5 Electric Potential (V) and Electric Field Strength () due to a Charged Spherical Conductor If: Figure Thenc potential all points isgiven by the expression: 04QVAs for the electric field strength (E), it is zero at all points flow of charge inside a charged conductor. This implies that there is no inside (i.e. V) at a point (P) outside the sphere is given by the expression: 01V………. (i)1and the magnitude of electric field strength (Eoutside214………. (ii)2Note that Eq. (ii) can also be used to determine the magnitude of electric field strength (E) at the spherical conductor: Er V=………. (iiiUsing the above information and Eqs. (i) and (ii), we can draw the following E: Figure 17..10ELECTRICFof

Relationship between Electric Field Strength (EV) at a PointE) at a point in the field is equal to the negativeofthe[2]=………. (iv)(Note that graphs of Figs. 17.2(a) and (b) are consistent with the above relationship). ELECTRICIELDSPage 52

6 Combined Electric Field Strength (E) and Electric Potential (V) due to Two Charged Bodies If: Figure 17.(the magnitude ofE11 20 2=4 4q q ………. (i)whereas the combined echarges (q1 and q) is given by: = +1 21 2q q r r………. (ii)1q into Eq. (i). Note that in the above example, values of 1 into Eq. (ii) (i.e.with their ‘+’ or ‘–’ signs). Note that in the above example, Example Figure 17ectric field strength at point P due to both point charges (q11 20 1q qE(where q1 and 2 are both +)V V V1 21 2= +4 4 (where q1 is ‘+’ and q2 is ‘–‘ELECTRIC132

ELECTRICFIELDSGraphs two(2 (a) Electric fieldstrengtharged sphere) (b) Electric fieldstrengthdue to twocharged spheres ()dueto a charged sphere (to two charged spheres ∆?? ൌ??ଵଶ?ଶ?2??଴4?଴???଴93?? ൌ?????4?଴?3???9?Field is ?++++++++++ +++?++++ + +- - - - +-Field is lines) ?+ + +???2?3? ൌ??-----???+ +++++- - - - -?+++++++൅?ଵ൅?ଶ??--- ----െ?െ?ଶ?+ ++ --

7 Analogy between Different Aspects of Electric Fields and Gravitational Fields SimilaritiesFor a point charge/point mass: radialfield strength at a point 21(distance of the point from the point charge/point mass)[1]potential at a point 1[1]DifferencesThe gravitational force is always attractive; whereas the electric force may be attractive orrepulsive. [1][1]ELECTRICIELDS) Electric potential energy ??????, ??୔???, ?increases as ?Page of 132

CAPACITANCE 1 Capacitance (C) Parallel-Figure or19.1 is closed, the battery moves electrons from left plate of the capacitor, through its interior, to the right plate. As a result, the left plate becomes Ethat point, the current in the circuit becomes zero (and remains zero thereafter). charge stored on a parallel-plate capacitor is zero. However, the work done (by the source) in separatingcharges is stored in the capacitor in the form of energy (electric PE). [4] =QCV.capacitor,capacitance =or an capacitance =charge stored on the conductorelectric potential at the surfaceunit faradconnected T 1 21 1 1C C C+1=2W QV=2W CVQ=Q0e0et/RCI/Theory suggests that when a capacitor is charged (or discharged): (where Q=where =The time constant of the circuit:in parallel:T = C1 + C21 farad =

Isolated Conductor(e.g. metal sphere placed on an insulating stand)If: Figure .2QVCSo, for an isolated conductor, capacitance may be defined as: charge stored on the conductor[1 or 2]Note Uses of Capacitors in Electrical CircuitsIn electrical circuits, capacitors are used to: block d.c.So, for a parallel-plate capacitor, capacitance may be defined as: charge stored on one platecapacitance =The SI unit of capacitance is 1 coulomb1 farad =1 voltCQ=CVVC57 of 132

2 Combined Capacitance (CTIf: Figure 19then the total capacitance of the combination of T=QCV………. (i)where, by principle of charge conservation: = 1+ Q2Plugging QQ2 into Eq. (i) gives:1 2=V1 2T= +Q QV V(where: 11and 2QCT 1 2[2 or 3]CAPACITANCEof

CAPACITANCE3 Combined Capacitance (CT) of Capacitors Connected in Series If: Figure 19.4then the total capacitance of the combination of the capacitors is given by: TQCV………. (i)where= V1+ V into Eq. (i) gives:T1 2V VC Q1 21= +T 1 2C Q V Q V/ /(where: 11=Qand 2T 1 21 1 1Notes When capacitors are connected in series, the adjacent plates become charged he  capacitors connected in series (as shown in Fig. 1 2TCLikewise, if there are only 2C Cand:12+

