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      Class 11 CHEMISTRY – JEE

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      • Chemistry
      • Class 11 CHEMISTRY – JEE
      CoursesClass 11ChemistryClass 11 CHEMISTRY – JEE
      • 1. Stoichiometry 1
        13
        • Lecture1.1
          Introduction & POAC 30 min
        • Lecture1.2
          Mole Stoichiometric relationship 29 min
        • Lecture1.3
          Successive reaction & Limiting reagent 25 min
        • Lecture1.4
          Gas Stoichiometry 29 min
        • Lecture1.5
          Important Types of Reactions 25 min
        • Lecture1.6
          Avogadro’s No.1 30 min
        • Lecture1.7
          Mole & Number 28 min
        • Lecture1.8
          Atomic, Molecular Wt 26 min
        • Lecture1.9
          Ionic wt, Avg. At. Wt. 15 min
        • Lecture1.10
          Molar wt. 27 min
        • Lecture1.11
          Molar Volume & Gas Analysis 30 min
        • Lecture1.12
          Gas Analysis 17 min
        • Lecture1.13
          Empirical Formula Determination 26 min
      • 2. Stoichiometry 2
        18
        • Lecture2.1
          Acid Base definition 23 min
        • Lecture2.2
          Acidity & Basicity 32 min
        • Lecture2.3
          Acidic Strength 30 min
        • Lecture2.4
          Acidic Strength 23 min
        • Lecture2.5
          Conjugate Acid-Base pair, Basic Strength 48 min
        • Lecture2.6
          Oxidation & Reduction 50 min
        • Lecture2.7
          Calculation of Oxidation Number 46 min
        • Lecture2.8
          O.A. & R.A., Balancing by Oxidation Number Method 01 hour
        • Lecture2.9
          Balancing by Ion Electron Method. 35 min
        • Lecture2.10
          Eq. Wt. 1 – n factor & Eq. Wt. Concept 47 min
        • Lecture2.11
          Eq. Wt. 2 – Eq. Concept 35 min
        • Lecture2.12
          Volumetric Analysis 43 min
        • Lecture2.13
          Volumetric analysis 44 min
        • Lecture2.14
          Titration – Acid Base Titration 49 min
        • Lecture2.15
          Titration – Acid Base Titration, Indicator 56 min
        • Lecture2.16
          Titration – Redox Titration-8 58 min
        • Lecture2.17
          Titration – Redox Titration, volume Strength of H2O2 50 min
        • Lecture2.18
          Titration – Redox Titration, Iodometry, Oleum, Bleaching Powder 49 min
      • 3. Thermodynamics & Thermochemistry
        19
        • Lecture3.1
          Zeroth Law 55 min
        • Lecture3.2
          1st law – System, Properties, State 40 min
        • Lecture3.3
          1st Law, Process, Internal energy, Work 43 min
        • Lecture3.4
          Work done in Irreversible process, Isobaric Process 49 min
        • Lecture3.5
          Isochoric Process & problems TD 42 min
        • Lecture3.6
          Isothermal irreversible Process, Problems on TD 46 min
        • Lecture3.7
          Adiabatic Process 49 min
        • Lecture3.8
          Problems on TD 44 min
        • Lecture3.9
          Thermochemistry & Enthalpy 38 min
        • Lecture3.10
          Hess’s Law, Kirchhoff’s Law 43 min
        • Lecture3.11
          Enthalpy of Formation, combustion 39 min
        • Lecture3.12
          Enthalpy of Hydrogenation, Hydration, dissolution, lattice energy 40 min
        • Lecture3.13
          Enthapy of Neutralisation, atomisation, Bond Energy 47 min
        • Lecture3.14
          Resonance energy & problems 45 min
        • Lecture3.15
          2nd Law, Entropy-positional 40 min
        • Lecture3.16
          TD Entropy, 3rd Law, Entropy change in a reaction 45 min
        • Lecture3.17
          Gibb’s free energy 43 min
        • Lecture3.18
          Efficiency, engine, pump & Carnot engine 39 min
        • Lecture3.19
          Chapter Notes – Thermodynamics & Thermochemistry
      • 4. Atomic Structure
        22
        • Lecture4.1
          Introduction, Cathode rays & Anode rays 41 min
        • Lecture4.2
          J.J. Thomson Model, Millikan Oil Drop Experiment 38 min
        • Lecture4.3
          Rutherford Experiment 51 min
        • Lecture4.4
          Quantum Mechanics, BlackBody Radiation Experiment 41 min
        • Lecture4.5
          Wave 44 min
        • Lecture4.6
          Photoelectric Effect 46 min
        • Lecture4.7
          Problems on Photoelectric Effect 35 min
        • Lecture4.8
          Atomic Structure 44 min
        • Lecture4.9
          Bohr Theory 47 min
        • Lecture4.10
          H – Spectrum 49 min
        • Lecture4.11
          Problems on Bohr’s Theory 40 min
        • Lecture4.12
          Adv. Problems on Bohr Theory & Sommerfeld model 51 min
        • Lecture4.13
          Quantum Mechanical Model for Atomic Structure 47 min
        • Lecture4.14
          Schrodinger wave equation 54 min
        • Lecture4.15
          No. of Orbitals & Quantum no 45 min
        • Lecture4.16
          Orbital Curve, RPD curve, Definition of Node 46 min
        • Lecture4.17
          Calculation of Node, Orbital Picture 43 min
        • Lecture4.18
          Radial Probability curve, MPD, Avg. distance, Screening effect, Zeff 38 min
        • Lecture4.19
          Multielectron system, Electronic configuration 56 min
        • Lecture4.20
          Stability of Elec. Configuration 36 min
        • Lecture4.21
          Chapter Notes – Atomic Structure
        • Lecture4.22
          NCERT Solutions – Atomic Structure
      • 5. Chemical equilibrium
        9
        • Lecture5.1
          Introduction, Eqb constant & Eqb Position 51 min
        • Lecture5.2
          Types of Eqb Constant, Heterogeneous Eqb, Reaction Quotient 45 min
        • Lecture5.3
          Range of Eqb Constant 43 min
        • Lecture5.4
          Problems on Chemical Eqb 41 min
        • Lecture5.5
          Problems on Chemical Eqb 42 min
        • Lecture5.6
          Le-chatelier Principle 42 min
        • Lecture5.7
          Le-Chatelier Principle 35 min
        • Lecture5.8
          Eqb & 2nd Law of TD 26 min
        • Lecture5.9
          NCERT Solutions – equilibrium
      • 6. Ionic Equilibrium
        17
        • Lecture6.1
          Electrolyte, Dissociation of H2O, Nature of Solution 46 min
        • Lecture6.2
          PH scale, Log & Antilog 40 min
        • Lecture6.3
          PH of Strong Acid, Base Solution 51 min
        • Lecture6.4
          PH of Weak Acid, Base solution 41 min
        • Lecture6.5
          PH of mixture of Acids, Bases 46 min
        • Lecture6.6
          PH of Polybasic acids 40 min
        • Lecture6.7
          PH of Salt Solution 1 43 min
        • Lecture6.8
          PH of salt solution 2 52 min
        • Lecture6.9
          Common ion effect, Buffer solution 49 min
        • Lecture6.10
          Buffer Capacity 45 min
        • Lecture6.11
          Titration & PH Curve 1 40 min
        • Lecture6.12
