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      Class 12 PHYSICS – JEE

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      • Class 12 PHYSICS – JEE
      CoursesClass 12PhysicsClass 12 PHYSICS – JEE
      • 1.Electrostatics (1)
        8
        • Lecture1.1
          Charge, Coulombs Law and Coulombs law in Vector form 41 min
        • Lecture1.2
          Electric Field; Electric Field Lines; Field lines due to multiple charges 42 min
        • Lecture1.3
          Charge Distribution; Finding Electric Field due to Different Object 01 hour
        • Lecture1.4
          Solid angle; Area Vector; Electric Flux; Flux of closed surface; Gauss Law 47 min
        • Lecture1.5
          Finding E Using Concept of Gauss law and Flux 01 hour
        • Lecture1.6
          Chapter Notes – Electrostatics (1)
        • Lecture1.7
          NCERT Solutions – Electrostatics
        • Lecture1.8
          Revision Notes Electrostatics
      • 2.Electrostatics (2)
        7
        • Lecture2.1
          Work done by Electrostatic Force; Work done by man in E-Field; Electrostatic Potential Energy 49 min
        • Lecture2.2
          Finding Electric Potential, Equipotential Surface and Motion in Electric Field 01 hour
        • Lecture2.3
          Electric Dipole and Dipole in Uniform and Non-uniform Electric field 01 hour
        • Lecture2.4
          Analysis of charge on conductors; Potential due to induced charge 58 min
        • Lecture2.5
          Conductors with cavity- Case 1: Empty cavity, Case 2: Charge Inside Cavity 41 min
        • Lecture2.6
          Connecting Two Conductors; Grounding of conductor; Electric field just outside conductor; Electrostatic pressure; Self potential Energy 54 min
        • Lecture2.7
          Chapter Notes – Electrostatics (2)
      • 3.Current Electricity (1)
        9
        • Lecture3.1
          Current, Motion of Electrons in Conductor; Temp. Dependence of Resistor 26 min
        • Lecture3.2
          Circuit Theory and Kirchoffs Laws 31 min
        • Lecture3.3
          Some Special Circuits- Series & Parallel Circuits, Open Circuit, Short Circuit 26 min
        • Lecture3.4
          Wheatstone Bridge, Current Antisymmetric 21 min
        • Lecture3.5
          Equivalent Resistance- Series and parallel, Equipotential Points, Wheatstone Bridge 25 min
        • Lecture3.6
          Current Antisymmetric, Infinite Ladder, Circuit Solving, 3D circuits 20 min
        • Lecture3.7
          Chapter Notes – Current Electricity
        • Lecture3.8
          NCERT Solutions – Current Electricity
        • Lecture3.9
          Revision Notes Current Electricity
      • 4.Current Electricity (2)
        4
        • Lecture4.1
          Heating Effect of Current; Rating of Bulb; Fuse 19 min
        • Lecture4.2
          Battery, Maximum power theorem; Ohmic and Non Ohmic Resistance; Superconductor 31 min
        • Lecture4.3
          Galvanometer; Ammeter & Voltmeter and Their Making 44 min
        • Lecture4.4
          Potentiometer and its applications ; Meter Bridge; Post Office Box; Colour Code of Resistors 32 min
      • 5.Capacitor
        6
        • Lecture5.1
          Capacitor and Capacitance; Energy in Capacitor 38 min
        • Lecture5.2
          Capacitive Circuits- Kirchoff’s Laws; Heat Production 01 hour
        • Lecture5.3
          Equivalent Capacitance; Charge on both sides of cap. Plate 52 min
        • Lecture5.4
          Dielectric Strength; Polar and Non-Polar Dielectric; Equivalent Cap. with Dielectric 01 hour
        • Lecture5.5
          Inserting and Removing Dielectric- Work (Fringing Effect), Force; Force between plates of capacitor 38 min
        • Lecture5.6
          Revision Notes Capacitor
      • 6.RC Circuits
        3
        • Lecture6.1
          Maths Needed for RC Circuits, RC circuits-Charging Circuit 19 min
        • Lecture6.2
          RC circuits-Discharging Circuit, Initial & Steady State, Final (Steady) State, Internal Resistance of Capacitor 44 min
        • Lecture6.3
          Revision Notes RC Circuits
      • 7.Magnetism and Moving Charge
        16
        • Lecture7.1
          Introduction, Vector Product, Force Applied by Magnetic Field, Lorentz Force, Velocity Selector 40 min
        • Lecture7.2
          Motion of Charged Particles in Uniform Magnetic Field 40 min
        • Lecture7.3
          Cases of Motion of Charged Particles in Uniform Magnetic Field 56 min
        • Lecture7.4
          Force on a Current Carrying Wire on Uniform B and its Cases, Questions and Solutions 59 min
        • Lecture7.5
          Magnetic Field on Axis of Circular Loop, Magnetic field due to Moving Charge, Magnetic Field due to Current 52 min
        • Lecture7.6
          Magnetic Field due to Straight Wire, Different Methods 40 min
        • Lecture7.7
          Magnetic Field due to Rotating Ring and Spiral 41 min
        • Lecture7.8
          Force between Two Current Carrying Wires 36 min
        • Lecture7.9
          Force between Two Current Carrying Wires 58 min
        • Lecture7.10
          Miscellaneous Questions 55 min
        • Lecture7.11
          Solenoid, Toroid, Magnetic Dipole, Magnetic Dipole Momentum, Magnetic Field of Dipole 54 min
        • Lecture7.12
          Magnetic Dipole in Uniform Magnetic Field, Moving Coil Galvanometer, Torsional Pendulum 01 hour
        • Lecture7.13
          Advanced Questions, Magnetic Dipole and Angular Momentum 56 min
        • Lecture7.14
          Chapter Notes – Magnetism and Moving Charge
        • Lecture7.15
          NCERT Solutions – Magnetism and Moving Charge
        • Lecture7.16
          Revision Notes Magnetism and Moving Charge
      • 8.Magnetism and Matter
        10
        • Lecture8.1
          Magnetic Dipole, Magnetic Properties of Matter, Diamagnetism; Domain Theory of Ferro 47 min
        • Lecture8.2
          Magnetic Properties of Matter in Detail 39 min
