PhysicsKund Physics class 9 to 12 NCERT, JEE - NEET

Definition of Nuclear Fission  Nuclear fission is the process in which a heavy nucleus splits into two or more lighter nuclei when bombarded by a neutron. A large amount of energy and neutrons are released during this process. Example of Nuclear Fission When a slow neutron strikes a Uranium-235 nucleus, it first forms an unstable Uranium-236 nucleus. This unstable nucleus then splits into two lighter nuclei. Nuclear Reaction $${}^{1}_{0}n + {}^{235}_{92}U \rightarrow {}^{236}_{92}U$$ $${}^{236}_{92}U \rightarrow {}^{144}_{56}Ba + {}^{89}_{36}Kr + 3\,{}^{1}_{0}n + \text{Energy}$$ Important Points About Nuclear Fission Uranium-235 is a fissile material. Nuclear fission is a neutron-induced reaction. Two or more lighter nuclei are formed. 2 to 4 neutrons are released. The fission fragments are radioactive. They emit beta particles and eventually become stable. Energy Released in Nuclear Fission The energy released during fission is called the Q-valu...

Notes : CBSE Class 12 Physics Chapter 13 Nuclei - Physics Kund Composition of the Nucleus of an Atom Atomic Mass Unit Proton: Definition, Discovery and Properties Neutron: Definition, Discovery and Properties Define Isotopes, Isobars and Isotones with Example Size of the Nucleus: Derivation of Radius and Nuclear Density Einstein Mass-Energy Relation Mass Defect and Nuclear Binding Energy Binding Energy per Nucleon and its Variation with Mass Number Nuclear Force Radioactivity Nuclear Fission Nuclear Fusion

Definition of Nuclear Fusion :  Nuclear fusion is the process in which two light nuclei combine to form a heavier nucleus. During this process, a large amount of energy is released because the resulting nucleus is more tightly bound. Examples of Fusion Reactions 1. Proton-Proton Reaction 1 H + 1 H → 2 H + e + + ν + 0.42 MeV Two hydrogen nuclei (protons) combine to form deuterium, a positron, and a neutrino, releasing 0.42 MeV energy. 2. Deuterium-Deuterium Reaction 2 H + 2 H → 3 He + n + 3.27 MeV Two deuterium nuclei combine to form helium-3 and a neutron, releasing 3.27 MeV energy. 3. Another Deuterium Fusion Reaction 2 H + 2 H → 3 H + 1 H + 4.03 MeV Two deuterium nuclei combine to form tritium and a proton, releasing 4.03 MeV energy. Why High Temperature is Required for Fusion? For fusion to occur, two nuclei must come very close so that the attractive nuclear force can act between them. Nuclear force is a short-range force and acts only at very...

Radioactivity is a nuclear phenomenon in which an unstable atomic nucleus spontaneously emits radiation and transforms into a more stable nucleus. This process is known as radioactive decay. Discovery of Radioactivity Radioactivity was discovered accidentally by A. H. Becquerel in 1896. While studying the fluorescence and phosphorescence of uranium compounds, he observed that uranium emitted invisible radiations capable of penetrating black paper and silver, causing the blackening of a photographic plate. What is Radioactivity? Radioactivity is the spontaneous disintegration of an unstable atomic nucleus accompanied by the emission of energetic particles or electromagnetic radiations. Key Characteristics of Radioactivity It is a spontaneous process. It is a nuclear phenomenon. It is independent of temperature and pressure. It is unaffected by chemical reactions. It results in the emission of α, β, or γ radiations. Types of Radioactive Decay 1. Alpha (α) Decay ...

Learn about Nuclear Force, its characteristics, short-range nature, charge independence, and potential energy curve with easy Class 12 Physics notes Definition: Nuclear Force  Nuclear force is the force that binds protons and neutrons (collectively called nucleons) inside the nucleus. Since positively charged protons repel each other due to the Coulomb force, a much stronger attractive force is required to hold the nucleus together. This force is known as the nuclear force . The binding energy per nucleon for medium-mass nuclei is approximately 8 MeV , which is much larger than the binding energy found in atoms. This indicates that nuclear force is extremely strong and plays a crucial role in maintaining nuclear stability. Characteristics of Nuclear Force 1. Nuclear Force is Very Strong Nuclear force is much stronger than both the Coulomb force and the gravitational force. It is strong enough to overcome the electrostatic repulsion between protons and keep nucleons bound ...

