Ttisagainst to law of twas failed to explain stability ofatom According to classical electromagnetic theo: the revolving electron should lose a woe avoslyandtravel inspiral all Finally ‘mst fallinto the nucleus. But it doesnot happen. The hydrogen atomic spectrum should be sontinvous band spect due to conttiouS Joss ofenergy but itis aline spectrum. > Itcan’texplainthe electronic structure ofatom and energies ofelectrons. ‘ATOMIC NUMBER & MAS' UMBER > “Themumberofprotos present ise the nucleus ofan atom of an element is called its atomic eee . > Moseley gives the relation between atomic number (Z) and frequency(v) of the characteristic X-rays of the element by the equation /v = a(Z-b) Where ‘a’ and ‘b’ are constants and depends onnature of the elements. > ‘Aneutralatom contains equal number ofelectrons and protons. For Cation : Number ofprotons Number ofelectrons= z— no.of electrons lost For Anion: Number ofprotons =z Number ofeléctrons=z+no.ofelectrons gained ye Slope=a vvv Slope-a Co > The sumofprotons and neutrons in an atom of an element is called its mass number(A). A=no.ofneutrons +no.of protons A=ntz Number ofneutrons= 4 -Z. Mass number is always a whole number, WES: What will be the percentage decrease in mass number if the number of neutrons halved and the number of electrons doubled in ?C Tagua aa ela ofprotons and neutrons) ‘Mass no is the sum In2C Initial Final 6 protons : 6 ° ‘Neutrons: 6 % ss NO + % feed the decrease in mass nois25.0% — WE.6: Calculate the n0.0f] Fprotons,neutrons ‘and electrons in {, Cl Ans: No.ofprotons=Atomic number(Z)=17 mass number(A)=37 No.ofneutrons~+ -J-31-17=20 No.of ‘electrons=17 ’ we. 7; Calculate the no.of protons, neutrons and electrons in 33Br ‘Ans: No.ofprotons = Atomic number(Z)=35 ‘mass number(A)=80 sutral. So, no.of electrons=35 The species isnet No.of neutrons=A-Z-80-35=45. WE.8: The number of electrons, protons and, neutrons in species are equal to 18, 16 and, 16 respectively. Assign the proper symbol to the species. Ans: No.of] protons=Atomic number(Z)=16 ‘The element is Sulphur(S). No.ofneutrons=16 ‘Mass number(A) = No. ofProtons+No. of Neutrons= 16+16=32 The species is not neutral as no.ofprotons is not equal to no, of electrons. It is anion with charge equal to excess electrons = 18- 16=2. Symbol is #25? W.E.9: Calculate the no.of protons, neutron and electron in \N*ion No.of protons=atomic number(Z)=7 \ No.of neutrons =(A-Z)=14-7=7 ! No.ofelectrons in an ion =Z+ magnitude of charge=7+3=10 W.E.10: The no.of electrons, protons and ORD in a species are equal to 10,11,12 respectively. Assign proper symbol to the species: Sol. No.of protons=11, hence atomic no.=11 so the element is Na. : It has one eletron less than the no.of protons, Pea eas No.ofneutrons=12 lass number=no. of protons + niit 293 a Therefore the symbol of that species is 3? Na” Sol4 Bienes DIFFERENT TYPES NU Isotopes ; Sue > en ee element having same atomic ut bei a ‘erent mass numbers are called > Isotopes show similar chemical properties but Z different Physical and radioactive properties. Fractional atomic masses of elements is due to the presence of Isotopes, eg. 1) Isotopes of hydrogen: Protium (|), Deuterium (?#7) or 2D era Tritium (}H) or 37°. Protium (9.985%), Deuterium (0.015%) 2) Isotopes ofchlorine: 31 and}? C1 3) isotopes of uranium: 2°U and 3'U _ DAs > Average atomic mass = vx 7 ‘A=Atomic mass ofisotope X=percentage abundance of isotope |B. 11: The mass number of three isotopes of | an element are 10,12,14 units, Their percentage abundance is 80,15 and 5 respectively. What is the atomic weight of | the element? Percentage abundances of Isotopes=80,15,5 ratio of percentage abundances of Tsotopes=16:3:1 The total ratio= 16+3+1=20 Sol: Y ratio of Yeabundancex Atomic weight ppiicalio ofa reee a Ayg.Awt= ie Total ratio @ 10x16+12x3+14x1 =105 20 The average Atomic weight =10.5. (2: Naturally occuring boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron? Ans: Let the: percentage of isotope with atomic weight 10.01 =x The percentage of isotope with atomic weight 11.01=(100-x) fi JEE-ADV CHEM-VOL-I wera 4 joe 4+% 10.81= x(10.01 eqn 720% +z. The % ofisotope with At.wt10.01 =20% The % of isotope with At.wt 11.01= 80.0% Tsobars: > The atoms of different elements having same mass number but different atomic numbers are called isobars. > [sobars show similar physical properties but different chemical properties. eg i4C, NN Die Arg Xm Ca 3): BXe, ye Ba Isotones : > The nuclides of the different elements with different atomic number and mass number but having same number of neutrons(A-Z) are called isotones. > Isotones show different physical and chemical properties. i eg!) 251,217,282) 'C,¢0 3) ftNa,iaMe Isodiaphers : > The nuclides having same isotopic number(A- 2Z) are