        • Lecture8.3
          Magnetization and Magnetic Intensity, Meissner Effect, Variation of Magnetization with Temperature 55 min
        • Lecture8.4
          Hysteresis, Permanent Magnet, Properties of Ferro for Permanent Magnet, Electromagnet 31 min
        • Lecture8.5
          Magnetic Compass, Earth’s Magnetic Field 20 min
        • Lecture8.6
          Bar Magnet, Bar Magnet in Uniform Field 49 min
        • Lecture8.7
          Magnetic Poles, Magnetic Field Lines, Magnetism and Gauss’s Law 32 min
        • Lecture8.8
          Chapter Notes – Magnetism and Matter
        • Lecture8.9
          NCERT Solutions – Magnetism and Matter
        • Lecture8.10
          Revision Notes Magnetism and Matter
      • 9.Electromagnetic Induction
        14
        • Lecture9.1
          Introduction, Magnetic Flux, Motional EMF 01 min
        • Lecture9.2
          Induced Electric Field, Faraday’s Law, Comparison between Electrostatic Electric Field and Induced Electric Field 43 min
        • Lecture9.3
          Induced Current; Faraday’s Law ; Lenz’s Law 56 min
        • Lecture9.4
          Faraday’s Law and its Cases 50 min
        • Lecture9.5
          Advanced Questions on Faraday’s Law 37 min
        • Lecture9.6
          Cases of Current Electricity 59 min
        • Lecture9.7
          Lenz’s Law and Conservation of Energy, Eddy Current, AC Generator, Motor 01 hour
        • Lecture9.8
          Mutual Induction 53 min
        • Lecture9.9
          Self Inductance, Energy in an Inductor 34 min
        • Lecture9.10
          LR Circuit, Decay Circuit 01 hour
        • Lecture9.11
          Initial and Final Analysis of LR Circuit 38 min
        • Lecture9.12
          Chapter Notes – Electromagnetic Induction
        • Lecture9.13
          NCERT Solutions – Electromagnetic Induction
        • Lecture9.14
          Revision Notes Electromagnetic Induction
      • 10.Alternating Current Circuit
        8
        • Lecture10.1
          Introduction, AC/DC Sources, Basic AC Circuits, Average & RMS Value 46 min
        • Lecture10.2
          Phasor Method, Rotating Vector, Adding Phasors, RC Circuit 35 min
        • Lecture10.3
          Examples and Solutions 21 min
        • Lecture10.4
          Power in AC Circuit, Resonance Frequency, Bandwidth and Quality Factor, Transformer 51 min
        • Lecture10.5
          LC Oscillator, Question and Solutions of LC Oscillator, Damped LC Oscillator 53 min
        • Lecture10.6
          Chapter Notes – Alternating Current Circuit
        • Lecture10.7
          NCERT Solutions – Alternating Current Circuit
        • Lecture10.8
          Revision Notes Alternating Current Circuit
      • 11.Electromagnetic Waves
        4
        • Lecture11.1
          Displacement Current; Ampere Maxwell Law 14 min
        • Lecture11.2
          EM Waves; EM Spectrum; Green House Effect; Ozone Layer 36 min
        • Lecture11.3
          Chapter Notes – Electromagnetic Waves
        • Lecture11.4
          Revision Notes Electromagnetic Waves
      • 12.Photoelectric Effect
        5
        • Lecture12.1
          Recalling Basics; Photoelectric Effect 50 min
        • Lecture12.2
          Photo-electric Cell 35 min
        • Lecture12.3
          Photon Flux; Photon Density; Momentum of Photon; Radiation Pressure- Full Absorption, Full Reflection; Dual nature 52 min
        • Lecture12.4
          Chapter Notes – Photoelectric Effect
        • Lecture12.5
          Revision Notes Photoelectric Effect
      • 13.Ray Optics (Part 1)
        12
        • Lecture13.1
          Rays and Beam of Light, Reflection of Light, Angle of Deviation, Image Formation by Plane Mirror 01 hour
        • Lecture13.2
          Field of View, Numerical on Field of Line, Size of Mirror 42 min
        • Lecture13.3
          Curved Mirrors, Terms Related to Curved Mirror, Reflection of Light by Curved Mirror 40 min
        • Lecture13.4
          Image Formation by Concave Mirror, Magnification or Lateral or Transverse Magnification 01 hour
        • Lecture13.5
          Ray Diagrams for Concave Mirror 45 min
        • Lecture13.6
          Image Formation by Convex Mirror; Derivations of Various Formulae used in Concave Mirror and Convex Mirror 01 hour
        • Lecture13.7
          Advanced Optical Systems, Formation of Images with more than one Mirror 24 min
        • Lecture13.8
          Concept of Virtual Object, Formation of Image when Incident ray are Converging, Image Characteristics for Virtual Object, 55 min
        • Lecture13.9
          Newton’s Formula, Longitudinal Magnification 23 min
        • Lecture13.10
          Formation of Image when Two Plane Mirrors kept at an angle and parallel; Formation of Image by two Parallel Mirrors. 43 min
        • Lecture13.11
          Chapter Notes – Ray Optics
        • Lecture13.12
          NCERT Solutions – Ray Optics
      • 14.Ray Optics (Part 2)
        13
        • Lecture14.1
          Refractive Index, Opaque, Transparent, Speed of Light, Relative Refractive Index, Refraction and Snell’s Law, Refraction in Denser and Rarer Medium 42 min
        • Lecture14.2
          Image Formation due to Refraction; Derivation; Refraction and Image formation in Glass Slab 57 min
        • Lecture14.3
          Total Internal Reflection, Critical Angle, Principle of Reversibility 01 hour
        • Lecture14.4
          Application of Total Internal Reflection 45 min
        • Lecture14.5
          Refraction at Curved Surface, Image Formation by Curved Surface, Derivation 56 min
        • Lecture14.6
          Image Formation by Curved Surface, Snell’s Law in Vector Form 01 hour
        • Lecture14.7
          Lens, Various types of Lens, Differentiating between various Lenses; Optical Centre, Derivation of Lens Maker Formula 01 hour
        • Lecture14.8
          Lens Formula, Questions and Answers 39 min
        • Lecture14.9
