleus in detail.

15.7.1 Constituents of a Nucleus

1. Protons and Neutrons (Nucleons):

• The nucleus is composed of two types of particles: protons and neutrons.
• Together, protons and neutrons are called nucleons.
• Protons are positively charged particles. The charge of a proton is equal in magnitude to the charge of an electron but opposite in sign.
• Neutrons are electrically neutral particles, meaning they have no charge.
• The mass of a proton is approximately \(1836\) times the mass of an electron.
• The mass of a neutron is slightly greater than that of a proton.

2. Atomic Number (Z):

• The number of protons in the nucleus of an atom is called the atomic number (Z).
• The atomic number determines the identity of an element. For example:
  • Hydrogen has 1 proton, so its atomic number is 1.
  • Carbon has 6 protons, so its atomic number is 6.
• The number of electrons in a neutral atom is also equal to the atomic number (Z). This ensures that the atom is electrically neutral because the positive charge of protons is balanced by the negative charge of electrons.

3. Neutron Number (N):

• The number of neutrons in the nucleus is denoted by N.
• Neutrons contribute to the mass of the nucleus but do not affect the atomic number or the chemical properties of the element.

4. Mass Number (A):

• The total number of nucleons (protons + neutrons) in the nucleus is called the mass number (A).
• It is calculated as: \[ A = Z + N \]
• The mass number determines the mass of the nucleus and the atom.

5. Representation of Atoms:

• Atoms are represented using the symbol of the element, along with their atomic number (Z) and mass number (A). The general notation is: \[ ^A_Z X \]
  • X = Symbol of the element
  • A = Mass number
  • Z = Atomic number
• Examples:
  • Hydrogen: \(^1_1 H\)
  • Carbon: \(^{12}_6 C\)
  • Oxygen: \(^{16}_8 O\)

Isotopes, Isobars, and Isotones

1. Isotopes:

• Isotopes are atoms of the same element that have the same number of protons (same Z) but different numbers of neutrons (different N).
• Example: Hydrogen has three isotopes:
  • \(^1_1 H\) (Hydrogen): 1 proton, 0 neutrons
  • \(^2_1 H\) (Deuterium): 1 proton, 1 neutron
  • \(^3_1 H\) (Tritium): 1 proton, 2 neutrons
• Isotopes have the same chemical properties but different physical properties (e.g., mass).

2. Isobars:

• Isobars are atoms of different elements that have the same mass number (A) but different atomic numbers (Z).
• Example:
  • \(^3_1 H\) (Tritium) and \(^3_2 He\) (Helium-3) are isobars because both have a mass number of 3.

3. Isotones:

• Isotones are atoms of different elements that have the same number of neutrons (N) but different atomic numbers (Z).
• Example:
  • \(^3_1 H\) (Tritium) and \(^4_2 He\) (Helium-4) are isotones because both have 2 neutrons.

Units for Measuring Masses of Atoms and Subatomic Particles

The masses of atoms and subatomic particles are extremely small, so special units are used to measure them:

1. Kilogram (kg):

• The standard unit of mass is the kilogram, but it is not convenient for measuring atomic masses.
• Masses of subatomic particles in kg:
  • Electron (\(m_e\)): \(9.109383 \times 10^{-31} \, \text{kg}\)
  • Proton (\(m_p\)): \(1.672623 \times 10^{-27} \, \text{kg}\)
  • Neutron (\(m_n\)): \(1.674927 \times 10^{-27} \, \text{kg}\)

2. Unified Atomic Mass Unit (u):

• A more convenient unit for atomic masses is the unified atomic mass unit (u).
• 1 u is defined as \(\frac{1}{12}\)th of the mass of a carbon-12 atom (\(^{12}_6 C\)).
• 1 u = \(1.6605402 \times 10^{-27} \, \text{kg}\).
• Masses of subatomic particles in u:
  • Electron (\(m_e\)): \(0.00055 \, \text{u}\)
  • Proton (\(m_p\)): \(1.007825 \, \text{u}\)
  • Neutron (\(m_n\)): \(1.008665 \, \text{u}\)