called isodiaphers. They posses same difference of neutrons and protons (n-p). eg. Fi Na Isosters : > The molecules with same atomicity and same. number of electrons are called isosters. Eg(1):CO,,N,O (2): CoH, BsNsHy (Atomicity = total no.ofatoms in molecule) Iso-electronic species > The molecules or ions with same number of electrons are called iso electronic species Eg 1: N*,07,F”, Ne, Na’, Mg” , Al” Eg?: P?,S?,CI”, Ar,K*,Ca™ Sc” NATURE OF LIGHT & WAVE PROPERTIES: > Itcanbe explained by two theories a) corpuscular theory of light b) wave theory of lightCorpuscular theory : According to Newion, light and other form of radiant energies are propagated in the form of Corpuscules(simple invisible particle). Different colors of the light is due to different sizes of Corpuscules. This theory does not explain diffraction, interference, polarisation of light This theory explains photo electric effect, compton effect After the wave theory of light, the corpuscular theory lost significance Wave theory : It was proposed by Huygens ‘According to Huygens Light and other forms of radiant energy propagate through space inthe form of waves. Different colors of light is due to different wave lengths. This theory explains diffraction, interference, polarisation of light This theory doesnot explain photo eketric effect, compton effect, Black body radiations. Electromagnetic theory: Maxwell proposed when electrically charged particle moves under acceleration, alternating electrical and magnetic fields are produced and transmitted. These are transmitted in the form of wave and are associated with electric ‘and magnetic fields are called electromagnetic radiation(EMR) or electromagnetic waves. Both electric and magnetic fields are mutually . perpendicular and, perpendicular to the direction of the propagation. Allelectromagnetic waves have same velocity but have different energies, frequencies, wave numbers and wave lengths. These electromagnetic waves do not require any ‘medium and can travel in vaccum. Intensity oc maximum magnetic field x maximum electric field. 1 o EyBy > Electromagnetic wave “Magnetic Field(®) Electric pea Propagation Des Characteristic properties of wave Wave Length ( ae e The distance between two similar points ing wn as Wave length. wave is kno\ ngth are m, cm, A®, om, The units of wave le y, mpor pm. 1A°=10%em=10'°m Inm= 107cm=10° m=Imp= 10A° Ipm=10"em= 10m Frequency (v): The number of waves that pass through a given: point in one second is called frequency. Units : The Sl units are sec! cycles per: second (cps) or Hertz (Hz). leps = 1 Hz=sec" Frequency(v) ‘and wavelength (2 ) are related as V= Fi c= Velocity oflight = 3 x 108 m/sec Wave Number (v) : The number of wave lengths per unit length is called wave number. The reciprocal of wave length is called wave number, Units: cm™ or m™' Wave number Y= The relation between v and v is = ev Amplitude (a): The height ofthe crest or depth of the trough of awave is called amplitude. Units: m, cm, pm Amplitude is a measure of the intensity or brightness ofa beam of light.Velocity (c); The distance travelle is called its velocity, Units: m/sec; cm/sec, Alltypes ofelectro; same velocity which is light. c=3x108 m5"! d bya wave in one second ‘etic radiations have the qual to the velocity of like A= 10 "en 10%m Inm=10’cm=10'm=10A Ipm=10 “em =10°"m “Velocity (c) c=vh | Lo WE. TI ‘he vividh bharati station of All india Radio, Dethi,broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to? c Sol: a where c= 3x10° msec v = frequency =1368kHz =1368x10° sec”! WE.14: The wavelength range of the visible spectrum extends from violet(400 nm) to red (750 nm). Express these wavelengths in ‘frequencies (Hy).(Inm=10"'m) Sol: Frequency of violet light ¢ _ 3.00x10°ms " =F q00x108 m= 750x10" Hz frequncy ofred light c _ 3.00x10%ms" 4 ~F > F50x10°m 400x10 Hz ‘The range of visible spectrum is from 4.0x10" t07.5x10"* Hz interms of frequency W.E.15: Calculate (a) Wavenumber and (b) frequency of yellow radiation having wavelength 5800’. Sol: a) Calculation of wavenumber (v) A= 5800A° = 5800x10%cm=5800 x10" m 1 1 A 580010" m =1.724x10%em™ v= =1.724x108m"! © _3.00x108 ms b) Calculation of the frequency (Vv) =~ == 219.3m vy 1368x10°s c axl0'ms! i is i crn =f 2 _e5.17ax10"s This is a characteristic radiowave wavelength. 7 3800x10m BS Increasing wavelength ——> —— Decreasing frequency ——> 10% 10% 10 10" 10" 10" to" 1010" tt 10 | | | ! Lo eee \ : Tm facmne [rll] i \ (a) lee yrays X-rays]UV || IR [sore [we Long radio waves | a | bo | 1 r Ty Vr wv = v v I ea) HO Ge 10g 0 103 10, 10 10 10 10 10. 0, 1G; - | eae (b) Violet E Blue | Green | Yellow Orange | Red |" | | | (A) 3000 4300 4500 4900 5500-5900 6500 7600 A (a) The spectrum of electromagnetic radiation and (b) the visible spectrum