          Titration & PH curve 2 46 min
        • Lecture6.13
          Acid Base indicator 35 min
        • Lecture6.14
          Solubility Equilibrium 47 min
        • Lecture6.15
          Precipitation of Solid, Qualitative analysis of cation 44 min
        • Lecture6.16
          Complex ion equilibrium 23 min
        • Lecture6.17
          Chapter Notes – Equilibrium
      • 7. Introduction & Development of Org. Chemistry
        3
        • Lecture7.1
          Introduction & Development Of Organic Chemistry 44 min
        • Lecture7.2
          Introduction & Syllabus 36 min
        • Lecture7.3
          NCERT Solutions – Org. Chemistry
      • 8. Nomenclature of Org. Compounds
        16
        • Lecture8.1
          Alkane 59 min
        • Lecture8.2
          Alkane 31 min
        • Lecture8.3
          Alkyl Group & Types Of Hydrogen 01 hour
        • Lecture8.4
          Alkene 54 min
        • Lecture8.5
          Alkenyl 32 min
        • Lecture8.6
          Alkyne & Alkenyl 47 min
        • Lecture8.7
          Cycloalkane 43 min
        • Lecture8.8
          Cycloalkene 35 min
        • Lecture8.9
          Bicycloalkane & Spirane 35 min
        • Lecture8.10
          Acid & Aldehyde 45 min
        • Lecture8.11
          Ester & Acid Halides 28 min
        • Lecture8.12
          Amide & Nitrile 28 min
        • Lecture8.13
          Alcohol & Sulphonic Acid 37 min
        • Lecture8.14
          Isonitrile, Amine, Nitroalkane, Halo Compounds 39 min
        • Lecture8.15
          Ketone, Anhydride & Ether 34 min
        • Lecture8.16
          Polyfunctional Group Compounds 41 min
      • 9. GOC 1- Hybridisation, Resonance, Aromaticity
        16
        • Lecture9.1
          Concept Of Hybridisation 42 min
        • Lecture9.2
          Sp3, Sp2 Hybridisation 44 min
        • Lecture9.3
          Sp Hybridisation, Relative Study Of Sp3, Sp2, Sp Orbitals 46 min
        • Lecture9.4
          Effect Of Hybridisation On Bond Length, Planar Nature 59 min
        • Lecture9.5
          Concept Of Resonance 39 min
        • Lecture9.6
          Doing Resonance 18 min
        • Lecture9.7
          Resonance Hybrid, Cannonical St. , Resonance Energy 44 min
        • Lecture9.8
          Condition Of Resonance 40 min
        • Lecture9.9
          Writing Cannonical St. 39 min
        • Lecture9.10
          Relative Stability Of Cannonical St. 37 min
        • Lecture9.11
          Resonance Energy 45 min
        • Lecture9.12
          Effect Of Resonance On Bond Length, Enthalpy Of Hydrogenation 43 min
        • Lecture9.13
          Introduction To Aromaticity 43 min
        • Lecture9.14
          Introduction To Aromaticity 39 min
        • Lecture9.15
          Unsaturation Factor 31 min
        • Lecture9.16
          Chapter Notes – GOC General Organic chemistry
      • 10. GOC 2 - Substituent effect
        5
        • Lecture10.1
          Substituent Effect, Hyperconjugation 48 min
        • Lecture10.2
          Substituent Effect, Hyperconjugation 43 min
        • Lecture10.3
          Substituent Effect, Mesomeric Effect 47 min
        • Lecture10.4
          Substituent Effect, Inductive Effect 46 min
        • Lecture10.5
          Substituent Effect, Electromeric Effect, Staric Effect, Relative M & I Effect 41 min
      • 11. GOC 2 - Reactive Intermediate
        6
        • Lecture11.1
          Reactive Intermediate, Carbocation 45 min
        • Lecture11.2
          Reactive Intermediate, Carbocation, Carbonium Ion Rearrangement 42 min
        • Lecture11.3
          Reactive Intermediate, Carbonium Ion Rearrangement 41 min
        • Lecture11.4
          Reactive Intermediate, Carbanion 36 min
        • Lecture11.5
          Reactive Intermediate, Free Radical 47 min
        • Lecture11.6
          Reactive Intermediate, Carbene & Nitrene 42 min
      • 12. GOC 2 - Acid, base, Electrophile, Nucleophile
        3
        • Lecture12.1
          Acid Base, Electrophile Nucleophile 50 min
        • Lecture12.2
          Acid Base, Electrophile Nucleophile 47 min
        • Lecture12.3
          Hard Acid Base, Electrophilic Nucleophilic Strength 40 min
      • 13. Isomerism
        20
        • Lecture13.1
          Structural Isomers 39 min
        • Lecture13.2
          Tautomerism 37 min
        • Lecture13.3
          Stability Of Tautomers 43 min
        • Lecture13.4
          Factors Affecting Stability, Catalysis In Tautomerism 39 min
        • Lecture13.5
          Geometrical Isomerism 41 min
        • Lecture13.6
          E-z Nomenclature, Properties Of G.i. 43 min
        • Lecture13.7
          No. Of G.i., Interconversion Of G.i. 48 min
        • Lecture13.8
          Optical Isomerism & Its Conditions 50 min
        • Lecture13.9
          Different Types Of Projections, R-s Configuration 57 min
        • Lecture13.10
          Relationship Between Optical Isomers 45 min
        • Lecture13.11
          Dissymmetry In A Molecule 44 min
        • Lecture13.12
          Enantiomers, Mesomers, Diastereomers 39 min
        • Lecture13.13
          Special Case Of Optical Isomerism 47 min
        • Lecture13.14
          No. Of Optical Isomers, Stereoisomers 45 min
        • Lecture13.15
          D,l Configuration, Retention & Inversion 36 min
        • Lecture13.16
          Measurement Of Optical Activity 45 min
        • Lecture13.17
          No. Of Isomers 35 min
        • Lecture13.18
          Resolution Of Optical Isomers, Syn, Anti Addition, Elimination. 28 min
        • Lecture13.19
          Conformational Isomers 51 min
        • Lecture13.20
          Conformers Of Propane, Butane, Cyclohexane & Problems 44 min
      • 14. Reaction Mechanism
        21
        • Lecture14.1
          Introduction, Types Of Organic Reactions 35 min
        • Lecture14.2
          Nucleophilic Substitution Reaction 40 min
        • Lecture14.3
          Sn1 & Sn2 Reaction, Sni Pathway 53 min
        • Lecture14.4
          Reactivity In Sn1 & Sn2 Path 42 min
        • Lecture14.5
          Reactivity In Sn1 & Sn2 Path 36 min
        • Lecture14.6
          Reactivity In Sn1 & Sn2 Path 30 min
        • Lecture14.7
          Reactivity In Sn1 & Sn2 Path 41 min
        • Lecture14.8
          Elimination Reaction 53 min
        • Lecture14.9
          E1 & E2 Reaction, Isotopic Effect 46 min
        • Lecture14.10
          Orientation In Elimination Reaction 45 min
        • Lecture14.11
          Problems On Elimination Reaction 48 min
        • Lecture14.12
          Elimination Vs Substitution 34 min
        • Lecture14.13
          Addition Reaction 51 min
        • Lecture14.14
          Problems On Addition Reaction 46 min
        • Lecture14.15
          Electrophilic Aromatic Substitution Reaction 49 min
        • Lecture14.16
          Orientation In Electrophilic Aromatic Substitution 53 min