3. Energy Equivalent (MeV/c²):

• According to Einstein’s mass-energy equivalence formula (\(E = mc^2\)), mass can be expressed in terms of energy.
• The unit used is MeV/c², where:
  • 1 u = \(931.5 \, \text{MeV/c}^2\).
• Masses of subatomic particles in MeV/c²:
  • Electron (\(m_e\)): \(0.511 \, \text{MeV/c}^2\)
  • Proton (\(m_p\)): \(938.28 \, \text{MeV/c}^2\)
  • Neutron (\(m_n\)): \(939.57 \, \text{MeV/c}^2\)

Key Points to Remember

• The nucleus is made up of protons and neutrons, collectively called nucleons.
• The atomic number (Z) is the number of protons in the nucleus.
• The mass number (A) is the total number of nucleons (protons + neutrons).
• Isotopes have the same Z but different N.
• Isobars have the same A but different Z.
• Isotones have the same N but different Z.
• Masses of atoms and subatomic particles are measured in kg, u, or MeV/c².

Practice Questions

1. Define atomic number and mass number. How are they related?
2. What are isotopes? Give two examples.
3. Calculate the number of protons, neutrons, and electrons in \(^{16}_8 O\).
4. Convert the mass of a proton from kg to MeV/c².
5. Explain the difference between isobars and isotones with examples.

Sizes of Nuclei and Nuclear Forces - Detailed Notes

15.7.2 Sizes of Nuclei

The size of an atom is determined by the sizes of the electron orbits. Similarly, the size of a nucleus depends on the number of nucleons (protons and neutrons) present in it, i.e., its mass number \(A\).

Radius of a Nucleus:

• The radius \(R_X\) of a nucleus \(X\) is related to its mass number \(A\) by the formula: \[ R_X = R_0 A^{1/3} \] where \(R_0 = 1.2 \times 10^{-15} \, \text{m}\).
• This formula shows that the radius of a nucleus increases with the cube root of its mass number.

Density of a Nucleus:

• The density \(\rho\) inside a nucleus is given by: \[ \rho = \frac{\text{Mass of Nucleus}}{\text{Volume of Nucleus}} = \frac{mA}{\frac{4}{3} \pi R_X^3} \] where \(m\) is the average mass of a nucleon (proton or neutron).
• Substituting \(R_X = R_0 A^{1/3}\), the density becomes: \[ \rho = \frac{3m}{4 \pi R_0^3} \]
• This shows that the density of a nucleus is constant and does not depend on the atomic number \(A\).
• The calculated nuclear density is approximately \(2.3 \times 10^{17} \, \text{kg/m}^3\), which is extremely high.
• For comparison, the density of osmium (the densest known element) is only \(2.2 \times 10^4 \, \text{kg/m}^3\), which is 13 orders of magnitude smaller than nuclear density.

15.7.3 Nuclear Forces

Nuclear forces are the strong forces that hold protons and neutrons together in the nucleus. These forces are fundamentally different from electrostatic and gravitational forces.

Properties of Nuclear Forces:

1. Strength:
   • Nuclear forces are the strongest among subatomic particles.
   • They are about 50-60 times stronger than electrostatic forces.
2. Range:
   • Nuclear forces have a very short range, typically a few femtometers (fm).
   • Beyond this range, the force becomes negligible.
3. Charge Independence:
   • Nuclear forces are independent of the charge of nucleons.
   • The force between two neutrons is the same as that between two protons or between a neutron and a proton at the same separation.

Role of Nuclear Forces:

• Protons in the nucleus repel each other due to their positive charges (electrostatic repulsion).
• Nuclear forces counteract this repulsion and bind nucleons together, ensuring the stability of the nucleus.
• Nuclear forces are much stronger than electrostatic forces at short distances, which is why nuclei remain stable despite proton repulsion.

Nuclear Binding Energy

Nuclear binding energy is the energy required to separate the nucleons (protons and neutrons) in a nucleus or the energy released when nucleons come together to form a nucleus.