4 Energy Stored (W) in a Capacitor Theory suggests that for both types of capacitor: QVSo, when a capacitor is charged (or discharged) we always a straight-line potential-charge graph that passes through the origin as shown below: Figure 19.5Note that the area under the potential-charge W12(where Q22W CV………. (i)Example1 to W222=(as WV2)CAPACITANCE60

5 Discharge of a Capacitor through a Resistor A capacitor may be discharged by connecting a resistor across it. If at a given instant: Figure 19.6then:IV(where Q is the charge on each plate of the capacitor)I(where IQ/∆ttQ1QQ………. (i)where RC=  of the circuit. From Eq. (i), it can be shown that:/t RCQ Q e………. (ii)where Q is the initial charge on (each plate of) the capacitor at t = 0 (i.e. when the switch is closed)and Q. Eq. (ii) gives the following Q vs. tgraph: Figure 19exponential decay curve and is described by the equation: y= y is the dependent variable, x is the independent variable, kand 0 are constants and Euler’s number.7); however, inVso the p.d. (itor and the current (t RCCAPACITANCE61

CAPACITANCE/=0t RCI I ewhere V0 is the initial p.d. across the capacitor and 0From the expression for time constant of a capacitor discharging through a resistor:it follows that the discharging time of a capacitor increases with the resistance (R) and/or theRIthe chargeflows off the plates of the capacitor more slowly, and hence the discharging time of the capacitor Likewise, when the capacitance (C) increases, the initial charge (Qthe same initial p.d. (V) across it, also increases (as Q= CV0the same. As there is now more charge on the plates of the capacitor, the discharging time of the Worked Example62 132

Classwork/homework1 In the following circuit: the resistor has a resistance of 1000 , the capacitor has a capacitance of 2000 F and the battery has an e.m.f. of 10 (a) (i)(ii)(iii) the current in the resistor when the switch is first connected to terminal Q. (b)switch is first connected to terminal Q. ( Based on your answer to , explain what effect this has on: (i)(ii)2 A 200 F capacitor is charged using a 10 V battery. It is then connected across a 500 (a) Calculate the charge stored on the capacitor. (b)(c) Calculate the time it takes for the charge on the capacitor to fall to 1.0 mC. (d)Answers1 (a) (i)(ii) 0.020 C (iii)0.01 A (c) (i) It reduces p.d. across the capacitor CAPACITANCEPage of 132

MAGNETICIELDS 1 Magnetic Field Figure 20.1.1 (a) Figure 20.1.1 (b)Magnetic field is a field of force produced by a (permanent) magnet or by a current-carryingconductor (or moving charge).[2] is the region of space where a particle experiences a force[1 or 2]Notes at an anglences a force.Magnetic field lines always form closed loops.Inside a (permanent) magnet, the field lines run from south (S) pole to north (N) pole.of force on a small north[2]ofa Long Straight Wire[3]The direction of magnetic field produced by a straight current-carrying wire is determined byright-hand grip rule. For a straight current-carrying wire, the rule states that if grip the wirewith our right hand such that the thumb points in the direction of the current, then the fingers curlin the direction of the magnetic field (see Fig. 2Force on a current-carrying conductorLenz’s Law of Electromagnetic Induction: the direction of the induced e.m.f. is such that it tendsto oppose the change causing it.F=BILsinarea perpendicularto the magnetic fieldFaraday’s 

As we move away from the wire, the distance between the concentric rings of the magnetic fieldlines increases, which implies that the strength of the field decreases.FieldPattern ofaolenoid Figure 20.1.3 (a) Figure 20.1.3 (b) 20.1.3 (c)When a soft-iron core (ferrous core) is placed inside the(a) also determined byright-hand grip rulein the direction of current, then the thumb pointsin the direction of the magnetic field (i.e. towards the N pole of solenoid) as shown in Fig. MAGNETICIELDSPage 132

MAGNETICFIELDS2 Force on a Current-Carrying Conductor Placed in a Magnetic Field If: Figure 20.1.4then the magnitude of the agnetic force on the conductor is given by: F BIL= sin………. (i) where BI) and flux density (sinFBILFor  = 90):FBIL(as sin 90 = 1) So, the (Bat right-angles if a (long straight) wire carrying a current of 1 A perpendicular to the magnetic field experiences a force per unit length of 1 N / m. Fleming’s Left- Figure 20.1.5 (a) (b) stretch the first two fingers thumb of our left hand such that they are mutually perpendicular to each other, with the first finger pointing in the direction of the magnetic field, and the second finger, in the direction of the magnetic force (see Figs. 22.5 (a)).Page 66 of 132