        • Lecture14.17
          Reactivity In Electrophilic Aromatic Substitution Reaction 30 min
        • Lecture14.18
          Examples Of Electrophilic Aromatic Substitution Reaction 37 min
        • Lecture14.19
          Examples Of Electrophilic Aromatic Substitution Reaction 37 min
        • Lecture14.20
          Nucleophilic Aromatic Substitution 44 min
        • Lecture14.21
          Benzyne Pathway 27 min
      • 15. Alkane
        7
        • Lecture15.1
          Alkane Preparation 49 min
        • Lecture15.2
          Alkane Preparation & Selective Hydrogenation 31 min
        • Lecture15.3
          Alkane Preparation 40 min
        • Lecture15.4
          Alkane Preparation 38 min
        • Lecture15.5
          Alkane Preparation 32 min
        • Lecture15.6
          Alkane Properties 55 min
        • Lecture15.7
          Alkane Properties & Problems 39 min
      • 16. Alkene
        7
        • Lecture16.1
          Alkene Preparation 45 min
        • Lecture16.2
          Alkene Preparation 36 min
        • Lecture16.3
          Alkene Properties 53 min
        • Lecture16.4
          Alkene Properties 40 min
        • Lecture16.5
          Alkene Properties 42 min
        • Lecture16.6
          Alkene Properties & Ozonolysis 41 min
        • Lecture16.7
          Alkene Properties, Oxidation, Substitution 38 min
      • 17. Alkyl Halides
        4
        • Lecture17.1
          Preparation 38 min
        • Lecture17.2
          Properties 49 min
        • Lecture17.3
          Haloform Reaction 28 min
        • Lecture17.4
          Grignard Reagent 29 min
      • 18. Chemical Bonding
        32
        • Lecture18.1
          Introduction, definition, Concept & Type of Bonding 53 min
        • Lecture18.2
          Ionic Bonding, covalent bonding 50 min
        • Lecture18.3
          Ionic Character in Covalent Bonding, Electronegativity 34 min
        • Lecture18.4
          Dipole Moment 42 min
        • Lecture18.5
          Fajan’s Rule 34 min
        • Lecture18.6
          Model for Covalent Compound, V.B.T. – Lewis St. Model 56 min
        • Lecture18.7
          Lewis Structure Model 45 min
        • Lecture18.8
          Formal Charge 46 min
        • Lecture18.9
          Formal Charge Rule 44 min
        • Lecture18.10
          Resonance 43 min
        • Lecture18.11
          Merits & Demerits of Lewis St. Model 44 min
        • Lecture18.12
          Drawing Lewis St. 30 min
        • Lecture18.13
          VSEPR 1 49 min
        • Lecture18.14
          VSEPR 2 51 min
        • Lecture18.15
          VSEPR 3 51 min
        • Lecture18.16
          VSEPR 4 33 min
        • Lecture18.17
          BackBonding 38 min
        • Lecture18.18
          Bond Angle determination 47 min
        • Lecture18.19
          Concept of Hybridisation 44 min
        • Lecture18.20
          Sp3, Sp2 Hybridisation 44 min
        • Lecture18.21
          SP hybridisation, Relative study of SP, SP2, SP3 Hybridisation 46 min
        • Lecture18.22
          Hybridsation involving D-orbitals 39 min
        • Lecture18.23
          Hybridsation with D-orbitals, Limitation of Hybridisation 41 min
        • Lecture18.24
          Calculation of Hybridisation of Central Atom, Problems 43 min
        • Lecture18.25
          Merits & demerits of VBT, Introduction to MOT 33 min
        • Lecture18.26
          MO formation, Bond Order 43 min
        • Lecture18.27
          MO with P-orbitals, B2, Magnetic Character 43 min
        • Lecture18.28
          MO of Diatomic Species, Hetroatomic Species 51 min
        • Lecture18.29
          Secondary Bondings 39 min
        • Lecture18.30
          H Bonding 37 min
        • Lecture18.31
          Metallic Bonding 52 min
        • Lecture18.32
          Chapter Notes – Chemical Bonding
      • 19. Periodic Table
        10
        • Lecture19.1
          Development of P.T. 43 min
        • Lecture19.2
          Mandeelev P.T. & Mosley, Modern P.T. 43 min
        • Lecture19.3
          Modern P.T. & Periodic Properties 27 min
        • Lecture19.4
          Atomic Volume & Radius 49 min
        • Lecture19.5
          Atomic Radius, Ionisation Energy 28 min
        • Lecture19.6
          Ionisation Energy 48 min
        • Lecture19.7
          Electron Affinity, Hydration Energy 52 min
        • Lecture19.8
          Electronegativity, Lattice Energy 46 min
        • Lecture19.9
          Oxidising & Reducing Power, Nature of oxides 38 min
        • Lecture19.10
          M.P. & B.P., Density, Bond Energy, Diagonal relationship, Inert Pair Effect 25 min
      • 20. Metallurgy
        7
        • Lecture20.1
          Introduction, Concentration of ore 49 min
        • Lecture20.2
          Roasting, Calcination, smelting 41 min
        • Lecture20.3
          Refining of metal 29 min
        • Lecture20.4
          Pyrometallurgy, electrometallurgy, Hydrometallurgy 32 min
        • Lecture20.5
          Ellingham Diagram 43 min
        • Lecture20.6
          Extraction of Cu & Fe 22 min
        • Lecture20.7
          Extraction of Al & Zn 26 min
      • 21. Hydrogen and its Compounds
        7
        • Lecture21.1
          preparation, properties & Type of Hydrogen 57 min
        • Lecture21.2
          Compounds of Hydrogen, Hydrides, Water, Hydrates 56 min
        • Lecture21.3
          Hardness of Water, H2O2 56 min
        • Lecture21.4
          Problems 48 min
        • Lecture21.5
          Problems 28 min
        • Lecture21.6
          Chapter Notes – Hydrogen and its Compounds
        • Lecture21.7
          NCERT Solutions – Hydrogen
      • 22. S block metals
        8
        • Lecture22.1
          IA 1 – elemental Properties of Alkali metals& its Compounds 57 min
        • Lecture22.2
          IA 2 – Na & its compounds 01 hour
        • Lecture22.3
          IA 3 – Na & its Compounds, Use of Na & K 27 min
        • Lecture22.4
          IIA 1 – Elemental Properties 41 min
        • Lecture22.5
          IIA 2 – Compounds of IIA Metals 53 min
        • Lecture22.6
          IIA 3 – Compounds of Ca 48 min
        • Lecture22.7
          Chapter Notes – S block metals
        • Lecture22.8
          NCERT Solutions – S block metals
      • 23. p block elements
        8
        • Lecture23.1
          Introduction to P – Block & IIIA – elemental properties 51 min
        • Lecture23.2
          IIIA – General properties of compounds & B-compounds 40 min
        • Lecture23.3
          IIIA – Boron compounds, Use of B and Al 35 min
        • Lecture23.4
          IVA – Elemental Properties of C family 46 min
        • Lecture23.5
          IVA – Allotropes of C & compounds of C 01 hour
        • Lecture23.6
          IVA – Compounds of Si 48 min
        • Lecture23.7
          Chapter Notes – p block elements
        • Lecture23.8
          NCERT Solutions – p block elements