Mass Defect:

• The mass of a nucleus is always less than the sum of the masses of its individual nucleons.
• This difference in mass is called the mass defect (\(\Delta M\)): \[ \Delta M = (Zm_p + Nm_n) - M \] where:
  • \(Z\) = Number of protons
  • \(N\) = Number of neutrons
  • \(m_p\) = Mass of a proton
  • \(m_n\) = Mass of a neutron
  • \(M\) = Mass of the nucleus

Binding Energy Formula:

• The binding energy (\(E_B\)) of a nucleus is given by: \[ E_B = \Delta M \cdot c^2 \] where \(c\) is the speed of light.
• This energy is equivalent to the mass defect according to Einstein's mass-energy relation (\(E = mc^2\)).

Binding Energy per Nucleon:

• The binding energy per nucleon (\(E_B/A\)) is a measure of the stability of a nucleus: \[ \frac{E_B}{A} = \frac{\text{Binding Energy}}{\text{Mass Number}} \]
• Nuclei with higher binding energy per nucleon are more stable.
• The binding energy per nucleon varies with the mass number \(A\):
  • It is minimum for deuterium (\(^2_1 H\)).
  • It reaches a peak around \(A = 56\) (iron), making iron one of the most stable nuclei.
  • For \(A > 80\), the binding energy per nucleon decreases gradually, making heavier nuclei less stable.

Key Points to Remember

• The radius of a nucleus is proportional to the cube root of its mass number: \(R_X = R_0 A^{1/3}\).
• Nuclear density is constant and extremely high (\(2.3 \times 10^{17} \, \text{kg/m}^3\)).
• Nuclear forces are the strongest forces, with a short range and charge independence.
• Binding energy is the energy required to separate nucleons in a nucleus.
• Binding energy per nucleon determines the stability of a nucleus.

Practice Questions

1. Calculate the radius of a nucleus with \(A = 27\).
2. Explain why nuclear density is constant for all nuclei.
3. Compare the strength and range of nuclear forces with electrostatic forces.
4. What is mass defect? How is it related to binding energy?
5. Why is iron considered one of the most stable nuclei?

Radioactive Decays

Many nuclei in nature are stable, meaning they can remain unchanged for a very long time. These nuclei have a specific ratio of the mass number (\(A\)) to the atomic number (\(Z\)). However, some nuclei are unstable and undergo changes in their structure by emitting particles. This process is called radioactive decay or radioactivity. It was discovered by Henri Becquerel in 1876.

During radioactive decay, the unstable parent nucleus transforms into a more stable daughter nucleus by emitting particles. The parent nucleus is the original nucleus, and the daughter nucleus is the resulting nucleus after the decay. There are three main types of radioactive decay:

1. Alpha Decay

In alpha decay, the parent nucleus emits an alpha particle, which is essentially the nucleus of a helium atom (\(^4_2\text{He}\)). The parent nucleus loses two protons and two neutrons in this process. The general equation for alpha decay is:

\[ ^A_Z\text{X} \rightarrow ^{A-4}_{Z-2}\text{Y} + ^4_2\text{He} \]

Here, \(X\) is the parent nucleus, and \(Y\) is the daughter nucleus. All nuclei with \(A > 210\) undergo alpha decay because they have a large number of protons. The electrostatic repulsion between these protons is very strong, and the attractive nuclear forces cannot balance it. This makes the nucleus unstable, and it ejects an alpha particle to become more stable.

Example: The alpha decay of bismuth (\(^{212}_{83}\text{Bi}\)) can be written as:

\[ ^{212}_{83}\text{Bi} \rightarrow ^{208}_{81}\text{Tl} + ^4_2\text{He} \]

The total mass of the products (daughter nucleus and alpha particle) is always less than the mass of the parent nucleus. The difference in mass is converted into kinetic energy of the products. This energy is called the Q-value of the decay and is calculated as:

\[ Q = [m_X - m_Y - m_{\text{He}}]c^2 \]

Here, \(m_X\), \(m_Y\), and \(m_{\text{He}}\) are the masses of the parent nucleus, daughter nucleus, and helium nucleus, respectively. \(c\) is the speed of light.