Forces between Parallel Current-Carrying Conductors Figure 20.1.6 (a) Figure 20.1.6 (b)Notes By Newton’s 3rdmagnitude of F1 = magnitude of F2 (even if I1I) Explanation[3].7 Both wires, 1 and 2, carry currents into the plane of the page, and produce magnetic fields in theonly) The magnetic flux density at wire 2 due to current in wire 1 is 1) of wire 1 is perpendicularforce 2 due to the magnetic field () of wire 1, which is directed rightwards MAGNETICof 132

4 Force on a Charge Moving in a Magnetic Field If: Figure 20.1.8then the magnitude by: F Bqvwhere B is the magnetic flux density, and enters the field). For  (i.e. when the direction of motion is perpendicular to the fieldB=(as sin 90 = 1) As for the direction of magnetic force (FFleming’s left-hand rulepositively-charged particle (e.g. proton), thedirection of current is the same as thmotion of the particle; whereas for a negativelythparticle, as shown below.Figure 20.1.9 (a) Figure 20.1.9ticle due to and always normalotion of the particle (see Fig. 20.1.8); so it provides the centripetal force, i.e. = 2=(where r is the radius of the semicircular path followed by the particle) =m[2]Note that the above equation is a very useful equation; it can also be used to derive expressions the to the direction of motion work the KE (or[2 or 3]MAGNETIC

Worked ExampleIf a charged particle of charge +qm, initially at rest, is accelerated through a p.d. V in a Figure 20.1.10then use the particle. [2 or 3]Solution By energy conservation: 210 =22=(where: mv= ) 22qV= 2MAGNETICFIELDS

5 Deflection of Beams of Charged Particles by Uniform Magnetic and Uniform Electric Fields Figure 20.1.11 (a) Figure NotesThe magnetic force is constant and always normalFig. 20.1.11 )inside the field is circularThe electric force to the direction of motion of theparticle (see Fig. 20.1.11 (b)), so the path of the particle parabolic). FB and FFFigure 20.1.12The velocity vFor the particle to pass undeviated (i.e. straight) through the region:FB= FEBqv qE(where B is the magnetic flux density and E=v[2 or 3]Any other charged particle moving the same velocity (vEthrno matter what its charge orF) of the particle, so changing mass has no FF[2 or 3]q forces increase (or decrease) MAGNETIC70

 When polarity (i.e. sign) of the charge on the particle is changed, the directions of both forces are reversed but their magnitudes remain the same, hence, no effect on the path of the particle. vFdeF[2 or 3]MAGNETICPage

6 Hall Effect Figure 20.1.13The charge carriersFup(VHFEB. At this point, the p.d. becomesconstant.) across a current-carrying conductor by a the , and the developed p.d. (VHis known as Hall voltage. (holes).MAGNETICFIELDSPage 72 of

Derivation of the Expression for Hall Voltage (VH)14hen VH= FB (where E is the electric field strength, and BE= (here:HHBvV= om the expression for current in a conducting slice: I=Plugging the above expression for vH (here: A= )HBdInqdtHBI[5]Note, is the dimension of the slice to theMAGNETICPage of

Hall ProbeFigure 20.1.15Hall probe is a device used to measure the flux density of magnetic field. Its main component is normal to the front face of its (current-carrying) sliceH=Vnqtit follows that VHBH) of the field. NotesVVHB, where is the component of flux density normalfront face of the slice of the a large (andVH[3]IIthe H1n, and the number of charge carriers per unit volume (n) is very H[2]1, where t is the thickness of the slice; so the slice of tobtain a large VHMAGNETICFPage

7)If: Figure 20.2.1 BA………. (i) In the above expression, 'A' is the area of the surface perpendicular to the magnetic field. So, magnetic flux may be defined as: magnetic flux = flux density × area perpendicular = 90BA (as sin 90 = 1) So, the magnetic flux through a surface is said to be 1passes perpendicularly area of the surface. If: Figure 20.2.2total  Τ (where  is flux through 1 turn of the coil) = sinwhere 'Asin' is the area of each turn perpendicular to the magnetic field. So, magnetic flux linkage no. of turns × flux density × area perpendicular Τ=NBA(as sin 90 = 1) MAGNETICIELDS75

