SS. ee H: STRUCTURE OF ATOMS, AND ‘Atom models As far back as 1803, Dalton, an English | teacher, showed that the matter is made up of | extremely small particles called atoms. Prout * | (1815), suggested that all elements ate made.up of atoms of hydrogen. i: \ CY | since many of the elements were fobndito Have | atomic weights that were‘hot exact multiples of _that of hydrogen, the uggeStion Was not accepted VA aa | thomson etam inoad col the study of discharge _ of electricity through gases, it became clear that an [210M cOnsists Of Positive and negative charges |J. Thomsen tried to explain the arrangement of order of 107m, The listributed over the ‘Ons are embedded in | Positive charge is uniformly di entire sphere and the electr the sphere of | Positive charge | 85 shown in Fig jlike seeds in a watermelon. |The total | positive charge j inside tne atorn jis equal to the | total negative - | charge carried by the electrons, so that every atom | is electrically neutral {nis model is also known as plum pudding model Drawbacks: | () According to electromagnetic theory, the | vibrating electron should radiate energy and the frequency of the emitted spectral line should be the _Same as the electron. in the case of hydrogen atom, Thomson's model gives only one spectral line of about 1300 A. But the experimental observations | reveal that hydrogen spectrum consists of five different series with several lines in each series (wt could not account for the scattering of | a particles through large angles, Que: Explain Geiger-Marsden experiment: 4 | In this experiment a narrow beam of u-particle from, radioactive source was incident on a gold foil. The scattered «particle were detected by the datactor fnxed ‘on rotating stand. Detector contains zinc sulphide vereen and microscope. «particle produced scintillations on screen which could be observed through in evacuated fu particle hamber The ob: fed number PIVATE Cine Ni 0 seal cahittecle otk function of scattering angle, from its original direction is cae. They observed that most it, passed undeviated, onty fay, ap hte more than 1°. Few were def, deflected by more than 99° Some back with 180° lead bricks Source of @— particle Neen B0ld foil detector ; ee je Ze) 1 | N=Nons ( a) | No=n0 of & ~particies incident obo “particles scattered atan angle # vicknessof the foil (7B= nd-of atoms pr unit volume Mz Tnass of x —particl Explain the Rutherford’s Atomic Model 1. The atom has tiny positively charged core called nucleus. Nucleus contains all positive charge. 2. 99.3% mass of at 4. Atom asa whole is electrically neutral | 5. Total no. of electron in atom is equal to total no. of ++ve charge in nucleus. ©. The size of the nucleus is 10-"° m, about the 100000 time smaller than size of atom. 7. Most of space in atom is empty space. Drawbacks of Rutherfordaipmic mode! 1. Fails to explain stabytty of aterm 2. electron revolve’ épmtinuous then its enerey must goes on di raga aur nga should st dheonv leus but it revolve only in fixed circular*ert ular vatocity and frequency of electro: uncréase if its energy emitted continuously 4. fh shOws line spectra of its energy instead of continuous spectra1 1 1 trl 3 99234 This series of lines is observed in ultraviolet-region. When n=2 then d = 1216 and n=3 then 1 = 1025 A series limit in lymen series is given by Pategories as emission spectra and absorption Emission spectra is divided as | 1 Continuous spectra: v ER a> Aimit = 91.175 nm | I ‘en metallic body is heated it abst ir | ee meee arieine Maximum wavelenth is obtained by | radiation of different Wavelengths. If this Pek zi Amax = 121.57 nm | radiation is passed through*rism continuous spectra is obtain, 90 nm 275am iat nm _0nm 120 : 1 HL | 2. Line spectra: When hydrogen gas is heated in glass tube and its emitted radiation is passed through prism it | shows some specific wavelengths and there is 2. Balmer an : : | no other emission between any two successive This series of a is emitted by transition or from outef orbit of quantum number = 3,4 ‘the second orbit of quantum number p = 2 emission lines (wavelengths) e.g. wavelengths in Visible range are as follows 410, 434, 486 and | 656 nm as shown in below figure (Ry. ee 4° Pv ‘These lines in emission spectra are called tus series of lines is observed invisible region. long with these visible ines. "\"S Whenn=3 then A = 6563 A e wavelengthsin UW ang naa then 2 = 4868 A i). ForHa line, Le. first member of Bal ed 1 1] ‘Thus — = = |.n=34,5 list = R| oe oa| emission line. hydrogen also emits th region and IR region. When these lines ate ined jes, Some known Series are as grouped called series. Som Serie sarees mer series, | \ follows n= 3. 