          Property of Image by Convex and Concave Lens; Lens Location, Minimum Distance Between Real Image and Object 01 hour
        • Lecture14.10
          Power of Lens, Combination of Lens, Autocollimation 35 min
        • Lecture14.11
          Silvering of Lens 44 min
        • Lecture14.12
          Cutting of Lens and Mirror, Vertical Cutting, Horizontal Cutting 49 min
        • Lecture14.13
          Newton’s Law for Lens and Virtual Object 01 hour
      • 15.Ray Optics (Part 3)
        6
        • Lecture15.1
          Prism, Angle of Prism, Reversibility in Prism 51 min
        • Lecture15.2
          Deviation in Prism, Minimum and Maximum Deviation, Asymmetric, Thin Prism, Proof for formula of Thin Prism 59 min
        • Lecture15.3
          Dispersion of Light, Refractive Index, Composition of Light, Dispersion through Prism 01 hour
        • Lecture15.4
          Rainbow Formation, Scattering of Light, Tyndall Effect, Defects of Image, Spherical Defect, Chromatic Defect, Achromatism. 57 min
        • Lecture15.5
          Optical Instruments, The Human Eye, Defects of Eye and its Corrections 01 hour
        • Lecture15.6
          Microscope & Telescope 02 hour
      • 16.Wave Optics
        21
        • Lecture16.1
          Introduction to Wave Optics 11 min
        • Lecture16.2
          Huygens Wave Theory 14 min
        • Lecture16.3
          Huygens Theory of Secondary Wavelets 10 min
        • Lecture16.4
          Law of Reflection by Huygens Theory 10 min
        • Lecture16.5
          Deriving Laws of Refraction by Huygens Wave Theory 10 min
        • Lecture16.6
          Multiple Answer type question on Huygens Theory 41 min
        • Lecture16.7
          Conditions of Constructive and Destructive Interference 22 min
        • Lecture16.8
          Conditions of Constructive and Destructive Interference 06 min
        • Lecture16.9
          Conditions of Constructive and Destructive Interference 23 min
        • Lecture16.10
          Incoherent Sources of Light 38 min
        • Lecture16.11
          Youngs Double Slit Experiment 12 min
        • Lecture16.12
          Fringe Width Positions of Bright and Dark Fringes 15 min
        • Lecture16.13
          Numerical problems on Youngs Double Slit Experiment 11 min
        • Lecture16.14
          Numerical problems on Youngs Double Slit Experiment 19 min
        • Lecture16.15
          Displacement of Interference Pattern 19 min
        • Lecture16.16
          Numerical problems on Displacement of Interference Pattern 28 min
        • Lecture16.17
          Shapes of Fringes 37 min
        • Lecture16.18
          Colour of Thin Films 59 min
        • Lecture16.19
          Interference with White Light 32 min
        • Lecture16.20
          Chapter Notes – Wave Optics
        • Lecture16.21
          NCERT Solutions – Wave Optics
      • 17.Atomic Structure
        6
        • Lecture17.1
          Thomson and Rutherford Model of Atom; Trajectory of Alpha particle; Bohr’s Model ; Hydrogen Like Atom; Energy Levels 58 min
        • Lecture17.2
          Emission Spectra, Absorption Spectra; De Broglie Explanation of Bohr’s 2nd Postulate; Limitations of Bohr’s Model 37 min
        • Lecture17.3
          Momentum Conservation in Photon Emission, Motion of Nucleus, Atomic Collision 58 min
        • Lecture17.4
          Chapter Notes – Atomic Structure
        • Lecture17.5
          NCERT Solutions – Atomic Structure
        • Lecture17.6
          Revision Notes Atomic Structure
      • 18.Nucleus
        6
        • Lecture18.1
          Basics- Size of Nucleus, Nuclear Force, Binding Energy, Mass Defect; Radioactive Decay 01 hour
        • Lecture18.2
          Laws of Radioactive Decay 36 min
        • Lecture18.3
          Nuclear Fission; Nuclear Reactor; Nuclear Fusion- Reaction Inside Sun 30 min
        • Lecture18.4
          Chapter Notes – Nucleus
        • Lecture18.5
          NCERT Solutions – Nucleus
        • Lecture18.6
          Revision Notes Nucleus
      • 19.X-Ray
        4
        • Lecture19.1
          Electromagnetic Spectrum, Thermionic Emission; Coolidge Tube – Process 1 22 min
        • Lecture19.2
          Coolidge Tube – Process 2; Moseley’s Law; Absorption of X-rays in Heavy Metal 39 min
        • Lecture19.3
          Chapter Notes – X-Ray
        • Lecture19.4
          Revision Notes X-Ray
      • 20.Error and Measurement
        2
        • Lecture20.1
          Least Count of Instruments; Mathematical Operation on Data with Random Error 18 min
        • Lecture20.2
          Significant Digits; Significant Digits and Mathematical Operations 30 min
      • 21.Semiconductors
        9
        • Lecture21.1
          Conductor, Semiconductors and Insulators Basics Difference, Energy Band Theory, Si element 21 min
        • Lecture21.2
          Doping and PN Junction 01 hour
        • Lecture21.3
          Diode and Diode as Rectifier 01 hour
        • Lecture21.4
          Voltage Regulator and Zener Diode and Optoelectronic Jn. Devices 01 hour
        • Lecture21.5
          Transistor, pnp, npn, Modes of operation, Input and Output Characteristics, , Current Amplification Factor 01 hour
        • Lecture21.6
          Transistor as Amplifier, Transistor as Switch, Transistor as Oscillator, Digital Gates 01 hour
        • Lecture21.7
          Chapter Notes – Semiconductors
        • Lecture21.8
          NCERT Solutions – Semiconductors
        • Lecture21.9
          Revision Notes Semiconductors
      • 22.Communication Systems
        5
        • Lecture22.1
          Basic working and terms; Antenna; Modulation and Types of Modulation 32 min
        • Lecture22.2
          Amplification Modulation, Transmitter, Receiver, Modulation index 40 min
        • Lecture22.3
          Chapter Notes – Communication Systems
        • Lecture22.4
          NCERT Solutions – Communication Systems
        • Lecture22.5
          Revision Notes Communication Systems