8Electromagnetic Induction Figure 20.2.3 (a) Figure 20.2.3 When the magnetic flux linked to a conductor (e.g. a coil of wire) is changed (as shown in Fig. 2), an e.m.f. is induced in it; this phenomenon is known as the electromagnetic induction. When the magnetic flux is ut by a conductor (as shown in Fig. 20.2.3 (b)this phenomenon is known as Faraday’s Law of Electromagnetic InductionThe law states thE) is proportional to the rate of change of magnetic flux Τ ΤE k(where in SI: k = 1) ΤFrom Faraday’s law, it follows that the (magnitude of) induced e.m.f. can be increased by: The law states that the direction of the causing it. [2] Figure 20.2.4 (a) In the above examples, it is actually the work done in moving (pushing or pulling) the magnet that is transformed into electrical energy (generated in the circuit). [1] MAGNETICIELDSof 132

Mutual InductionFigure 20.2.5When the current (IPS) is induced in the nearby; this phenomenon is known as theS PtMAGNETICFIELDSPage 77 of 132

9CurrentsFigure 20.2.6These currents vary in magnitude and direction, so they are called eddy currents.The eddy currents produce heat in the conductor which flows into the surroundings; so eddycurrents cause loss of energy. shown below: MAGNETICof 132

BBMagnetic flux density B is varying with time. In an incomplete coil: (1ed presented. no current in coil (2) it produces an induced current. B B A rotating metal disc in a magnetic field: Itreated as a rotating conductor (or , I ): Φ, Be.m.f. tHall voltage Hall voltage: (or , ). : (1) The induced e.m.f. is the gradient of graph of e.m.f. (2) It would be perfectly correct if the “-” of e.m.f. is .ൌെΔ?Δ?Φe.m.f. Φ

ALTERNATING CURRENTS1 Alternating Current (a.c.).1 (a)21Direct current (d.cperiodically (i.e. at regular intervals of time). Sinusoidal Alternating Current and VoltageFigure ) alternating current (I) vs. time (tFigure The above sinusoidal alternating voltage and current are given by the expressions: = sin[1]where V0and I are the peak values of the expressions: =2π and π=where f is the frequency and TtheSinusoidal alternating current/voltage:x02=I2xrmsMaximum power

ALTERNATING CURRENTSRoot--IandMeanSquareVrmsmean of squared values of alternating current (or voltage) over the whole cycle. Let’s now calculate antheFigure 21ternating current, the I graph is as shown below:Figure Now, the mean value of 2I2=20+2= 2Finally, the root-mean-square value of the sinusoidal alternating current is calculated as: Irms2I=202 = 02So, for the sinusoidal alternating current and voltage, the root-mean-square values are given by: 0………. (i) 0rmsV………. (ii) Now consider the following circuits: Figure .6 (a) Figure 216 Note that, in the above circuits, heat dissipation occurs at the same rate in the identical resistors (R) the a.c. ammeter (the d.c. ammeter (ISo, the root-mean-square value of an a.c. (Irms) is equal to the value of steadythe same heating effect in the same (or identical) resistor. Noteor alternating current and voltage. Page 81 of

2 Power Dissipation in a Resistive Load Carrying Sinusoidal AC For the circuit shown below: Figure 21.7Rt(where I is current in Rthe maximum power dissipated in the resistor is given by: P I R………. (i)and the mean power dissipated in the resistor (over the whole cycle of a.c.) is given by: rmsPutting 0= 2 ×I Iinto Eq. (i) gives:2 2= ( 2 × ) = 2× = 2 < >rmsSo, the maximum power and the mean power dissipated in a resistor carrying sinusoidalrelated by: P PNotesThe power dissipated in a resistor depends on current2 (as = 2 is always[2]rms rms2rmsVPALTERNATING CURRENTS

LTERNATING CURRENTSWorked show V-t graphs (respectively) for the resistor in Fig. Figure 21On Fig. 21.8 (c), show the varit) of the power (P[3]Solution (At a given instant of time, the power corresponding voltage and current (as P= ), so V-t-P-t graph:) Notesower () reaches its peak value twice; so frequency of = 2 × frequency ofPage 83