2. Beta Decay

In beta decay, the nucleus emits an electron (\(e^-\)) or a positron (\(e^+\)). There are two types of beta decay:

a) Beta Minus Decay (\(\beta^-\))

In \(\beta^-\) decay, a neutron in the nucleus is converted into a proton, emitting an electron and an antineutrino (\(\bar{\nu}\)). The general equation is:

\[ n \rightarrow p + e^- + \bar{\nu} \]

The mass number (\(A\)) remains unchanged, but the atomic number (\(Z\)) increases by 1. The decay can be written as:

\[ ^A_Z\text{X} \rightarrow ^A_{Z+1}\text{Y} + e^- + \bar{\nu} \]

b) Beta Plus Decay (\(\beta^+\))

In \(\beta^+\) decay, a proton in the nucleus is converted into a neutron, emitting a positron (\(e^+\)) and a neutrino (\(\nu\)). The general equation is:

\[ p \rightarrow n + e^+ + \nu \]

The mass number (\(A\)) remains unchanged, but the atomic number (\(Z\)) decreases by 1. The decay can be written as:

\[ ^A_Z\text{X} \rightarrow ^A_{Z-1}\text{Y} + e^+ + \nu \]

Note: In beta decay, the total mass of the products is less than the mass of the parent nucleus. The difference in mass is converted into kinetic energy of the products. The Q-value for beta decay is calculated as:

\[ Q = [m_X - m_Y - m_e]c^2 \]

Here, \(m_e\) is the mass of the electron or positron. The mass of the neutrino is negligible and is ignored in the calculation.

Key Points to Remember

• Radioactive decay occurs because unstable nuclei transform into more stable nuclei by emitting particles.
• Alpha decay involves the emission of an alpha particle (\(^4_2\text{He}\)).
• Beta decay involves the emission of an electron (\(\beta^-\)) or a positron (\(\beta^+\)).
• The Q-value represents the energy released during the decay and is calculated using the mass difference between the parent and daughter nuclei.

Gamma Decay and Radioactive Decay

Gamma Decay

In gamma decay, gamma rays are emitted by the parent nucleus. A gamma ray is a high-energy photon. Unlike alpha and beta decay, the daughter nucleus in gamma decay is the same as the parent nucleus because no particles are emitted. However, the daughter nucleus has less energy as some energy is released in the form of the emitted gamma ray.

Just like electrons in an atom are arranged in different energy levels (orbits), nucleons (protons and neutrons) in a nucleus also occupy energy levels with different energies. A nucleon can transition from a higher energy level to a lower energy level, emitting a photon (gamma ray) in the process.

The key difference between atomic and nuclear energy levels lies in their energies:

• Atomic energy levels are of the order of a few \( \text{eV} \) (electron volts).
• Nuclear energy levels are of the order of a few \( \text{keV} \) to a few \( \text{MeV} \) (kilo-electron volts to mega-electron volts).

As a result, radiations emitted by atoms are in the ultraviolet to radio region, while radiations emitted by nuclei are in the gamma-ray range.

Note: Nucleons in a nucleus are usually in the lowest possible energy state. Excitation requires a large amount of energy (in \( \text{keV} \) or \( \text{MeV} \)). However, a nucleon may end up in an excited state after the parent nucleus undergoes alpha or beta decay.

Example of Gamma Decay

Consider the decay of \( ^{57}\text{Co} \) (Cobalt-57). It undergoes beta-plus decay to form the daughter nucleus \( ^{57}\text{Fe} \) (Iron-57), which is in an excited state with an energy of \( 136 \, \text{keV} \). There are two ways it can transition to its ground state:

1. Emit a gamma ray of energy \( 136 \, \text{keV} \) directly to the ground state.
2. Emit a gamma ray of energy \( 122 \, \text{keV} \), transitioning to an intermediate state, and then emit another gamma ray of energy \( 14 \, \text{keV} \) to reach the ground state.

Both these emissions have been observed experimentally.

Series Decay

Often, the daughter nucleus is not stable and undergoes further decay. A chain of decays may occur until a stable nucleus is formed. For example, \( ^{238}\text{U} \) (Uranium-238) undergoes a series of 14 alpha and beta decays to finally reach a stable daughter nucleus of \( ^{206}\text{Pb} \) (Lead-206).