        NCERT Solutions – Nucleus

        13.1. (a) Two stable isotopes of lithium  and have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

        (b) Boron has two stable isotopes, and. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of  and .

        Answer

        (a) Mass of lithium isotope , m1 = 6.01512 u

        Mass of lithium isotope , m2 = 7.01600 u

        Abundance of , η1= 7.5%

        Abundance of , η2= 92.5%

        The atomic mass of lithium atom is given as:

        (b) Mass of boron isotope , m1 = 10.01294 u

        Mass of boron isotope , m2 = 11.00931 u

        Abundance of , η1 = x%

        Abundance of , η2= (100 − x)%

        Atomic mass of boron, m = 10.811 u

        The atomic mass of boron atom is given as:

        And 100 − x = 80.11%

        Hence, the abundance of  is 19.89% and that of is 80.11%.

        13.2. The three stable isotopes of neon: and have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

        Answer

        Atomic mass of , m1= 19.99 u

        Abundance of , η1 = 90.51%

        Atomic mass of , m2 = 20.99 u

        Abundance of , η2 = 0.27%

        Atomic mass of , m3 = 21.99 u

        Abundance of , η3 = 9.22%

        The average atomic mass of neon is given as:

        13.3. Obtain the binding energy (in MeV) of a nitrogen nucleus, given =14.00307 u

        Answer

        Atomic mass of nitrogen, m = 14.00307 u

        A nucleus of nitrogen  contains 7 protons and 7 neutrons.

        Hence, the mass defect of this nucleus, Δm = 7mH + 7mn − m

        Where,

        Mass of a proton, mH = 1.007825 u

        Mass of a neutron, mn= 1.008665 u

        ∴Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307

        = 7.054775 + 7.06055 − 14.00307

        = 0.11236 u

        But 1 u = 931.5 MeV/c2

        ∴Δm = 0.11236 × 931.5 MeV/c2

        Hence, the binding energy of the nucleus is given as:

        Eb = Δmc2

        Where,

        c = Speed of light

        ∴Eb = 0.11236 × 931.5 

        = 104.66334 MeV

        Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

        13.4. Obtain the binding energy of the nuclei  and in units of MeV from the following data:

         = 55.934939 u = 208.980388 u

        Answer

        Atomic mass of, m1 = 55.934939 u

         nucleus has 26 protons and (56 − 26) = 30 neutrons

        Hence, the mass defect of the nucleus, Δm = 26 × mH + 30 × mn − m1

        Where,

        Mass of a proton, mH = 1.007825 u

        Mass of a neutron, mn = 1.008665 u

        ∴Δm = 26 × 1.007825 + 30 × 1.008665 − 55.934939

        = 26.20345 + 30.25995 − 55.934939

        = 0.528461 u

        But 1 u = 931.5 MeV/c2

        ∴Δm = 0.528461 × 931.5 MeV/c2

        The binding energy of this nucleus is given as:

        Eb1 = Δmc2

        Where,

        c = Speed of light

        ∴Eb1 = 0.528461 × 931.5 

        = 492.26 MeV

        Average binding energy per nucleon 

        Atomic mass of, m2 = 208.980388 u

         nucleus has 83 protons and (209 − 83) 126 neutrons.