4 Rectification Rectification is the process of converting an alternating voltage (or current) into direct voltage (or It is done by using diode(s). [1] Figure Figure 21NotesDiode is said to be forward biased when its p-type side is connected to the high-potential point (or+ terminal) and n-type side, to the low-potential point (or – terminal).forward biased.In forward bias, the resistance (R) of diode is almost zero. So, all the applied voltage (E) drops21Diode is said to be reverse biased when its n-type side is connected to the high-potential point (or+ terminal) and p-type side, to the low-potential point (or – terminal).Dthe the p.d. () across R (see Fig. 21).Half-Wave Rectification by a Single Diode Figure 21.13 (a) Figure .13 (b)ycle of alternating voltage (in), terminal A is ‘+’ and B is ‘–’ (i.e. diode is forward biased).And, for the other half of the cycle, terminal A is ‘–’ and B is ‘+’ (i.e. diode is reverse biased).When the diode is forward biased (Fig. 21.13 (a)Voutineverse biased (Fig. .13 Fig. outis a half-wave rectified voltage, as it has only one polarity (i.e. it is always +).[3]14ALTERNATING CURRENTS132

CURRENTSFullWave RectificationDiodes (or Bridge Rectifier) Figure 21Figure 21.15 (b)When terminal A is ‘+’ and B is ‘–’ (Fig. (a)21.15 ), only diodes D2 and D4 conduct current.so the current in R is rectified voltage (as shown in Fig. .16).Figure 21.16unction (P) towards which diodes point is always at the higher potential (i.e. + endIn one cycle of alternating voltage, the corresponding full-wave rectified voltage reaches the peakvalue twice, so:Compared with the half-wave rectified current, the full-wave rectified current has greater r.m.s.Irms2of

VPower is varying all the time. Maximum power is the same in the above circuits: ଴଴, ????ଵfull-wave rectification, 〉?଴〈?〉ଵଶoutVinVinoutVVinVinVoutVoutVoutPoutPVHalf-Wave Rectification Full-Wave Rectification0ALTERNATINGURRENTSPower in r

Smoothing a Rectified Voltage by a CapacitorSmoothing is the process of reducing the variationin the rectified voltage (or current). 1].17 (a).[Figure 21.17Notes[1]RAs the resistance of the discharging loop is greater than that of the charging loop (see Fig. ), so the capacitor charges up fast and discharges slowly (see Fig. [1][2]Figure ALTERNATING CURRENTSPage of

Appendix: Mathematic Solutions for average and r.m.s. The formula for average: ?୫ୣୟ୬ൌଵ்?ൌଵ׬ଶሺ?For a sinusoidal signal ଴and ? ൌଶగ, ሺቀଶగ୰୫ୱൌ??ൌ1න ?଴ଶ??்ඨ଴ଶ2r.m.s. value √2?଴?଴√30tttx0--xPower dissipated in a resistor: ?ሺ?ሻ ൌ ?ሻsin଴ሺሻൌ ?sin??Mean power: 〈??ሻ??ൌଵ்׬଴଴LTERNATING

QUANTUM PHYSICS 1 Particulate Nature of Electromagnetic Radiation Figure 22.1photon (or quantum) of energy of EM radiation. [2] of electromagnetic energy) [2]=;=hcwhere h (= 6.63 × 1034 JPlank constant (= 3.0 × 10=;=hfc;=hpEnergy of a photon:Energy conservation in photoelectric effect:E=hf====MAX0==Absorption or emission of a photon by an isolated atom:2 1energy of photon = differencetwo electron energy levels

Power (P) of a Beam of EM RadiationGenerally, power is given by the expression: =tand intensity is given by the expressions: =I;=EtAwhere E is the energy transferred in time Figure 22.2then the power of the incident beam of EM radiation is given by the expressions: P hf;and the intensity of the incident beam of EM radiation is given by the expressions: = .NtA;= .where ‘N/tN/tA’ is the no. of photons Note beam of EM radiation, without changing its frequency (or wavelength), the no. of photons incident on QUANTUM PPage 90

UANTUM PHYSICS2 Photoelectric Effect is illuminated by EM radiation (see Fig. 22[2]f0), where:Threshold frequency0) is the minimumemission of[2]Different metals have different threshold frequencies.Page 91 of

ectric EffectFigure 22.4Photoelectric effect is an instantaneousThere is no delay between illumination and emission of electrons.[1] of incident EM radiation required for the emission of electrons. [1]incident EM radiation is (see Figs. 25.4 (a) (b))The maximum KE of emitted electrons (also known as the ) does not depend onthe intensity of incident EM radiation (see Figs. 25.4 (c) (d)). [1]The maximum KE of photoelectrons depends on the frequency of incident EM radiation (see Figs.(d) (e)). [1]Increasing the intensity of incident EM radiation at constant frequency increases the rate ofphotoelectric current& [1]QUANTUM PHYSICS92 of