Law of Radioactive Decay

Materials that undergo alpha, beta, or gamma decay are called radioactive materials. The nature of radioactivity is such that we cannot predict exactly when a single atom will decay. However, we can describe the decay process statistically.

Let \( N_0 \) be the number of radioactive atoms at time \( t = 0 \). At any time \( t \), the number of remaining parent nuclei is \( N(t) \). The number of decays \( dN \) in a small time interval \( dt \) is proportional to \( N(t) \) and \( dt \):

\[ dN = -\lambda N(t) \, dt \]

Here, \( \lambda \) is the decay constant. The negative sign indicates that \( N(t) \) decreases with time.

Integrating this equation gives the law of radioactive decay:

\[ N(t) = N_0 e^{-\lambda t} \]

The activity \( A(t) \), which is the rate of decay, is given by:

\[ A(t) = \frac{dN}{dt} = \lambda N(t) = \lambda N_0 e^{-\lambda t} \]

At \( t = 0 \), the activity is \( A_0 = \lambda N_0 \). Thus:

\[ A(t) = A_0 e^{-\lambda t} \]

Units of Activity:

• Becquerel (Bq): 1 decay per second.
• Curie (Ci): \( 1 \, \text{Ci} = 3.7 \times 10^{10} \, \text{Bq} \).

Detailed Notes on Atoms

Table of Contents

• 15.10.1 Half-life of Radioactive Material
• 15.10.2 Average Life of a Radioactive Species
• 15.11 Nuclear Energy
• 15.11.1 Nuclear Fission

15.10.1 Half-life of Radioactive Material

The time taken for the number of parent radioactive nuclei of a particular species to reduce to half its value is called the half-life (\( T_{1/2} \)) of the species. This can be obtained from the equation:

\[ N = N_0 e^{-\lambda t} \]

At \( t = T_{1/2} \), the number of nuclei \( N \) becomes \( \frac{N_0}{2} \). Substituting this into the equation, we get:

\[ \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \]

Simplifying, we find:

\[ T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda} \]

The interesting thing about half-life is that even though the number of nuclei goes down from \( N_0 \) to \( \frac{N_0}{2} \) in time \( T_{1/2} \), after another time interval \( T_{1/2} \), the number of parent nuclei will not go to zero. It will go to half of the value at \( t = T_{1/2} \), i.e., to \( \frac{N_0}{4} \). Thus, in a time interval equal to half-life, the number of parent nuclei reduces by a factor of \( \frac{1}{2} \).

Key Points:

• Half-life (\( T_{1/2} \)) is the time taken for half of the radioactive nuclei to decay.
• It is calculated using \( T_{1/2} = \frac{\ln 2}{\lambda} \).
• After each half-life, the number of nuclei reduces by half.

15.10.2 Average Life of a Radioactive Species

We have seen that different nuclei of a given radioactive species decay at different times, i.e., they have different lifetimes. We can calculate the average lifetime (\( \tau \)) of a nucleus of the material using the equation:

\[ \tau = \frac{1}{\lambda} \]

The relation between the average life and half-life can be obtained using the equation:

\[ T_{1/2} = \tau \ln 2 \approx 0.693 \tau \]

Key Points:

• Average lifetime (\( \tau \)) is the reciprocal of the decay constant (\( \lambda \)).
• It is related to half-life by \( T_{1/2} = 0.693 \tau \).

15.11 Nuclear Energy

You are familiar with the naturally occurring, conventional sources of energy. These include the fossil fuels, i.e., coal, petroleum, natural gas, and firewood. The energy generation from these fuels is through chemical reactions. It takes millions of years for these fuels to form. Naturally, the supply of these conventional sources is limited and with indiscriminate use, they are bound to get over in a couple of hundred years from now. Therefore, we have to use alternative sources of energy. The ones already in use are hydroelectric power, solar energy, wind energy, and nuclear energy, nuclear energy being the largest source among these.