        Hence, the mass defect of this nucleus is given as:

        Δm‘ = 83 × mH + 126 × mn − m2

        Where,

        Mass of a proton, mH = 1.007825 u

        Mass of a neutron, mn = 1.008665 u

        ∴Δm‘ = 83 × 1.007825 + 126 × 1.008665 − 208.980388

        = 83.649475 + 127.091790 − 208.980388

        = 1.760877 u

        But 1 u = 931.5 MeV/c2

        ∴Δm‘ = 1.760877 × 931.5 MeV/c2

        Hence, the binding energy of this nucleus is given as:

        Eb2 = Δm‘c2

        = 1.760877 × 931.5

        = 1640.26 MeV

        Average bindingenergy per nucleon = 

        13.5. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of atoms (of mass 62.92960 u).

        Answer

        Mass of a copper coin, m’ = 3 g

        Atomic mass of atom, m = 62.92960 u

        The total number of atoms in the coin

        Where,

        NA = Avogadro’s number = 6.023 × 1023 atoms /g

        Mass number = 63 g

        nucleus has 29 protons and (63 − 29) 34 neutrons

        ∴Mass defect of this nucleus, Δm‘ = 29 × mH + 34 × mn − m

        Where,

        Mass of a proton, mH = 1.007825 u

        Mass of a neutron, mn = 1.008665 u

        ∴Δm‘ = 29 × 1.007825 + 34 × 1.008665 − 62.9296

        = 0.591935 u

        Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022

        = 1.69766958 × 1022 u

        But 1 u = 931.5 MeV/c2

        ∴Δm = 1.69766958 × 1022 × 931.5 MeV/c2

        Hence, the binding energy of the nuclei of the coin is given as:

        Eb= Δmc2

        = 1.69766958 × 1022 × 931.5 

        = 1.581 × 1025 MeV

        But 1 MeV = 1.6 × 10−13 J

        Eb = 1.581 × 1025 × 1.6 × 10−13

        = 2.5296 × 1012 J

        This much energy is required to separate all the neutrons and protons from the given coin.

        13.6. Write nuclear reaction equations for

        (i) α-decay of  (ii) α-decay of 

        (iii) β−-decay of  (iv) β−-decay of 

        (v) β+-decay of  (vi) β+-decay of 

        (vii) Electron capture of 

        Answer

        α is a nucleus of helium and β is an electron (e− for β− and e+ for β+). In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β−-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

        For the given cases, the various nuclear reactions can be written as:

        13.7. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

        Answer

        Half-life of the radioactive isotope = T years

        Original amount of the radioactive isotope = N0

        (a) After decay, the amount of the radioactive isotope = N

        It is given that only 3.125% of N0 remains after decay. Hence, we can write:

        Where,

        λ = Decay constant

        t = Time

        Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.

        (b) After decay, the amount of the radioactive isotope = N

        It is given that only 1% of N0 remains after decay. Hence, we can write:

        Since, λ = 0.693/T

        Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

        13.8. The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive  present with the stable carbon isotope . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

        Answer

        Decay rate of living carbon-containing matter, R = 15 decay/min

        Let N be the number of radioactive atoms present in a normal carbon- containing matter.

        Half life of, = 5730 years

        The decay rate of the specimen obtained from the Mohenjodaro site:

        R‘ = 9 decays/min

        Let N’ be the number of radioactive atoms present in the specimen during the Mohenjodaro period.

        Therefore, we can relate the decay constant, λand time, t as:

        Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.

         

        13.9. Obtain the amount of necessary to provide a radioactive source of 8.0 mCi strength. The half-life of is 5.3 years.

        Answer

        The strength of the radioactive source is given as:

        Where,

        N = Required number of atoms

        Half-life of,  = 5.3 years

        = 5.3 × 365 × 24 × 60 × 60

        = 1.67 × 108 s

        For decay constant λ, we have the rate of decay as:

        Where, λ 

        For:

        Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

        ∴Mass of atoms 

        Hence, the amount of  necessary for the purpose is 7.106 × 10−6 g.

        13.10. The half-life of is 28 years. What is the disintegration rate of 15 mg of this isotope?

        Answer

        Half life of , = 28 years

        = 28 × 365 × 24 × 60 × 60

        = 8.83 × 108 s

        Mass of the isotope, m = 15 mg

        90 g of atom contains 6.023 × 1023 (Avogadro’s number) atoms.

        Therefore, 15 mg of  contains:

        Rate of disintegration, 

        Where,

        λ = Decay constant 

        Hence, the disintegration rate of 15 mg of the given isotope is
        7.878 × 1010 atoms/s.

        13.11. Obtain approximately the ratio of the nuclear radii of the gold isotope  and the silver isotope.

        Answer

        Nuclear radius of the gold isotope = RAu

        Nuclear radius of the silver isotope = RAg

        Mass number of gold, AAu = 197

        Mass number of silver, AAg = 107

        The ratio of the radii of the two nuclei is related with their mass numbers as:

        Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

        13.12. Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) and (b).