9 1/1] { | Lymen series, Balmer series, Paschen sefies, 2 =R iF +3 | Bracket series and Pfund series. Names of series . | ii) For Hp line, Le. second member of Balmer is given after the scientist. r Let us discuss the first three series stared from series, we have p=2,n=4 | tower wavelength side fat] | L416 | ii) For Hy line, Le. 3° member of Balmer series P | =2,n=5. | - uy , y s10nm 786mm 560m R —_ 434 | series limit: The shortest (minimum) wavelength in led series limit. We can calculate the | series is call 2 | series limit by putting n=o>. ie. using (E| WN { ‘ side - d limit series limit tn Balmer series 1s given by On shorter wavelength side or towards series lit Citi spertra ines comes closer ERE gal? Ainae = 364.60 Maximum wavelength in series is obtained by 4 maximum wavelength is obtained by | putting n= pet . a) tek iE = A> Ajax = 656.3 nm 3, PaschenSeries :-1his series of lines 1s emitted by 1. Lyman Series : | This series of lines is ernitte from outer orbit of quantum number 364.6 400 nm <0 nm ayia jrst orbit of quantum number ST we aD Jd by transition of electron n= 2,3,4—~ to the fi Balin fimit | a 1 | ok SHUT = transition of electron from outer orbits of — _ BH a thus 5=R [ps RIVATE CIRCULATION ONLYquantum number n= 4,5,6———~to the inner energy level it radiates enereyin ‘ | orbit of quantum number p = 3. o photon or quanta. Radiated energy, ‘\ a difference of the energy of two levels ta "\\_ transition takes place. a | al} - ]9=45.05 o 9 Let En be the energy of n® Orbit whic | this series of line is observed inthe ifr ater. series limit in lymen series is given by | nm by, — 2 hime = 820-41 St ptiinesty vA Agia = 1875 nm a 15) 2.um um wa Line emitted by each element is unique which helps to | identify the element. | Rutherford's atomic model fails to explain the atomic | spectra. \ que: State and explain the Bohr’s postulate. And: 1* postulate: The electron in the oe | atom revolves around the nucleus in circular | with nucleus at center, the necessary centripetal force is provided by clectrostatic force between | tvly charged nueleus(proton) atid “vely charged electron. a if m is mass of electron ake . isslelocity and ris | radius of Bohr's orbit then. Yet q be charge on | electron. | According to 1* postulate centripetal force = | electrostatic force mv? ie | r 4TtE9 r | 2" postulate : Electron revolves around nucleus without radiating energy only in those orbits for | which the angular momentum of electron is equal ‘toe 2 | {iis postulate explains the stability of atom. | The orbit is known as stationary, stable, quantized, | permitted or Bohr's orbit According to 2" postulate ‘ h anguiar omentum (== : an |mvr =r where n is called principle quantum | no. of sheli no. | 3" postulate: Third postulate states that when electron jumps from higher enaray lovel to lower PRIVATE CIRCULATION ONLY. the higher level, Let Ep be the energy of p>..." which is atthe lower level. Thus according ny Postulate when electron jump from E,to Ey, radiates energy (hv) which is given by hy=E,— Ep \ Que: Obtain the expression for radiusof Behr’s orbit. Ans: According to 1* postulate we have | 2eahn® | Comparing eqn 1 and 2 we get | neh? | 4m?r2i? This is the required, \\\_ equation of n™ ; 4 0.053nm x 2° | orbit \ | NOTE: Fromyabove | equation * 4 oC 1 85h, eo, €, m and mare constant If we put the values of constants and as for \\bidrogen Z= 1, for first orbit ie. for n=1 we get \ | = 0.053 nm or 0.53.A. This radius is known as Bohr's radius and denoted by ag ay ( “s) = 0.053nm mne Expression for angular speed ‘a’ of electron in n™ orbit Accor election can have only those or seuont ing to Bolts posiui for which its2 gular momentur nh AN m,mvr= Ey =2,2,3—— . AN . = (22) ame \2 \ eo h AY But linear speed v=ro.