Binding Energy per Nucleon and the Binding Energy Curve Definition: Binding Energy per Nucleon The binding energy per nucleon is defined as the total binding energy of a nucleus divided by its mass number . $E_{bn}=\frac{E_b}{A}$ where:  $E_{b}$= Total binding energy of the nucleus  A= Mass number of the nucleus The binding energy per nucleon is a measure of the stability of a nucleus. A larger value of indicates a more stable nucleus. Binding Energy per Nucleon Curve The variation of binding energy per nucleon with mass number shows the following features: 1. The binding energy per nucleon increases rapidly for light nuclei. 2. It remains nearly constant at about 8 MeV per nucleon for nuclei with 30 < A < 170 3. The maximum value of binding energy per nucleon is about 8.75 MeV for A = 56 4. For very heavy nuclei, the binding energy per nucleon decreases gradually and becomes about 7.6 MeV for A = 238 5. Thus, light nuclei and heavy nuclei are less tightly bound tha...

Definition nuclear binding energy The total energy required to disintegrate the nucleus into its constituent particles (i.e. nucleons) is called nuclear binding energy or binding energy of the nucleus. Or Binding energy is basically the energy required to hold the nucleons in a nucleus.  Thus, the energy equivalent to mass defect is the binding energy of the nucleus. i.e. $\text{Binding energy, } E_b=\Delta mc^2$ where $ \Delta m$ is mass defect and c is velocity of light in vacuum. Since $\Delta m=\left[\{m_p Z+m_n (A-Z)\}-M\right]$ where $m_p$ is mass of proton, $m_n$ is mass of neutron, Z is atomic number and A is mass number. Therefore, $E_b=\left[\{m_pZ+m_n(A-Z)\}-M\right]c^2$ Binding energy is expressed in joule, if mass defect is measured in kg. When mass defect $\Delta m$ is expressed in a.m.u., then the binding energy is written as, $E_b=\left[\{m_pZ+m_n(A-Z)\}-M\right]\text{ a.m.u.}$ Since $1\ \text{a.m.u.}=931\ \text{MeV}$ $E_b=\left[\{m_pZ+m_n(A-Z)\}-M\right]\times93...

Learn Einstein's Mass Energy Equivalence Equation (E=mc²), its derivation , class 12 physics , chapter 13 nuclei Physicskund  Einstein's Mass Energy Equivalence Equation  Using Einstein's mass–energy equivalence relation: $E_T=mc^2$ where: $E_T$ = Total energy of the particle $m_0$ = Relativistic mass c = Speed of light in vacuum The total energy consists of rest-mass energy and kinetic energy: $E_T = m_0c^2 + K.E.$ Therefore, $mc^2 = m_0c^2 + K.E.$ From this, $K.E. = mc^2 - m_0c^2$ $K.E. = (m-m_0)c^2$ Let, $\Delta m = m-m_0$ Hence, $K.E.=\Delta m\,c^2$ Result: Kinetic energy of a particle = Change in mass × (speed of light) Einstein's mass–energy equivalence relation is verified by nuclear reactions. Numerical Example Calculate the energy equivalent to 1 g in eV. Given: $m = 1\,\text{g} = 10^{-3}\,\text{kg}$ $c = 3 \times 10^8 \,\text{m s}^{-1}$ Using $E = mc^2$ $E = 10^{-3}\times (3\times10^8)^2$ $E = 10^{-3}\times 9\times10^{16}$ $E = 9\times10^{13}\ \text{J}$ Since,...

Learn the derivation of the nuclear radius formula and nuclear density. Complete Class 12 Physics notes with formulas and explanations. SIZE OF THE NUCLEUS Expression For Nucleus Radius  The first experimental determination of the size of a nucleus was made from the results of Rutherford scattering of α-particles. Distance of closest approach was found to be $4 \times 10^{-14}$ m for 5.5 MeV energetic α-particles. This fact indicated that the size of nucleus should be less than $4 \times 10^{-14}$ m. For α-particles having kinetic energy more than 5.5 MeV, the distance of closest approach will be smaller. For kinetic energy more than 5.5 MeV, attractive nuclear forces start affecting the Coulomb's repulsive force between α-particles and gold nucleus. Size of nucleus can be measured by using fast electrons instead of α-particles for the scattering experiment. The size of nucleus was found to vary linearly with mass number (A). Since nucleus is supposed to be spherical, having radius...

Isotopes, Isobars and Isotones Isotopes Definition Isotopes are atoms of the same element having the same atomic number (Z) but different mass numbers (A) . Therefore, isotopes have the same number of protons and electrons but different numbers of neutrons. Example Hydrogen has three isotopes: $$^{1}_{1}\mathrm{H}$$ $$^{2}_{1}\mathrm{H}$$ $$^{3}_{1}\mathrm{H}$$ For Protium: $$p = 1,\quad e = 1,\quad n = 1 - 1 = 0$$ For Deuterium: $$p = 1,\quad e = 1,\quad n = 2 - 1 = 1$$ For Tritium: $$p = 1,\quad e = 1,\quad n = 3 - 1 = 2$$ Observation: All isotopes have the same number of protons and electrons but different numbers of neutrons. Isobars Definition Isobars are atoms of different elements having the same mass number (A) but different atomic numbers (Z) . Example Argon-40: $$^{40}_{18}\mathrm{Ar}$$ Calcium-40: $$^{40}_{20}\mathrm{Ca}$$ For Argon-40: $$p = 18,\quad e = 18,\quad n = 40 - 18 = 22$$ For Calcium-40: $$p = 20,\qu...