        Given  = 226.02540 u,  = 222.01750 u,

        = 220.01137 u, = 216.00189 u.

        Answer

        (a) Alpha particle decay of emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.

        Q-value of

        emitted α-particle = (Sum of initial mass − Sum of final mass) c2

        Where,

        c = Speed of light

        It is given that:

        Q-value = [226.02540 − (222.01750 + 4.002603)] u c2
        = 0.005297 u c2

        But 1 u = 931.5 MeV/c2

        ∴Q = 0.005297 × 931.5 ≈ 4.94 MeV

        Kinetic energy of the α-particle 

        (b) Alpha particle decay of  is shown by the following nuclear reaction.

        It is given that:

        Mass of = 220.01137 u

        Mass of = 216.00189 u

        ∴Q-value = 

        ≈ 641 MeV

        Kinetic energy of the α-particle 

        = 6.29 MeV

        13.13. The radionuclide 11C decays according to

        The maximum energy of the emitted positron is 0.960 MeV.

        Given the mass values:

        calculate Q and compare it with the maximum energy of the positron emitted

        Answer

        The given nuclear reaction is:

        Atomic mass of = 11.011434 u

        Atomic mass of 

        Maximum energy possessed by the emitted positron = 0.960 MeV

        The change in the Q-value (ΔQ) of the nuclear masses of the  nucleus is given as:

        Where,

        me = Mass of an electron or positron = 0.000548 u

        c = Speed of light

        m’ = Respective nuclear masses

        If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case ofand 5 me in the case of.

        Hence, equation (1) reduces to:

        ∴ΔQ = [11.011434 − 11.009305 − 2 × 0.000548] c2

        = (0.001033 c2) u

        But 1 u = 931.5 Mev/c2

        ∴ΔQ = 0.001033 × 931.5 ≈ 0.962 MeV

        The value of Q is almost comparable to the maximum energy of the emitted positron.

        13.14. The nucleus decays byemission. Write down the decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

        = 22.994466 u

        = 22.989770 u.

        Answer

        In  emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.

         emission of the nucleus  is given as:

        It is given that:

        Atomic mass of = 22.994466 u

        Atomic mass of = 22.989770 u

        Mass of an electron, me = 0.000548 u

        Q-value of the given reaction is given as:

        There are 10 electrons in  and 11 electrons in. Hence, the mass of the electron is cancelled in the Q-value equation.

        The daughter nucleus is too heavy as compared to and . Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.

        13.15. The Q value of a nuclear reaction A + b → C + d is defined by

        Q = [ mA+ mb− mC− md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

        (i) 

        (ii) 

        Atomic masses are given to be

        Answer

        (i) The given nuclear reaction is:

        It is given that:

        Atomic mass 

        Atomic mass 

        Atomic mass 

        According to the question, the Q-value of the reaction can be written as:

        The negativeQ-value of the reaction shows that the reaction is endothermic.

        (ii) The given nuclear reaction is:

        It is given that:

        Atomic mass of 

        Atomic mass of 

        Atomic mass of 

        The Q-value of this reaction is given as:

        The positive Q-value of the reaction shows that the reaction is exothermic.

        13.16. Suppose, we think of fission of a nucleus into two equal fragments,. Is the fission energetically possible? Argue by working out Q of the process. Given  and.

        Answer

        The fission of can be given as:

        It is given that:

        Atomic mass of  = 55.93494 u

        Atomic mass of 

        The Q-value of this nuclear reaction is given as:

        The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.

        13.17. The fission properties of are very similar to those of.

        The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure undergo fission?

        Answer

        Average energy released per fission of, 

        Amount of pure, m = 1 kg = 1000 g

        NA= Avogadro number = 6.023 × 1023

        Mass number of= 239 g

        1 mole of contains NA atoms.

        ∴m g of contains

        ∴Total energy released during the fission of 1 kg ofis calculated as:

        Hence,  is released if all the atoms in 1 kg of pure undergo fission.

         

        13.18. A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of and that this nuclide is consumed only by the fission process.

        Answer

        Half life of the fuel of the fission reactor, years

        = 5 × 365 × 24 × 60 × 60 s

        We know that in the fission of 1 g of nucleus, the energy released is equal to 200 MeV.

        1 mole, i.e., 235 g of contains 6.023 × 1023 atoms.

        ∴1 g  contains

        The total energy generated per gram ofis calculated as:

        The reactor operates only 80% of the time.

        Hence, the amount of consumed in 5 years by the 1000 MW fission reactor is calculated as:

        ∴Initial amount of = 2 × 1538 = 3076 kg

        13.19. How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

        Answer

        The given fusion reaction is:

        Amount of deuterium, m = 2 kg

        1 mole, i.e., 2 g of deuterium contains 6.023 × 1023 atoms.

        ∴2.0 kg of deuterium contains 

        It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

        ∴Total energy per nucleus released in the fusion reaction:

        Power of the electric lamp, P = 100 W = 100 J/s

        Hence, the energy consumed by the lamp per second = 100 J

        The total time for which the electric lamp will glow is calculated as:

        13.20. Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

        Answer

        When two deuterons collide head-on, the distance between their centres, d is given as:

        Radius of 1st deuteron + Radius of 2nd deuteron

        Radius of a deuteron nucleus = 2 fm = 2 × 10−15 m

        ∴d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m

        Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C

        Potential energy of the two-deuteron system:

        Where,

         = Permittivity of free space

        Hence, the height of the potential barrier of the two-deuteron system is

        360 keV.