-. angular speed @ 3 ~ 1 YN r nh 4t \ nh \ o= RE AES Qrgnt FD? me ay yh _ &, hin? But for nt oPbibraaiuy? : \ a me’ amet which is the required in) 2ehn® expression for angular speed of an electron. constant As m, m, €, 0, hare constant @ = — 5 n i 1 ed n | Thus the angular speed of electron inversely proportional to cube of principle quantum is number. AX Que: Obtain the expression for energy of an )/ electron In Bohr orbit. ah | ans: We know that Total energy K- | 4. tofind K. E. cE, + PE. From 1% postulate we hav ke Brey ¥ 2. To find P. €. of electron | P.€. = (Potential on unit charge } x charge on electron jane | PRIVATE CIRCULATION ONLY Ci -ve sign indicated that electron is bound to the nucleus by attractive nature of force. Since the quantity in the bracket is constant we can say 1 ATER ! Hence, for electron, energy of n" orbit is inversely proportional to the square of principle quantum no. ifwe put the values of constants then we have pr E, =-13.6 nm? eV | ‘As for hydrogen atom Z=1 and for n=1 | En 3218.6 eV | For zily'atom when n=1 it called ground state and | fépany other value (other than zero and 1)called | excited state. Therefore, when hydrogen atom i its atom (and hence electron) has enerey \ E= 13.6 eV. ground state | Note: As € x And so on Origin of Spectral Lines in anormal hydrogerratom, electron revolves in the innermost orbitgfor which the principal quantum number is 1, This Is tHe lower energy state of the electron. hg elertron is said to be in the ground | states if hydrogen gas is heated or an electric field is applied “fo the gas, it receives energy. If the energy | absorbed is equal to or greater than 13.6 eV, the | @lectron escapes from the atom. If the energy absorbed is less than 13.6 &V the electron jumps from the innermost orbit to an outer orbit of higher energy level and begins to revolve. In such a case, the atom is said to be in an excited state. In the excited state, returns almost immediately to its stable state. This isachieved by the electron returning to its ground state in one or more steps. During each transition, the electron radiates energy equal to the differerice | between the energy levels corresponding to thia two” | orbits. This energy is radiated in the forrof photon of definite frequency as given by thi | lonization energy: The minim | required by electron in the gréi | free from atom, eis abo(igo of hydrogen atom. ( \ \ 4 For hydrogen atartit is ediyal bd —13.6 eV Excitation energy*thé chery required by an electron to takét into excited state from ground a is obtained in the spectry, transition of the electron from “ quantum number n to the inna” number p. iy Energy level diagram for ‘The energy of electron in the found state 13.6 eV and the energy in otf : ther orbits fyb ev, Wheren=1,2,3-__ In this diagram the excited stat. Values ofn are plotted on vertical scale. The energy levels are represented by horizontal lin © for ditferen, 5 as shown late ; in figure, | When nsfielectfon is said to be in ground state The E,= 0ev =o x | energy of electron in the ground state is EL bye o5tek tit as |—13.6eV and that for other values of n eleactron is f° ssw a 4 said to in excited state byolse 1 egne2, B= 34 eV Hence for hydrogen atom minimum excitation energy is = E, — E,=~13.6 eV — -3.4 eV = 10.2 eV Let E, be the energy of electron in nt orbit, and E- be the energy in the p obit wheren>p, then, energy of electron in respective orbits can be written * 1 «4 ] ier ( sanz) bi n?] me* i 1 t=) Is-s eegh?) |p? ~ 12 If 2 is the wavelength of the radiation, then velocity of light in free space, C = vi c ory= = 2 * 1 ) Ip? ne 1 1 met constant. Its value is 1.097 ¥ 107 m=# "is the wave number 1u5 above relation explains the origin of ctral line: A line of characteristic waveleneth ‘7 ¢ CULATION ONLY As ‘n’ becomes iarger, the energy levei becomes progressively closer, forn = 00, En=0. ‘The transition of electrons which gives rise to Lyman, Balmer, Paschen and Brackett series are shown by vertical arrowed lines in