        NCERT Solutions – Atomic Structure

        2.1. (i) Calculate the number of electrons which will together weigh one gram.
        (ii) Calculate the mass and charge of one mole of electrons. 

        Answer

        (i) Mass of one electron = 9.11×10-31kg
        ∴ 1g = 10-3 kg =  (1/9.11×10-31)×10-3 electrons = 1.098×1027

        (ii) Mass of one electron = 9.11×10-31kg

        ∴ Mass of 1 mole of electrons = (9.11×10-31)×(6.022×1023) = 5.48×10-7 kg
        Charge on one electron = 1.602×10-19 coulomb
        ∴ Charge on 1 mole of electrons = (1.602×10-19)×(6.022×1023) = 9.65×104 coulombs. 

        2.2. (i) Calculate the total number of electrons present in one mole of methane.
        (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675×10-27kg).
        (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.
        Will the answer change if the temperature and pressure are changed ?

        Answer

        (i) Electrons present in 1 molecule of methane (CH4) = 6+4 = 10

        ∴ Electrons in  mol i.e. 6.022×1023 molecules = 6.022×1024 

        1 atom of 14C contains = 14-6 = 8 neutrons.

        ∴ The number of neutrons in 14 g of 14C = 6.022×1023×8 neutrons
        Number of neutrons in 7 mg = (6.022×1023×8×7)/14000 = 2.4088×1021 neutrons
        (b) Mass of one neutron = 1.674×10-27kg
        Mass of total neutrons in 7 g of 14C = (2.4088×1021) (1.675 ×10-27 kg) = 4.035 ×10-6 kg
        (iii) (a) 1 mol of NH3 = 17g NH3 = 6.022×1023 molecules of NH3 
        1 atom of NH3  contains = 7+3 = 10 protons
        ∴ The number of protons in 1 mol of NH3 = 6.022×1024 protons.
        Number of protons in 34 mg of NH3 = (6.022×1024×34)/17×1000 = 1.2044×1022 protons.
        (b) Mass of one proton = 1.6726×10-27kg
        ∴ Mass of 1.2044×1022 protons = (1.6726×10-27) ×(1.2044×1022) kg = 2.0145×10-5 kg.
        No, there will be no effect of temperature and pressure.
        2.3. How many neutrons and protons are there in the following nuclei ?
        13C6, 16O8, 24Mg12, 56Fe26, 88CSr38

        Answer

        Nucleus        Z           A         Protons(Z)        Neutrons(A-Z)

        13C6                 6           13             6                      13-6=7
        16O8                 8           16             8                      16-8=8
        24Mg12           12          24            12                     24-12=12
        56Fe26             26         56            26                     56-26=30
         88CSr38          38         88            38                     88-38=50

        2.4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
        (i) Z = 17 , A = 35.
        (ii) Z = 92 , A = 233.
        (iii) Z = 4 , A = 9.

        Answer
        (i) 35Cl17  
        (ii) 233U92

        (iii) 9Be4 

         

        2.5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number (ṽ) of the yellow light.

        Answer
        λ = 580 nm = 580×10-9 m
        frequency (ν) = c/λ = 3.0×108 ms-1/580×10-9 m = 5.17×1014 s-1

        wave number (ṽ) = 1/λ = 1/580×10-9 m = 1.72×106 m-1

         

        2.6. Find energy of each of the photons which
        (i) correspond to light of frequency 3×1015 Hz.
        (ii) have wavelength of 0.50 Å.

        Answer
        (i) ν = 3×1015 Hz
        E = hν = (6.626×10-34 Js)×(3×1015 s-1) = 1.988×10-18 J
        (ii) λ = 0.50×10-10 m
         E = hν = hc/λ = (6.626×10-34 Js)×(3×1015 s-1)/0.50×10-10 m = 3.98×10-15 J
        2.7. Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0×10-10 s.
        Answer
        Frequency (ν) =  1/Period = 1/2.0×10-10 s = 5×109 s-1.
        Wavelength (λ) = c/ν = 3.0×108 ms-1/5×109 s-1 = 6.0×102 m
        Wave number (ṽ) = 1/λ = 1/6.0×102 m = 16.66 m-1
         

        2.8. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?

        Answer
        λ = 4000 pm = 4000×10-12 m = 4×10-9 m
        E = Nhν = Nhc/λ
        ∴ N = E×λ/h×c = (1J×4×10-9 m)/(6.626×10-34 Js×3.0×108 ms-1) = 2.012×1016 photons. 

        2.9. A photon of wavelength 4×10-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate
        (i) the energy of the photon (eV),
        (ii) the kinetic energy of the emission, and
        (iii) the velocity of the photoelectron (1 eV = 1.6020×10-19 J).

        Answer

        (i) Energy of the photon (E) = hν = hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/4×10-7 m = 4.97×10-19 J
        = 4.97×10-19/1.602×10-19 eV
        (ii) Kinetic energy of emission (1/2 mv2) = hν- hνo = 3.10-2.13 = 0.97 eV
        (iii) 1/2 mv2 = 0.97 eV = 0.97×1.602×10-19 J
        ⇒ 1/2×(9.11×10-31 kg)×v2 = 0.97×1.602×10-19 J
        ⇒ v2 = 0.341×1012 = 34.1×1010
        ⇒ v = 5.84×105 ms-1

        2.10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.