Definition of Proton A proton is a positively charged subatomic particle present in the nucleus of every atom. It carries one unit of positive charge and, together with neutrons, forms the nucleus of an atom. Symbol: $p$ or $^1_1H$ Charge: $+1.602 \times 10^{-19}\ \text{C}$ Mass: $1.6726 \times 10^{-27}\ \text{kg}$ Relative Mass: $\approx 1\ \text{a.m.u.}$ Discovery of Proton The proton was discovered by Ernest Rutherford in 1917 during his experiments on the bombardment of nitrogen gas with alpha particles. Rutherford's Experiment Rutherford bombarded nitrogen nuclei with high-speed alpha particles. He observed the emission of hydrogen nuclei, which were positively charged particles. Nuclear Reaction: $^{14}_{7}N + ^{4}_{2}He \rightarrow ^{17}_{8}O + ^{1}_{1}H$ or ${}^{14}_{7}\mathrm{N}+{}^{4}_{2}\mathrm{He}\rightarrow{}^{17}_{8}\mathrm{O}+{}^{1}_{1}\mathrm{H}$ From this experiment, Rutherford concluded that the hydrogen nucleus is a fundamental constituent of atomic nuclei and na...

Neutron: Definition, Discovery and Properties Definition of Neutron A neutron is a fundamental subatomic particle present in the nucleus of an atom. It has no electric charge and is represented by the symbol ${}_{0}^{1}n$ Discovery of Neutron The neutron was discovered by James Chadwick in 1932. Discovery Experiment Chadwick bombarded a beryllium (Be) target with α-particles emitted from polonium. Highly penetrating neutral radiations were produced. These radiations struck a paraffin wax block and ejected high-speed protons. From the conservation of energy and momentum, Chadwick concluded that the radiations consisted of neutral particles having a mass nearly equal to that of a proton. He named these particles neutrons. Nuclear Reaction: ${}_{2}^{4}He+{}_{4}^{9}Be\rightarrow{}_{6}^{12}C+{}_{0}^{1}n$ Properties of Neutron 1. A neutron has no electric charge. 2. Its mass is 1.6748 × 10⁻²⁷ kg or 1.008665 u. 3. Being neutral, it is neither attracted nor repelled by atomic nuclei and can pe...

Question: Describe the composition of the nucleus of an atom. Solutions: The alpha-particle scattering experiment of Rutherford suggested the existence of a positively charged region within an atom. This positively charged region is known as the nucleus of the atom. Almost all the mass of an atom is concentrated within this small volume. Initially, the nucleus was assumed to contain positively charged particles called protons and negatively charged particles called electrons. However, it was later established that electrons cannot exist inside the nucleus. The discovery of the neutron (a neutral particle having no charge) led to the present concept of the nucleus being composed of protons and neutrons. Protons and neutrons in a nucleus are collectively known as nucleons. Define Atomic Number The number of protons present in the nucleus of an atom is called the atomic number. It is denoted by Z. Define Atomic Mass Number The sum of the number of protons (Z) and the number of neutrons (...

Define atomic mass unit. Express it in kg. Show that energy equivalent to 1 a.m.u. is 931 MeV. Solutions :  Atomic mass unit (amu or u) is defined as (1/12)th of the mass of a carbon-12 atom (i.e., ₆C¹²). The mass of the nucleus and its constituent particles is expressed by a very small unit called unified mass unit (u). This unit is also known as atomic mass unit (a.m.u.), which is equal to 1.66 × 10⁻²⁷ kg  According to Avogadro's hypothesis, 1 mole of a substance contains atoms equal to Avogadro's number: N = 6.0225 × 10²³ Thus, 1 mole of carbon atoms contains 6.0225 × 10²³ atoms. ∴ Mass of 6.0225 × 10²³ carbon (₆C¹²) atoms = 12 g or Mass of 1 atom of carbon (₆C¹²) = 12 / (6.0225 × 10²³) g According to the definition of a.m.u., 1 u or 1 a.m.u. = (1/12) × mass of 1 carbon atom (₆C¹²) = $\frac{1}{12} \times \frac{12}{6.0225 \times 10^{23}}$ = 1.6604 × 10⁻²⁴ g or 1 a.m.u. (i.e., 1 u) = 1.66 × 10⁻²⁷ kg Mass of electron , $m_{e}= 0.00055 amu = 9.11 \times 10^{-31}kg$ Mass of prot...