figure. As ‘n’ increases energy difference between successive energy levels decreases, therefore the spectral lines of higher frequencies in any series get Progressively together in spectrum. Asingle atom can'groduce only one spectral line at any given instan¢. However, since even a ‘small mass of gagcortains a very large number of atoms, differen transition takes place simult ly. Hence, all the spectral lines are observed y instant faniZation Potential The energy in electron equired to remaye the electron fram the ground stata to inf i.e. to the zero energy level, is called tion potential, or ionization energy for hyd The value for hydrogen is 413.6 eV me* lonization potential, W. = Gzjoglie’s Hypothesis : According to Bohr length rst orbit is equal to one de Broglie Fvelength(A) the that of the electron in secont, C ‘orbit has length equal to 2A . As show in fig the\ The total number of protons and neutrons in the nucleus is called the mass number, itis denoted by A. Hence, A = ZHN. The neutrons and protons are collectively called fi oa in ait orbithasits orbit length isequahto. nucleons. ; enigma eco aOR Nucleus is denoted as 4 | oar, = ; ; ; Tr : \ Where Xis the chemical symbol of the element, Ais According to de Broglie hypothesis yh the mass number (A=Z+N) and Zis the atomic nypothesis pa | Where p is momentumss)p>mv\ nee | h \4 Isotopes: Nuclei of same element having same number | | of protonsi.e. Same atomic no, (2) but different number | \ 4. of neutrons are called isotopes E.g. deuterium iH, | | / »\\ tritium3H are the isotopes of hydrogen. | \ ae e\ Isotopes are place inthe same placed inthe periodic | | ih i SK — table. | |. 2mm == is 7 | \ mv { } | h \ Isobars: Nuclei of ifferent elements having same mass amv, =277 - V umber (A) but diferent atomicno, arecalledisobars | 7 Eg, deuterium?tI, tritium tle are the isobars. | |: wo : | a 7 Iso iittei having same neutrons but different | | is equation is atom numnbérs are called ixotones | nothing but the second postulate of Bohs. Hence Aug bd "Au arethe eorones (*38,and BACH) | |assuming electron as a wave agrees with Bohr’s | | second postulate. AN 2 | | "Mass energy relation \ Limitations of Bohr atom model: Y By Einstein equation relation between mass and energy 7 isgivenas E= me! ,__|(e) Bohr atom model isa ont font Byer SETS egy miemmssinkgandctstheveoaty atom or hydrogen like-atoms aut not for oflightin air { Energy of an electron is given bY = mass of electron x (velocity of ligt)” | E29.1x10-% x (3 x 10°)*=273 x10 20.511x 108 eV or 0.511 x MeV Energy of proton is F~ 941.iMeV Energy of neutron is E,, =941.1MeV Energyis also measured in Unified ‘Atornc mass unit |_| {b) When the spectral lines are closely | | examined, individual ines of hydrogen | spectrum is accompanied by 2 number of faint | ines, These are often called fine structure. This | | \complex atoms. |i not explained by Bohr atom model. | (c} Bohr atom model fails to explain the e spectral lines. ‘Lamu =1 u= (1/12) th of mass of neutral carbon atom in its ground state Itis denoted'by u | intensity variations in th |[d) The distribution of electrons in ato | completely explained by Bohr atom model. i | Ime= 9.109383 m | | Ls ms is not = Units of mass 108! ky 000055 u-O.ST 938.2 Iiny~ 1672023X 10? 1007825 u 1674927 107" ke 008669 U Nucleus: | The nucleus mainly contains protons and neutrons. The protons have postive chars and its Fragntodesequalin to the charge ofthe eecon Sie of rules: The as of rules ety | Rtevtrons af electrically aeuirl no charaeen) Proportional te the cube roct of atomic no The total number of protons in the nucTeus {5 en called the atomic number and itis denoted 8Y Z- ne | |, the number of neutrons in the nucieus is called Pete in. and has value | nui Seah ca ele n number (). As density (plof nucleus s | fy 2931.5 MeV Mtmass of nircleus M du if we consicter average mass of proton and neutron & \_ equal to m then total mass of nucieus M= mA ma (RAN) mA BRT vor pucteus = ERY _ Ts shows that uigae depaiyi independent on | atomic no, of nucleus), \_ if we put values of allconstant we get | pe asc ie» | Le. densitVof nucleus is very high, | Atomic