        Answer

        E = Nhν = Nhc/λ = (6.022×1023 mol-1)×(6.626×10-34 Js×3.0×108 ms-1)/242×10-9 m
           = 4.945×105 Jmol-1 = 494.5 kJmol-1
         

        2.11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

        Answer
        Energy emitted by the bulb = 25 watt = 25 Js-1
        Energy of one photon (E) = hν = hc/λ
        λ = 0.57µm = 0.57×10-6 m
        E = (6.626×10-34 Js×3.0×108 ms-1)/0.57×10-6 m = 3.48×10-19 J
        ∴ No. of photons emitted per sec = 25 Js-1/3.48×10-19 J = 7.18×1019

        2.12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (νo) and work function (wo) of the metal.

        Answer
        c = νλ
        ∴ νo = c/λo = 3.0×108 ms-1/6800×10-10 m = 4.14×1014 s-1
        Work function (wo) = hνo = 6.626×10-34 Js×4.14×1014 s-1 = 2.92×10-19 J
        2.13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?
        Answer
        ṽ = R(1/n12 – 1/n22) = 109677(1/22 – 1/42) cm-1 = 20564.4 cm-1 
        λ = 1/ν = 1/20564.4 = 486×10-7 cm =  486×10-9 m = 486 nm 

        2.14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n=1 orbit).

        Answer

        En = -21.8×10-19/n2 Jatom-1 
        For ionization from 5th orbit, n1 = 5, n2 = ∞

        ∴ ΔE = E2 – E1 = -21.8×10-19×(1/n22 – 1/n12) = 21.8×10-19×(1/n12 – 1/n22)
           = 21.8×10-19×(1/52 – 1/∞) = 8.72×10-20 J
        For ionization from 1st orbit, n1 = 1, n2 = ∞
        ∴ ΔE’ = 21.8×10-19×(1/12 – 1/∞) = 21.8×10-19 J
        ΔE’/ΔE = 21.8×10-19/8.72×10-20 = 25

        Hence, 25 times less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.

        2.15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state? 

        Answer

        The number of spectral lines produced when an electron in the nthlevel drops down to the ground state is given by n(n-1)/2.

        Given, n=6
        ∴ Number of spectral lines = 6×5/2 = 15
        also given by, ∑(n2 – n1) = ∑(6-1) = ∑5 = 5+4+3+2+1 = 15
        2.16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18×10-18 J atom-1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
        Answer
        (i) En = -21.8×10-19/n2 J
        ∴ E5 = -21.8×10-18/52 J = 8.72×10-20 J
        (ii) For H atom, rn = 0.529×n2 Å
        ∴  r5 = 0.529×52 = 13.225 Å = 1.3225 nm
        2.17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
        Answer

        For the Balmer series, n1 = 2. Hence, ṽ = R(1/22 – 1/n22)

        ṽ = 1/λ  (inversely proportional)
        For λ to be maximum, ṽ should be minimum. This can be happened when n2 is minimum i.e. n2 = 3. Hence, ṽ = (1.097×107 m-1) (1/22 – 1/32) = 1.097×107×5/36 m-1 = 1.523×106 m-1
         

        2.18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18×10-11 ergs.

        Answer
        1erg = 10-7 J
        As ground state electronic energy is  –2.18×10-11 ergs, this means that En = -21.8×10-11/n2 ergs.
         ΔE = E5 – E1 = 2.18×10-11 (1/12 – 1/52) = 2.18×10-11(24/25) = 2.09×10-11 ergs = 2.09×10-18 J
        When electron returns to ground state (n=1), energy emitted = 2.09×10-11 ergs.
        As, E = hν = hc/λ
        ⇒ λ = hc/E = (6.626×10-27 erg sec) (3.0×1010 cm s-1)/2.09×10-11 ergs
                = 9.51×10-6 cm = 951×10-8 cm = 951 Å
        2.19. The electron energy in hydrogen atom is given by E n = (–2.18×10-18)/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
        Answer
        ΔE = E∞– E2 = 0 – (–2.18×10-18 J atom-1/22) = 5.45×10-19 J atom-1
        ΔE = hν = hc/λ
        ⇒ λ = hc/ΔE = (6.626×10-34 Js)×(3.0×108 ms-1)/5.45×10-19 J = 3.674×10-7 m = 3.674×10-5 cm 

        2.20. Calculate the wavelength of an electron moving with a velocity of 2.05×107 ms-1.

        Answer

        By de Broglie equation,
        λ  = h/mv = 6.626×10-34 Js/(9.11×10-31 kg) (2.05×107 ms-1) = 3.55×10-11 m
        2.21. The mass of an electron is 9.11×10-31 kg. If its K.E. is 3.0×10-25 J, calculate its wavelength.
        Answer

        K.E. = 1/2 mv2

        NCERT Solutions for Class 11th: Ch 2 Structure of Atom

        ∴ v = √2 K.E./m

        = 812 ms-1

        By de Broglie equation, λ = h/mv = 6.626×10-34 Js/(9.11×10-31 kg) (812 ms-1) = 8.967×10-7 m

        2.22. Which of the following are isoelectronic species i.e., those having the same number of electrons? 
        Na+ , K+ , Mg2+ , Ca2+ , S2- , Ar.
        Answer
        Notes:

        Isoelectronic are the species having same number of electrons.
        A positive charge means the shortage of an electron.

        A negative charge means gain of electron.
        Number of electrons in Na+ = 11-1 = 10
        Number of electrons in K+ = 19-1 = 18
        Number of electrons in Mg2+ = 12-2 = 10
        Number of electrons in Ca2+ = 20-2 = 18
        Number of electrons in S2- = 16+2 = 18
        Number of electrons in Ar = 18
        Hence, the following are isoelectronic species:
        1) Na+ andMg2+ (10 electrons each)
        2) K+, Ca2+, S2- and Ar (18 electrons each)
        2.23. (i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2- (d) F–
        (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5 ?
        (iii) Which atoms are indicated by the following configurations ? (a) [He]2s1 (b) [Ne]3s2 3p3 (c) [Ar] 4s2 3d1 .
        Answer
        (i) (a) 1H = 1s1 . A negative charge means gain of electron.
        ∴ electronic configuration of H– = 1s2
        (b) 11Na = 1s22s22p63s1 . A positive charge means the shortage of an electron.
        ∴ electronic configuration of Na+ = 1s22s22p6
        (c) 8O = 1s22s22p4
        ∴ electronic configuration of O2- = 1s22s22p6