Nuclear energy is the energy released when nuclei undergo a nuclear reaction, i.e., when one nucleus or a pair of nuclei, due to their interaction, undergo a change in their structure resulting in new nuclei and generating energy in the process. While the energy generated in chemical reactions is of the order of few eV per reaction, the amount of energy released in a nuclear reaction is of the order of a few MeV. Thus, for the same weight of fuel, the nuclear energy released is about a million times that released through chemical reactions. However, nuclear energy generation is a very complex and expensive process and it can also be extremely harmful. Let us learn more about it.

15.11.1 Nuclear Fission

We have seen that the binding energy per nucleon (\( E_B/A \)) depends on the mass number of the nuclei. This quantity is a measure of the stability of the nucleus. As seen from the figure, the middle-weight nuclei (mass number ranging from 50 to 80) have the highest binding energy per nucleon and are most stable, while nuclei with higher and lower atomic masses have smaller values of \( E_B/A \). The value of \( E_B/A \) goes on decreasing till \( A \approx 238 \), which is the mass number of the heaviest naturally occurring element, uranium. Many of the heavy nuclei are unstable and decay into two smaller mass nuclei.

Let us consider a case when a heavy nucleus, say with \( A \approx 230 \), breaks into two nuclei having \( A \) between 50 and 150. The \( E_B/A \) of the product nuclei will be higher than that of the parent nucleus. This means that the combined masses of the two product nuclei will be smaller than the mass of the parent nucleus. The difference in the mass of the parent nucleus and that of the product nuclei taken together will be released in the form of energy in the process. This process in which a heavy nucleus breaks into two lighter nuclei with the release of energy is called nuclear fission and is a source of nuclear energy.

One of the nuclei used in nuclear energy generation by fission is \( ^{236}_{92}U \). This has a half-life of \( 2.3 \times 10^7 \) years and an activity of \( 6.5 \times 10^{-5} \) Ci/g. However, it being fissionable, most of its nuclei have already decayed and it is not found in nature. More than 99% of natural uranium is in the form of \( ^{238}_{92}U \) and less than 1% is in the form of \( ^{235}_{92}U \). \( ^{238}_{92}U \) also decays, but its half-life is about \( 10^3 \) times higher than that of \( ^{236}_{92}U \) and is therefore not very useful for energy generation. The species needed for nuclear energy generation, i.e., \( ^{236}_{92}U \), can be obtained from the naturally occurring \( ^{235}_{92}U \) by bombarding it with slow neutrons. \( ^{235}_{92}U \) absorbs a neutron and yields \( ^{236}_{92}U \). This reaction can be written as:

\[ ^{235}_{92}U + n \rightarrow ^{236}_{92}U \]

\( ^{236}_{92}U \) can undergo fission in several ways producing different pairs of daughter nuclei and generating different amounts of energy in the process. Some of its decays are:

\[ ^{236}_{92}U \rightarrow ^{137}_{53}I + ^{97}_{39}Y + 2n \] \[ ^{236}_{92}U \rightarrow ^{140}_{56}Ba + ^{94}_{36}Kr + 2n \] \[ ^{236}_{92}U \rightarrow ^{133}_{51}Sb + ^{99}_{41}Nb + 4n \]

Some of the daughter nuclei produced are not stable and they further decay to produce more stable nuclei. The energy produced in the fission is in the form of kinetic energy of the products, i.e., in the form of heat which can be collected and converted to other forms of energy as needed.

Key Points:

• Nuclear fission involves the splitting of a heavy nucleus into smaller nuclei, releasing energy.
• \( ^{235}_{92}U \) is commonly used in nuclear reactors for energy generation.
• The energy released is in the form of kinetic energy of the fission products.

Nuclear Reactors and Fusion - Detailed Notes

1. Uranium Nuclear Reactor

A nuclear reactor is a device designed to carry out nuclear fission in a controlled manner. The primary purpose of a nuclear reactor is to produce energy in the form of heat, which is then converted into electricity. In a uranium reactor, the isotope Uranium-235 (²³⁵U) is used as the fuel. When Uranium-235 is bombarded by slow neutrons, it absorbs a neutron and becomes Uranium-236 (²³⁶U), which is highly unstable and undergoes fission.