density of osmium is highest density among atl the known elements and this atomic density is smaller than the nuclear density by thirteen orders of magnitude ‘Nuclear forces: | 4. Wis the strongest of all the basic forces of nature, tis short range, of the order of 10° m. This force is charge independent, pwr Do not obeys inverse square Rw! Now The provons im the nuckeby hcl one another | dtu to there similar charges, The nuclear formes bende the nucleons counter the forces of cloctrastane rapalsion As untea! faye As moat stronger than the clectrastaie force far the distances benwecn oucteoas in a typreal musteus tH overcomes the repulive force ard keep the muckuas tagetter he onetous stable: Mass Delt “re fererve BEDE ATA MAS nucleus 2st macs ofits comshtuent murtenes Mans defen xnen by YS footer a together in the nucieus of an ator. tt 2. Mis maamum &.75 MeV for A = SS ands 75 Me. A=238 3. Ris low for both ight nuciel ie for Ac 30 and heavy nuclei A>70. This indicates that nucleons of these atoms are oesety LEN 4. When heavy nucleus breats into lighter sucke: hee RE of nucieons get increased S. When lighter nucleons ane joined f form 2 Sezer pucieus, BE. gets increased Radioactivity: The Re OF SAY AMSG OF FARTS MAN ARTE SD calles padiowcey dooay cade Tris disrnverad iY Rago! in ! Tred Seay Te nactors 1 catty pares mes mse yi kya o ERO demgiter saviews \ Vat af eAD NIN gest ER Aor oye ts ou YY TA THQ ae Y0 - decay f the atoms for wh.ch mass no.(A) is greater / 120 undergo a — decay, The nuclei of these\ 7oms have large no. of protons and are unstable J x hence they emit protons in the fornyaf | « particle for their stability. “S\ Y | When a a particle is emitted from an atomithen its | mass no. decreases by and citomitnd: decheases |by2. Ny XA 124 + oat FH Endegy \ 4 ¥ e.g. a emission ; \e @ NA 2agy 5 23474 tHe » Here 7380 is parent nucleus and 234Th is daughter nucleus.(Note: $Héis not daughter nucleus) The total mass of the daughter nucleus and a particle ( $He ) is always less than that of the parent nucleus. The difference in mass Am = my — my —Mq | | is released as energy called disintegration | eneray (or Qvalue) and is given by Q= [my — (my + Ma) ]C? B decay | prays are fast moving electrons from nucleus |im air. There are two types of B decay f /1) Bminus decay (B-) jwnen aB decay takes pla |" remains same and atomi | tn beta-minus decay, a neutron | proton within the nucleus ‘ jap es antineutrino(v) iA ce, mags 10. oF atom icnovincreased by 1 transforms into a | & 3 in gen X* — iV + 16° + 9 + Energy 23tpa +e + + energy & = 32p 4 325+ Ye + 0+ energy LI B plus decay (B°) | When a [}* decay takes place, mass NO. of atom | remains same and atomic no. decreased by 1. | im beta-plus decay, a proton transforms into neutron pons ef > neutrino(#) ee Lin general X89 as¥* + 2° +0 + Energy eg. > fiSi+ ,fe +d + Energy + Energy PRIVATE CRCUFATION ONLY The total mass of the daughter nucleus and je is always less than that of the parent nucleus. The difference in mass ‘bm = my — my ~ mq is released as energy called disintegration energy {or Q value) andis given by Q= [my — (my + m,)]c? y-decay 1 y are not particle waves of very shor emitted from nucleus are called y - 12) 2, are neutral means 3. gamma decays usual after ator B decay atom goes in e? 7¢o undergoes beta Pl example, * the daughter nucleus © state having ene Experimental transitiGnto its ground state in {Ajby emitting a gamma ray of en: direct to ground state. And Bg) by emitting a gamma ray ,diate state first and then ‘gy 14 keV to reach the going to an interme emitting a photon of ener! ground state. Which type of decay depends on whic of is more stable. Often, the d not stable and it undergoes of decays may take place un nucleus is stable. ‘An example of su’ which undergoes @ a total of 14 times, to finally reach a stabl nucleus of Pb. 3) of but are the electromagnetic rt wavelength. Thus photons having no any charge. ily occur after a or B decays ition state. For | Jus decay to form 7Fe which isin an excited rgy of 136 keV. ily itis observed that it can make 2 two ways ergy 136 keV energy 122 keV and a nucleus will undergo the