        13.21.From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

        Answer

        We have the expression for nuclear radius as:

        R = R0A1/3

        Where,

        R0 = Constant.

        A = Mass number of the nucleus

        Nuclear matter density, 

        Let m be the average mass of the nucleus.

        Hence, mass of the nucleus = mA

        Hence, the nuclear matter density is independent of A. It is nearly constant.

        13.22. For the  (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K−shell, is captured by the nucleus and a neutrino is emitted).

        Show that if  emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.

        Answer

        Let the amount of energy released during the electron capture process be Q1. The nuclear reaction can be written as:

        Let the amount of energy released during the positron capture process be Q2. The nuclear reaction can be written as:

        = Nuclear mass of 

        = Nuclear mass of 

        = Atomic mass of 

        = Atomic mass of 

        me = Mass of an electron

        c = Speed of light

        Q-value of the electron capture reaction is given as:

        Q-value of the positron capture reaction is given as:

        It can be inferred that if Q2 > 0, then Q1 > 0; Also, if Q1> 0, it does not necessarily mean that Q2 > 0.

        In other words, this means that ifemission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically-allowed nuclear reaction.

        13.23. In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are (23.98504u), (24.98584u) and (25.98259u). The natural abundance of is 78.99% by mass. Calculate the abundances of other two isotopes.

        Answer

        Average atomic mass of magnesium, m = 24.312 u

        Mass of magnesium isotope, m1 = 23.98504 u

        Mass of magnesium isotope, m2 = 24.98584 u

        Mass of magnesium isotope, m3 = 25.98259 u

        Abundance of, η1= 78.99%

        Abundance of, η2 = x%

        Hence, abundance of, η3 = 100 − x − 78.99% = (21.01 − x)%

        We have the relation for the average atomic mass as:

        Hence, the abundance of is 9.3% and that of is 11.71%.

        13.24. The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei and from the following data:

        = 39.962591 u

        ) = 40.962278 u

        = 25.986895 u

        ) = 26.981541 u

        Answer

        For 

        For

        A neutron is removed from a nucleus. The corresponding nuclear reaction can be written as:

        It is given that:

        Mass = 39.962591 u

        Mass) = 40.962278 u

        Mass = 1.008665 u

        The mass defect of this reaction is given as:

        Δm = 

        ∴Δm = 0.008978 × 931.5 MeV/c2

        Hence, the energy required for neutron removal is calculated as:

        For, the neutron removal reaction can be written as:

        It is given that:

        Mass = 26.981541 u

        Mass = 25.986895 u

        The mass defect of this reaction is given as:

        Hence, the energy required for neutron removal is calculated as:

        13.25. A source contains two phosphorous radio nuclides (T1/2 = 14.3d) and (T1/2 = 25.3d). Initially, 10% of the decays come from. How long one must wait until 90% do so?

        Answer

        Half life of, T1/2 = 14.3 days

        Half life of, T’1/2 = 25.3 days

         nucleus decay is 10% of the total amount of decay.

        The source has initially 10% of nucleus and 90% of nucleus.

        Suppose after t days, the source has 10% of nucleus and 90% of  nucleus.

        Initially:

        Number of nucleus = N

        Number of nucleus = 9 N

        Finally:

        Number of 

        Number of 

        For nucleus, we can write the number ratio as:

        For, we can write the number ratio as:

        On dividing equation (1) by equation (2), we get:

        Hence, it will take about 208.5 days for 90% decay of .

        13.26. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

        Calculate the Q-values for these decays and determine that both are energetically allowed.

        Answer

        Take a  emission nuclear reaction:

        We know that:

        Mass of m1 = 223.01850 u

        Mass of m2 = 208.98107 u

        Mass of, m3 = 14.00324 u

        Hence, the Q-value of the reaction is given as:

        Q = (m1 − m2 − m3) c2

        = (223.01850 − 208.98107 − 14.00324) c2

        = (0.03419 c2) u

        But 1 u = 931.5 MeV/c2

        ∴Q = 0.03419 × 931.5

        = 31.848 MeV

        Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

        Now take a emission nuclear reaction:

        We know that:

        Mass of m1 = 223.01850

        Mass of m2 = 219.00948

        Mass of, m3 = 4.00260

        Q-value of this nuclear reaction is given as:

        Q = (m1 − m2 − m3) c2

        = (223.01850 − 219.00948 − 4.00260) C2

        = (0.00642 c2) u

        = 0.00642 × 931.5 = 5.98 MeV

        Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.

         

        13.27. Consider the fission of by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are  and. Calculate Q for this fission process. The relevant atomic and particle masses are

        m =238.05079 u

        m =139.90543 u

        m = 98.90594 u

        Answer

        In the fission of, 10 β− particles decay from the parent nucleus. The nuclear reaction can be written as:

        It is given that:

        Mass of a nucleusm1 = 238.05079 u

        Mass of a nucleus m2 = 139.90543 u

        Mass of a nucleus, m3 = 98.90594 u

        Mass of a neutronm4 = 1.008665 u

        Q-value of the above equation,

        Where,

        m’ = Represents the corresponding atomic masses of the nuclei

        = m1 − 92me

        = m2 − 58me

        = m3 − 44me

        = m4

        Hence, the Q-value of the fission process is 231.007 MeV.