        (d) 9F = 1s22s22p5

        ∴ electronic configuration of F– = 1s22s22p6

        (ii) (a) 3s1
        Completing the electron configuration of the element as 1s22s22p63s1
        ∴ Number of electrons present in the atom of the element = 2+2+6+1 = 11
        ∴ Atomic number of the element = 11
        (b) 2p3
        Completing the electron configuration of the element as 1s22s22p3
        ∴ Number of electrons present in the atom of the element = 2+2+3 = 7
        ∴ Atomic number of the element = 7
        (c) 3p5
        Completing the electron configuration of the element as 1s22s22p63s23p5
        ∴ Number of electrons present in the atom of the element = 2+2+6+2+5 = 17
        ∴ Atomic number of the element = 9

        (iii)  (a) [He]2s1
        electronic configuration = 1s22s1
        ∴ Atomic number of the element = 2+1 = 3
        Hence, the element with the electronic configuration [He]2s1 is lithium (Li).
        (b) [Ne]3s23p3
        electronic configuration = 1s22s22p63s23p3
        ∴ Atomic number of the element = 2+2+6+2+3 = 15
        Hence, the element with the electronic configuration  [Ne]3s23p3 is phosphorus (P).
        (c) [Ar] 4s23d1
        electronic configuration = 1s22s22p63s23p64s23d1
        ∴ Atomic number of the element = 2+2+6+2+6+2+1 = 21
        Hence, the element with the electronic configuration [Ar] 4s23d1 is scandium (Sc). 

        2.24. What is the lowest value of n that allows g orbitals to exist?

        Answer

        For g-orbitals, l = 4.

        For any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
        ∴ For l = 4, minimum value of n = 5
        2.25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
        Answer

        For the 3d orbital:
        Principal quantum number (n) = 3
        Azimuthal quantum number (l) = 2
        Magnetic quantum number (ml) = -2,-1,0,1,2

        2.26. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
        Answer
        (i) For  neutral atom,  number of protons = number of electrons.
        ∴ Number of protons in the atom of the given element = 29 = Atomic number
        (ii) The electronic configuration of the atom with Z=29 is  1s22s22p63s23p64s13d10
         

        2.27. Give the number of electrons in the species H2+ , H2 and O2+

        Answer 

        H2+ = 2-1 = 1 electron
        H2 = 1H + 1H = 2 electrons
        O2+ = 16-1 = 15 electrons

        2.28. (i) An atomic orbital has n = 3. What are the possible values of l and ml ?
        (ii) List the quantum numbers (ml and l ) of electrons for 3d orbital.
        (iii) Which of the following orbitals are possible?
               1p, 2s, 2p and 3f

        Answer

        (i) For a given value of n, l can have values from 0 to (n-1).
        ∴ For n = 3 , l = 0, 1, 2
        For a given value of l, ml can have (2l+1) values.
        When l = 0, m = 0
        l = 1, m = – 1, 0, 1
        l = 2, m = – 2, – 1, 0, 1, 2
        l = 3, m = -3, -2, -1, 0, 1, 2, 3
        (ii) For 3d orbital, n = 3, l = 2.
        ∴ For l = 2
        m2 = -2, -1, 0, 1, 2
        (iii) 1p is not possible because when n = 1, l = 0. (for p, l = 1)
        2s is possible because when n=2, l = 0,1 (for s, l=0)
        2p is possible because when n=2, l = 0,1 (for p, l=1)
        3f is not possible because when n=3, l = 0, 1 , 2 (for f, l=3)

        2.29. Using s, p, d notations, describe the orbital with the following quantum numbers.
                 (a) n=1, l=0; (b) n=3; l=1 (c) n=4; l=2; (d) n=4; l=3.

        Answer

        (a) 1s
        (b) 3p
        (c) 4d
        (d) 4f

        2.30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.
                (a) n = 0,           l = 0,         ml = 0,            ms = +½
                (b) n = 1,          l = 0,          ml = 0,            ms = –½
                (c) n = 1,          l = 1,          ml = 0,            ms = +½
                (d) n = 2,          l = 1,          ml = 0,            ms = –½
                (e) n = 3,          l = 3,          ml = –3,          ms = +½
                (f) n = 3,          l = 1,           ml = 0,            ms = +½

        Answer

        (a) Not possible because n≠0
        (b) Possible
        (c) not possible because when n=1, l≠1

        (d) Possible
        (e) Not possible because when n=3,  l≠3
        (f) Possible 

        2.31. How many electrons in an atom may have the following quantum numbers?
               (a) n = 4, ms = –½        (b) n = 3, l = 0

        Answer

        (i) The total number of electrons in n is given by 2n2
        n=4, Number of electrons = 2×42 = 32

        Half of 32 electrons will have spin quantum number ms = –½ i.e. 16 electrons
        (ii) n=3 and l=0 means it is 3s orbital which can have only 2 electrons.
        2.32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
        Answer
        According to Bohr postulate of angular momentum,
        mvr = nh/2π ⇒ 2πr = nh/mv   …(i)
        According to de Broglie equation, λ=h/mv   …(ii)
        Substituting the value of eqn (ii) in eqn (i) we get,
        2πr = nλ
        Thus, circumference of the Bohr orbit for the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit. 

        2.33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?

        Answer

        For H-like particles, ṽ = (2π2mZ2e4/ch3)×(1/n12 – 1/n22) = RZ2(1/n12 – 1/n22)
        ∴ For He+ spectrum, Balmer transition, n=4 to n=2
        ṽ = 1/λ = RZ2(1/22 – 1/42)  = R×4×3/16 = 3R/4
        For hydrogen spectrum ,
        ṽ = 1/λ = R(1/n12 – 1/n22) = 3R/4  ⇒ (1/n12 – 1/n22) = 3/4
        which can be true only for n1=1 and n2=2 i.e. transition from n=2 to n=1.

        2.34. Calculate the energy required for the process
                He+(g) → He2+(g) + e– 
        The ionization energy for the H atom in the ground state is 2.18×10-18 J atom-1.

        Answer

         
        For H-like particles, En = – (2π2mZ2e4/n2h2)
        For H-atom, I.E. = E-E1 = 0 – (-2π2m×22×e4/12×h2)
        = (4×2π2me4/h2) = 4×2.18×10-18 J = 8.72×10-18 J
        2.35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
        Answer

        Diameter of a carbon atom = 0.15 nm = 0.15×10-9 m = 1.5×10-10 m

        Length along which atoms are to be placed = 20 cm = 2×10-1 m
        ∴ No. of C-atoms which can be placed along the line = 2×10-1 m/1.5×10-10 m = 1.33×109
         

        2.36. 2×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

        Answer
        Total length = 2.4 cm
        Total number of atoms along the length = 2×108
        ∴ Diameter of each atom = 2.4 cm/2×108 = 1.2×10-8 cm
        ∴ Radius of the atom = Diameter/2 = 1.2×10-8 cm/2 = 0.6×10-8 cm
        2.37. The diameter of zinc atom is 2.6 Å.Calculate (a) radius of zinc atom in pm and (b)number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
        Answer
        (a) Radius of zinc atom = 2.6Å/2 = 1.3Å = 1.3×10-10 m = 130×10-12 pm
        (b) Given length = 1.6 cm = 1.6×10-2 m
        Diameter of one atom = 2.6 Å = 2.6×10-10 m
        ∴ No. of atoms present along the length = 1.6×10-2/2.6×10-10 = 6.154×107
         