QUANTUMHYSICSFailure of Wave Theory ofAccording to the wave theory of EM radiation: The bound electrons should continuously absorb energy from the incident EM radiation, and be[1]Figure 22.5Figure 22.6, surface per unit time per unit area. However, the energy (= hf) of each p[3]thebeam (ttedinstantaneously[3]the i.e. hf E= + MAX………. (i)MAXhc………. (ii)NotesElectrons lying at the surface are emitted with maximum kinetic energy (EMAX); whereas thosewith kinetic energies lessKE(seePhotoelectric effect provides evidence for the particulate nature of EM radiation.

Work Function Energy () and Threshold Wavelength (0)Figure 22.7 energy of the incident photon required for the electron from the surface of the metal (see Fig. 22.7). 0;0where 0f=c fThe threshold wavelength (QUANTUM PHYSICS94 132

QUANTUM PHYSICSEfand Evs. 1/GraphsRearranging Eq. (i):MAXgives: EMAXhf +) (which is a linear equation in EMAX vs. graph is straight-line with ‘gradient = hy’Figure 22.8Likewise, rearranging Eq. (ii):= + MAXgives: MAX1E hc (–) (which is a linear equation in Eand ‘y-intercept = : Figure 22.9of

QUANTUM PHYSICS3 Wave-Particle Duality In 1924, a French scientist Louis de Broglie proposed that: just like EM radiation can behave like particles, perhaps particles of matter, such as electrons, can also behave like waves (these waves were later called De Broglie Wavelength (=p;=where p is the momentum, mv is the velocity (or speed) of the moving particle. Electron Diffraction[4]nIn electron diffraction, the crystal (e.g. thin film of carbon) serves as the diffraction grating (for the Figure 221= h/d= n[2]As diffraction is a wave property, so electron diffraction provides evidence for the wave nature of96 of

QUANTUM PHYSICS4 Absorption and Emission of a Photon by an Isolated Atom Figure 22.12 conservation: energy of photon absorbed = increase in the energy of electron i.e. hf E EThe above equation can also be written as: 3 1hcNoteAbsorption of a photon by an atom can take place only if the energy of the photon is (exactly) equal to the energy difference between ANY two of its electron energy levels (i.e. neither smaller nor greater). If: Figure 22.13 conservation: decrease in the energy of electron = energy of photon emitted E E hf =or 97

5 Emission and Absorption Line Spectra Emission Line SpectrumFig. 22.14 shows a schematic diagram of the arrangement used to produce emission line spectrum of a Figure 22.shows a typical emission line spectrum of a hot gas. Figure 22.Line SpectrFig. 22.16 shows a schematic diagram of the arrangement used to produce absorption line spectrum of a Figure 22.16Fig. 22.17 sFigure 22.17orption line spectrum appears as a continuous spectrum crossed by dark lines at specific [2]QUANTUM P98 of 132

QUANTUM PHYSICSExplanationThe electrons in the gas atoms absorb, from the incident white light, photons of energies equal tohigher energy levels).When the electrons de-excite, photons of these energies (or wavelengths) are reThe emergent light has very small (i.e. negligible) proportion of these photons in the originalphotons’ wavelengths is very low in[3 or 4]Explain how line spectrum provides evidence for the Each line in the spectrum is associated with photons of a specific energy, and: Now, line spectrum implies that photons’ energies (LHS of the above equation) are discrete, and so are the energy levels (RHS of the above equation). [3 or 4]Page 132

NUCLEAR PHYSICS 1 Nuclear Binding Energy MassEnergy EquivalenceThe energy () and mass (m) of a body (or a system) are related by: where c (= 3.0 × 101) is the speed of light in a vacuum. From the above equation, it follows that:2where ∆Em is the corresponding change (increase or decrease) in the mass of the body (or system). mass-energy of an isolated system Nuclear Binding Energy (BMass Defect (∆23Figure in Fig. 23.1pn XSo, the mass defect of a nucleus is the difference between the mass of the nucleus and the sum of massesB m c= .that is, the binding energy of a nucleus is the energy equivalent to its mass defect. Mass-Energy Equivalence:Activity (A) and Decay Constant (λExponential nature of radioactive decay:E=mc) by:=1/2t( X)AZin Fig. 23.1 is given by: mass excess = mX–Auwhere u (= 1.66 × 10–27 kg) is the unified atomic mass unit.AZPage 100 of 132