        13.28. Consider the D−T reaction (deuterium−tritium fusion)

        (a) Calculate the energy released in MeV in this reaction from the data:

        = 2.014102 u

        = 3.016049 u

        (b)Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

        Answer

        (a) Take the D-T nuclear reaction: 

        It is given that:

        Mass of, m1= 2.014102 u

        Mass of, m2 = 3.016049 u

        Mass of m3 = 4.002603 u

        Mass of, m4 = 1.008665 u

        Q-value of the given D-T reaction is:

        Q = [m1 + m2− m3 − m4] c2

        = [2.014102 + 3.016049 − 4.002603 − 1.008665] c2

        = [0.018883 c2] u

        But 1 u = 931.5 MeV/c2

        ∴Q = 0.018883 × 931.5 = 17.59 MeV

        (b) Radius of deuterium and tritium, r ≈ 2.0 fm = 2 × 10−15 m

        Distance between the two nuclei at the moment when they touch each other, d = r + r = 4 × 10−15 m

        Charge on the deuterium nucleus = e

        Charge on the tritium nucleus = e

        Hence, the repulsive potential energy between the two nuclei is given as:

        Where,

        ∈0 = Permittivity of free space

        Hence, 5.76 × 10−14 J or of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.

        However, it is given that:

        KE

        Where,

        k = Boltzmann constant = 1.38 × 10−23 m2 kg s−2 K−1

        T = Temperature required for triggering the reaction

        Hence, the gas must be heated to a temperature of 1.39 × 109 K to initiate the reaction.

        13.29. Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that

        m (198Au) = 197.968233 u

        m (198Hg) =197.966760 u

        Answer

        It can be observed from the given γ-decay diagram that γ1 decays from the 1.088 MeV energy level to the 0 MeV energy level.

        Hence, the energy corresponding to γ1-decay is given as:

        E1 = 1.088 − 0 = 1.088 MeV

        hν1= 1.088 × 1.6 × 10−19 × 106 J

        Where,

        h = Planck’s constant = 6.6 × 10−34 Js

        ν1 = Frequency of radiation radiated by γ1-decay

        It can be observed from the given γ-decay diagram that γ2 decays from the 0.412 MeV energy level to the 0 MeV energy level.

        Hence, the energy corresponding to γ2-decay is given as:

        E2 = 0.412 − 0 = 0.412 MeV

        hν2= 0.412 × 1.6 × 10−19 × 106 J

        Where,

        ν2 = Frequency of radiation radiated by γ2-decay

        It can be observed from the given γ-decay diagram that γ3 decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

        Hence, the energy corresponding to γ3-decay is given as:

        E3 = 1.088 − 0.412 = 0.676 MeV

        hν3= 0.676 × 10−19 × 106 J

        Where,

        ν3 = Frequency of radiation radiated by γ3-decay

        Mass of = 197.968233 u

        Mass of = 197.966760 u

        1 u = 931.5 MeV/c2

        Energy of the highest level is given as:

        β1 decays from the 1.3720995 MeV level to the 1.088 MeV level

        ∴Maximum kinetic energy of the β1 particle = 1.3720995 − 1.088

        = 0.2840995 MeV

        β2 decays from the 1.3720995 MeV level to the 0.412 MeV level

        ∴Maximum kinetic energy of the β2 particle = 1.3720995 − 0.412

        = 0.9600995 MeV

         

        13.30. Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor.

        Answer

        (a) Amount of hydrogen, m = 1 kg = 1000 g

        1 mole, i.e., 1 g of hydrogen () contains 6.023 × 1023 atoms.

        ∴1000 g of  contains 6.023 × 1023 × 1000 atoms.

        Within the sun, four  nuclei combine and form one  nucleus. In this process 26 MeV of energy is released.

        Hence, the energy released from the fusion of 1 kg is:

        (b) Amount of  = 1 kg = 1000 g

        1 mole, i.e., 235 g of  contains 6.023 × 1023 atoms.

        ∴1000 g ofcontains 

        It is known that the amount of energy released in the fission of one atom of is 200 MeV.

        Hence, energy released from the fission of 1 kg of  is:

        ∴ 

        Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the energy released in the fission of 1 kg of uranium.

        13.31. Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV.

        Answer

        Amount of electric power to be generated, P = 2 × 105 MW

        10% of this amount has to be obtained from nuclear power plants.

        ∴Amount of nuclear power, 

        = 2 × 104 MW

        = 2 × 104 × 106 J/s

        = 2 × 1010 × 60 × 60 × 24 × 365 J/y

        Heat energy released per fission of a 235U nucleus, E = 200 MeV

        Efficiency of a reactor = 25%

        Hence, the amount of energy converted into the electrical energy per fission is calculated as:

        Number of atoms required for fission per year:

        1 mole, i.e., 235 g of U235 contains 6.023 × 1023 atoms.

        ∴Mass of 6.023 × 1023 atoms of U235 = 235 g = 235 × 10−3 kg

        ∴Mass of 78840 × 1024 atoms of U235

        Hence, the mass of uranium needed per year is 3.076 × 104 kg.

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