Key Points:

• Uranium-235 is the fuel used in nuclear reactors.
• Slow neutrons (thermal neutrons) are used to initiate the fission process.
• The fission of Uranium-236 releases a large amount of energy.

1.1 Nuclear Fission Reaction

The fission of Uranium-235 can be represented by the following reaction:

²³⁵U + n → ²³⁶U → Fission Products + 2-3 Neutrons + Energy

During this process, the Uranium-236 nucleus splits into two smaller nuclei (called fission products), along with the release of 2-3 neutrons and a tremendous amount of energy (approximately 200 MeV per fission).

1.2 Chain Reaction

The neutrons released during fission can initiate further fission reactions if they are absorbed by other Uranium-235 nuclei. This creates a chain reaction. On average, about 2.7 neutrons are produced per fission reaction. If this process is not controlled, it can lead to an exponential increase in the number of fissions, resulting in a massive explosion.

Chain Reaction: ²³⁵U + n → ²³⁶U → Fission Products + 2.7n + Energy

In a nuclear reactor, the chain reaction is carefully controlled using control rods (made of materials like boron or cadmium) to absorb excess neutrons. This ensures that the reaction proceeds at a steady rate, producing a constant amount of energy.

Note: More than 15 countries use nuclear reactors to generate power. India has 22 nuclear reactors, with the largest one located in Kudankulam, Tamil Nadu. The USA generates the maximum nuclear power globally.

2. Nuclear Fusion

Nuclear fusion is the process in which two light nuclei combine to form a heavier nucleus, releasing a significant amount of energy in the process. Fusion occurs because the binding energy per nucleon (EB/A) of the product nucleus is higher than that of the reactant nuclei. This difference in binding energy is released as energy.

Key Points:

• Fusion occurs in light nuclei (A < 40).
• Energy is released because the product nucleus has a higher binding energy per nucleon.
• For fusion to occur, nuclei must come within 1 femtometer (fm) of each other.

2.1 Conditions for Nuclear Fusion

For nuclear fusion to take place, two nuclei must overcome the electrostatic repulsion between their positive charges. This requires extremely high temperatures (millions of degrees Celsius) to provide the nuclei with enough kinetic energy to come close together. At such high temperatures, matter exists in the form of plasma, where electrons are stripped away from atoms, leaving behind bare nuclei.

Example of Fusion Reaction: ²H + ³H → ⁴He + n + 17.6 MeV

In this reaction, deuterium (²H) and tritium (³H) fuse to form helium (⁴He), releasing a neutron and 17.6 MeV of energy.

2.2 Fusion in Stars

Nuclear fusion is the primary source of energy in stars, including our Sun. At the core of the Sun, hydrogen nuclei (protons) fuse to form helium in a series of steps known as the proton-proton chain reaction. The overall reaction can be summarized as:

4p → ⁴He + 2e⁺ + 2ν + 26.7 MeV

Here, four protons (hydrogen nuclei) combine to form a helium nucleus, releasing two positrons (e⁺), two neutrinos (ν), and 26.7 MeV of energy. This reaction has been ongoing in the Sun for about 4.5 billion years and will continue for a similar period in the future.

Note: Elements heavier than boron up to iron are formed through nuclear fusion in stars.

3. Harmful Effects of Nuclear Energy

While nuclear energy is a powerful source of electricity, it also has harmful effects if not controlled properly. An uncontrolled chain reaction can lead to the release of an enormous amount of energy in a short time, resulting in a nuclear explosion. This principle is used in atomic bombs, which rely on either fission or a combination of fission and fusion.

Historical Context:

• The first atomic bombs were developed by the USA during World War II.
• Two atomic bombs were dropped on Hiroshima and Nagasaki in Japan in August 1945, causing massive destruction.
• Today, several countries, including India, have developed nuclear weapons.

It is essential to use nuclear energy responsibly to avoid catastrophic consequences. Proper safety measures and regulations are crucial in the operation of nuclear reactors and the handling of nuclear materials.