resulting daughter nucleus | laughter nucleus is also further decay. A chain the final daughter til ch a series decay is that of *°U, series of alpha and beta decays, le daughter ‘decay per unit time is State the decay ude that the number of nuclei underBoin proportignatp the num! present at that time. ber of unchanged nuclei J£N is number of atom unchanged at any instant time t andin dN is the number of nuclei that disintegrated in time dt then we can write -dN — aN dt dN —— = N; where d is decay constant -dN dt From the decay law we have dN w de =\N vIntegrating on both side we get | [pf loge N = —> t+ ¢-——-2 Where cis constant of integration To find the value of constant we used initial » | condition | At t=o N= Nowhere Nois the initial no: dF niclei. 210g, My = AX OFC + C= loge A | Put the value Ofcin\eduation 2 we get | log, N= —> B+ loge No logg Me loge No = > t N oN =—-xt BEN N las 2 N=N ee This is the required expression for no of atom present at any instant time t. From the equation it is seen that radioactive substance decay exponentially with time. | | i Expression for Decay constant. dN From the decay law we have ~~ = AN aN yea) | N Hence Decay constant is‘ the ratio of the amount of substance disintegrated per unit tine.to mount of substance present at that tiene, From the equation N =N,e7"" if we define time t 1/2 Then equation becomes none 2Ad PRIVATE CIRCE) ATION ONLY 1 N= 37% No Le. decay constants the reciprocal of time in vino, the substance remains 37 % of its initial vaiue Half-life period (1): halt-life period is the time in which substance disintegrate to half of its initial value.( or 50 % of its initial value) Expression for Half life period(T): From the equation we have N =n,e*t --1 By definition of half life period at t=T ; N= No ae Thus. wvetT=2 | AT = log. 2 aT = 0.693 0.693 x This is the required expression for half-life period of substance where is decay constant. ‘Average life of a radioactive element: | Mean life (1) The time-interval, during which the number of nuclei of a radioactive element becomes equal to the e™ part of its original number, 1s called | the mean life or averagealife tof the element. | ce | at | seocgty put t= T= mean life note: || Ikraction of radioactive | lubstance lett undecayed is & Qy \ Where n no of hulf lives net) ‘As we have N =0694 wo bad 7 , tye OR1,,, = 0.6937 0.694 Nuclear Energy: Energy releasedin nuclear reaction ts called nuclear reaction. Nuclear Energy released In following two process , lighter nuclei is called nuclear fission. Combined masses of the two product nuclei ill be inaller than thé mass of the parent nucleus 6g. "J5U is split into barium and Krypton when it bombarded by neutron. Reaction is a5 follows y tin > 736U > 84Ba + 22Kr + 3gn+Q. | Other examples are as follows 2264 140) 4 4 U > oBa + 36Kr + Zon 2350 +) > 755 , G2 2355 in > 2BU > BI + 3+ 2gn SU san + 2260 > 124Sb + 42ND +. 40m 733 “asus ain > SGU > Saxe + 3850+ Zon From above reaction it is seen that the original | nucleus has a greater mass ‘thar the sum of masses | of two daughter nuclei. The mass defect comes out in the form of energy This energy fist converted into Kinetic energy and | then into heat (or other needed form of energy) | This heat energy is used in Nuclear power plant for | | production of electricity. Uncontrolled fission process is used in atorn bomb. l | nuclear reactor is an apparatus OF 4 which nuclear fission is carried out in a contratled mannt to produce energy in the fort heat which 1s {then converted to electricity. In uraniuin reactor 740 5 used as the fuel. Ibis borrharded by slow [neutrons to produce “)ZU which undergoes fission. * |; Chain Reaction: | device in | Neutrons are produced in the fission reaction shown in tg. of fission as stated above Some | eactians produce 2 neutrons while others produce | or A neutrons The average number of neutrons tion can be shown te be 2 7. These Ltrons are in turn absorbed by other “2Uauclel PRIVATE CIRCULATION ONLY, mestereenmmememeens si IY NEA \ wh 1. Nucleas fission? Splitting of heavy nuclei into \\serces. itis very difficult for two atoms to come tl to produce “2 further 2.7 