        2.38. A certain particle carries 2.5×10-16 C of static electric charge. Calculate the number of electrons present in it.

        Answer
        Charge on one electron = 1.602×10-19 C
        ∴ Number of electrons carrying 2.5×10-16 C charge = 2.5×10-16/1.602×10-19 =1560
        2.39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282×10-18 C, calculate the number of electrons present on it.
        Answer
        As in the above question,
        Number of electrons present in oil drop = –1.282×10-18/1.602×10-19 = 8
        2.40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
        Answer
        Heavy atoms have nucleus carrying  large amount of positive charge. Therefor, some α-particles will easily deflected back. Also a number of α-particles are deflected through small angles because of large positive charge.
        If light atoms are used, their nuclei will have small positive charge, hence the number of α-particles getting deflected even through small angles will be negligible.
        2.41. Symbols 79Br35 and 79Br can be written, whereas symbols  35Br79 and 35Br are not acceptable. Answer briefly. 

        Answer

        35Br79 is not acceptable because atomic number should be written as subscript, while mass number should be written as superscript. 35Br is not acceptable because atomic number of an element is fixed. However, mass number is not fixed as it depends upon the isotopes taken. Hence, it is essential to indicate mass number.

        2.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

        Answer

        Mass number = protons+neutrons = p+n = 81 (given)
        Let p be x, then neutrons = x + (31.7/100)x = 1.317 x
        ∴ x + 1.317 x = 81
        ⇒ 2.317 x = 81
        ⇒ x = 81/2.317 = 35
        Thus, protons = 35 = atomic number.
        The symbol of the element is 81Br35 or 8135Br

        2.43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

        Answer

        Let the number of electrons in the ion = x
        Then, number of neutrons = x+(11.1 x/ 100) = 1.111 x
        Number of electrons in the neutral atom = x-1 (ion possesses one unit of negative charge)
        ∴ Number of protons = x-1
        Mass number = No. of protons + No. of neutrons
        ∴ 1.111 x + x – 1 = 37
        ⇒ 2.111x = 38
        ⇒ x = 18
        ∴ No. of protons = Atomic no. = x-1 = 18-1 = 17
        The symbol of the ion is 3717Cl-1

        2.44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

        Answer
        Let the number of electrons in the ion = x
        Then, number of neutrons = x+(30.4 x/ 100) = 1.304 x
        Number of electrons in the neutral atom = x+3 (ion possesses 3 units of positive charge)
        ∴ Number of protons = x+3
        Mass number = No. of protons + No. of neutrons
        ∴ 1.304 x + x +3 = 56
        ⇒ 2.304x = 53
        ⇒ x = 23
        ∴ No. of protons = Atomic no. = x+3 = 23+3 = 26
        The symbol of the ion is 5626Fe+3

        2.45. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
        Answer

        The increasing order of frequency is as follows:
        Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays

        2.46. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6×1024, calculate the power of this laser. 

        Answer

        E = Nhν = Nhc/λ = (5.6×1024)×(6.626×10-34 Js×3.0×108 ms-1)/337.1×10-9 m = 3.3×106 J

        2.47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

        Answer

        λ  = 616 nm = 616×10-9 m
        (a) Frequency, ν = c/λ = 3.0×108 ms-1/616×10-9 m = 4.87×1014 s-1

        (b) Velocity of the radiation = 3.0×108 ms-1

        ∴ Distance travelled in 30 s = 30×3×108 m = 9.0×109 m
        (c) E = hν = hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/616×10-9 m = 32.27×10-20 J
        (d) No. of quanta in 2J of energy = 2J/32.27×10-20 J = 6.2×1018

        2.48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15×10-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

        Answer
        Energy of one photon = hν = hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/600×10-9 m = 3.313×10-19 J
        Total energy received = 3.15×10-18 J
        ∴ No. of photons received = 3.15×10-18 J/3.313×10-19 J = 9.51 (approx 10)
        2.49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5×1015 , calculate the energy of the source.
        Answer
        Frequency = 1/2×10-19 s = 0.5×109 s-1

        Energy = Nhν = (2.5×1015)×(6.626×10-34 Js)×(0.5×109 s-1) = 8.28×10-10 J

        2.50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states.
        Answer
        λ1 = 589 nm = 589×10-9 m
        ∴ ν1 = c/λ1 = 3.0×108 ms-1/589×10-9 m = 5.093×1014 s-1
        λ2 = 589.6 nm = 589.6×10-9 m
        ∴ ν2 = c/λ2 = 3.0×108 ms-1/589.6×10-9 m = 5.088×1014 s-1
        ΔE = E2 – E1 = h(ν2 – ν1) = (6.626×10-34 Js)×(5.093-5.088)×1014 s-1 = 3.31×10-22 J 

        2.51. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

        Answer

        (a) Work function (W0) = hν0
        ∴ ν0 = W0/h = 1.9×1.602×10-19 J/6.626×10-34 Js = 4.59×1014 s-1            (1eV = 1.602×10-19 J)
        (b) λ0 = c/ν0 = 3.0×108 ms-1/4.59×1014 s-1 = 6.54×10-7 m = 654×10-9 m = 654 nm
        (c) K.E. of ejected electron = h(ν – ν0) = hc (1/λ – 1/λ0 )
        = (6.626×10-34 Js×3.0×108 ms-1)×(1/500×10-9 m – 1/654×10-9 m)
        = (6.626×3.0×10-26/10-9)×(154/500×654) J = 9.36×10-20 J
        K.E. = 1/2 mv2 = 9.36×10-20 J
        ∴ 1/2×(9.11×10-31 kg) v2 = 9.36×10-20 kgm2s-2
        ⇒ v2 = 2.055×1011 m2s-2 = 20.55×1010 m2s-2

        ⇒ v = 4.53×105 ms-1
         

        2.52. Following results are observed when sodium metal is irradiated with different wavelengths.         Calculate (a) threshold wavelength and, (b) Planck’s constant.

              λ (nm)                          500             450           400
            v×10-5(cm s-1)               2.55            4.35          5.35
        Answer
        Let the threshold wavelength to be λ0 nm = λ0×10-9 m . 
        Following equation holds true for photoelectric emission in given case:
        K.E. = 1/2 mv2 = h(ν – ν0)
        ⇒1/2 mv2 = hν – hν0
        ⇒ hν0 = hν – 1/2 mv2
        ⇒ hc/λ0 = hc/λ – 1/2 mv2 

        (a) Substituting the value of λ and v from the above given data, we get three values of λ0 as,
        λ0(1) = 541 nm
        λ0(2) = 546 nm
        λ0(3) = 542 nm
        Threshold frequency = λav = {λ0(1)+λ0(2)+λ0(3)}/3 = (541+546+542)/3 = 543 (approx 540)
        (b) Part of this question can’t be solved due to incorrect value of v i.e 5.35.
        Students can assume this value as 5.20 if they want to solve this question.