nestrons per fission This can have a cascadi ing effec a sumber of nestrons produced and number of "20 nuclei produced can increase quickly. This is called 2 chain reaction. Such 2 ‘reaction will lead to 2 fast increase in the nurnic fissions and thereby in a rapid increace amount of energy produced. This will lead to an explosion. in a nudear r22c methods are employed to stop a chain reaction from occurring and fission and energy generation is allowed to occur in a controlled fashion. The ene"B/ generated, which is in the form of heat, is carried away and converted to electricity by using turbines etc. More than 15 countries have nuch reactors and use nudear oower. India is on= 07 them. There are 22 nudear reactors in India, On being at Kudankulam, Tamil Neds tare ati ear power is generated by the 2: Nuclear fusion: When two lighter nucle’ are fused to form heavy nuclei this process is calles nudear fusion. um of masses of two parental nudet is atways -m (dough Agein ISA. | greater newly for: defect is converted into energy. This process is occurred at high temperature approximately of order of 10°°C. Energy generated in sun is due to fusion. eg, 1: When two hydrogen atoms are fused deuterons is formed with energy of 0.42 MeV. The reaction is as follows TH+1H > TH+ e* +042 MeV e.g 1: When two dapterons atoms are fused helium is formedwith energy of 24 MeV and a -particle The'reaction 's as follows 2H Bf tHe + 24Mev For a quciear reaction to tak’ fortwanticlel to come to within about 1 fm of each atherso that they can experience the nuclear e, itis necessary hat | close to each other due to the electrostatic ' repulsion between the electrons of the two atoms. This problem can be solved by stripping the atoms of their electrons and producing bare nucle! itis possible to do so by giving the electrons energies larger than the i potentials of the atoms by heating a gas 0 - a ==" oms Buteven afterthis, the two bare nuclei find it very difficult to go near each other due to the repulsive force between their positive charges. s | For nuclear fusion to occur, we have to hearitfie gas [to very high temperature thereby prdyiding-the nuclei with very high kinetic energies.\These high energies can help them fto, “overcome” the electrostatic repulsion and (tome, ctese* to one another. As the positive chargéof a nucleus goes on increasing with increase in'its atgmnic number, the kinetic energies of sels the temperature of the gas necessary for huclear fusion to occur goes jon increasing with ineréase in 7. ns Either fission alone or both fis, 8 are used in these bombs. The first such made towards the end of the second wor America. By now, several countries includ, have successfully made and tested such's. devices: America remains the only county t, % actually used two atom bombs which completa destroyed the cities of Hiroshima and Nagasaki Ht Japan in early August 1945, Note: Ifan uncontrolled chain reaction sets up in a nuclear fuel, an extremely large amount of energy can be generated in a very short time. This fact has | been used to produce what are called atom bombs | or nuclear devices. | de, ‘Note: In nuclear reactor energy is produced by using controlled nuclear process where as in atom bomb energy produced by uncontrolled process. sec | Controlled chain reaction | | Controlled by artificial method \ |_Allneutrons are absorbed except one |_Wsrateis slow \ \_ Reproduction factor k = 1 | Energy liberated in this type obgeaction iwalways “Uncontrolled chain reaction ] | No diny control over this type of nuclear reaction | | Mofe than one neutron takes part into reaction | ‘srateis Fast ae | | Reproduction factor k> 1 aca { ‘Allarge amount of energy is liberated in this type of | reaction Chain reaction is the paeiple ofinuclear reactors Uncontrolled chain reaction is the principle of atom bomb. | |_ less than expiosive energy | i N= a _ aN dt 6.023 x1024x M A dN _ 9.693 | 6.023 «1078; nana ae | PRIVATE CIRCUS TION ONLY yee AN; Where Nis no. of atom present at given activity As no. of atoms = mole of element no. of atoms in one mole no, of atoms in one mole x mass of element in gram, atomic mass no aia