        2.53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

        Answer

        Energy of the incident radiation = Work function + Kinetic energy of photoelectron
        E = hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/(256.7×10-9 m) = 7.74×10-19 J = 4.83 eV
        The potential applied gives kinetic energy to the electron.
        Hence, kinetic energy of the electron = 0.35 eV
        ∴ Work Function = 4.83 eV – 0.35 eV = 4.48 eV

        2.54. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5×107 ms-1 , calculate the energy with which it is bound to the nucleus.

        Answer

        Energy of the incident photon= hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/(150×10-12m) = 13.25×10-16 J
        Energy of the electron ejected = 1/2 mv2 = 1/2×(9.11×10-31kg)×(1.5×107ms-1)2 = 1.025×10-16 J
        Energy with which the electron was bound to the nucleus = 13.25×10-16 J – 1.025×10-16 J
        = 12.225×10-16 J = 12.225×10-16/1.602×10-19 eV = 7.63×103 eV

        2.55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as v = 3.29×1015(Hz) [1/32–1/n2] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

        Answer

        ν = c/λ = 3.0×108 ms-1/1285×10-9 m = 3.29×1015 (1/32 – 1/n2)
        ⇒ 1/n2 = 1/9 – (3.0×108 ms-1/1285×10-9 m)×(1/3.29×1015) = 0.111-0.071 = 0.04 = 1/25
        ⇒ n2 = 25
        ⇒ n = 5
        The radiation corresponding to 1285 nm lies in the infrared region.

        2.56. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

        Answer

        Radius of nth orbit of H-like particles = 0.529n2/Z Å = 52.9n2/Z pm
        r1 = 1.3225 nm = 1322.5 pm = 52.9n12
        r2 = 211.6 pm = 52.9n22/Z
        ∴ r1/r2 = 1322.5 pm/211.6 pm = n12/n22
        ⇒ n12/n22 = 6.25
        ⇒ n1/n2 = 2.5
        If n2 = 2, n1 = 5. Thus the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series.
        ṽ = 1.097×107 m-1 (1/22 – 1/52) = 1.097×107×21/100 m-1
        λ = 1/ṽ = 100/(1.097×21×107) m = 434×10-9 m = 434 nm
        It lies in visible range.

        2.57. Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6×106 ms-1 , calculate de Broglie wavelength associated with this electron.

        Answer

        λ = h/mv = 6.626×10-34 kgm2s-1/(9.11×10-31 kg) (1.6×106 ms-1) = 4.55×10-10 m = 455 pm

        2.58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

        Answer

        Mass of neutron = 1.675×10-27 kg
        λ = h/mv
        ⇒ v = h/mλ = 6.626×10-34 kgm2s-1/(1.675×10-27 kg) (800×10-12 m) = 4.94×104 ms-1

        2.59. If the velocity of the electron in Bohr’s first orbit is 2.19×106 ms-1, calculate the de Broglie wavelength associated with it.

        Answer
        λ = h/mv = 6.626×10-34 kgm2s-1/(9.11×10-31 kg) (2.19×106 ms-1) = 3.32×10-10 m =332 pm
        2.60. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37×105 ms-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity.
        Answer
        λ = h/mv = 6.626×10-34 kgm2s-1/(0.1 kg) (4.37×105 ms-1) = 1.516×10-28 m
        2.61. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is 
        h/(4π×0.05 nm), is there any problem in defining this value.
        Answer

        Δx = 0.002 nm = 2×10-12 m

        Δx × Δp = h/4π
        ∴ Δp = h/4πΔx = 6.626×10-34 kgm2s-1/(4×3.14×2×10-12 m) = 2.638×10-23 kgms-1
        Actual momentum = h/(4π×0.05 nm) = h/(4π×5×10-11 m)
        = 6.626×10-34 kgm2s-1/(4×3.14×5×10-11 m) = 1.055×10-24 kgms-1
        It cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty.
        2.62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
        1. n = 4, l = 2, ml  = –2 , m s = –1/2
        2. n = 3, l = 2, ml  = 1 , m s = +1/2
        3. n = 4, l = 1, ml  = 0 , m s = +1/2
        4. n = 3, l = 2, ml  = –2 , m s = –1/2
        5. n = 3, l = 1, ml  = –1 , m s = +1/2
        6. n = 4, l = 1, ml  = 0 , m s = +1/2
        Answer
        The orbitals occupied by the electrons are:
        (1) 4d
        (2) 3d
        (3) 4p
        (4) 3d
        (5) 3p
        (6) 4p
        Same orbitals will have same energy and higher the value of (n+l) higher is the energy,
        Their energies will be in order: (5)<(2)=(4)<(6)=(3)<(1)
        2.63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?
        Answer
        4p electrons, being farthest from the nucleus experience the lowest effective nuclear charge.
        2.64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
        Answer
        (i) 2s is closer to the nucleus than 3s. Hence, 2s will experience larger effective nuclear charge.
        (ii) 4d (Reason being the same as above)
        (iii) 3p (Reason being the same as above)
        2.65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
        Answer
        Silicon has greater nuclear charge (+14) than aluminium (+13). Hence, the unpaired 3p electron in case of silicon will experience more effective nuclear charge.
        2.66. Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
        Answer
        (a) 15P = 1s2 2s2 2p6 3s2 3 px1 py1 pz1. 3 unpaired electrons. (in 3p)
        (b) 14Si = 1s2 2s2 2p6 3s2 3 px1 py1. 2 unpaired electrons.(in 3p)
        (c) 24Cr = 1s2 2s2 2p6 3s2 3p6 3d5 4s1. 6 unpaired electrons. (5 in 3d and 1 in 4s)
        (d)  26Fe = 1s2 2s2 2p6 3s2 3p6 3d6 4s2. 4 unpaired electrons. (in 3d)
        (e) 36Kr = It is a Noble gas. All orbitals are filled. No unpaired electrons. 

        2.67. (a) How many sub-shells are associated with n = 4 ? (b) How many electrons will be present in the sub-shells having ms value of –1/2 for n = 4 ?

        Answer

        (a) n=4, l = 0, 1, 2, 3. 4 sub-shells are associated with n = 4
        (b) No. of orbitals in the shells = n2 = 42 = 16
        Each orbitals has one electron with ms = -1/2. Hence, there will be 16 electrons with ms = -1/2.

        Prev Chapter Notes – Atomic Structure
        Next Introduction, Eqb constant & Eqb Position

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