Modern Physics

 RADIOACTIVITYThis is the spontaneous disintegration of a radioactive nucleus emitting either α-particles, β-particles,or γ-rays .Atomics number (Z). This is the number of protons in the nucleus of an atom.Mass number (A). This is the number of neutrons and protons in the nucleus of an atom.Isotopes. These are atoms of the same element with the same number of protons but different number of neutrons. OR These are atoms of the same element with the same atomic number but different mass number.Radio-isotopes. These are radio-atoms of the same element having the same atomic number but different mass numbers. Artificial radioactive decay. This is the process through which a stable atom or isotope is changed into a radioactive isotope or atom.This can be done in three ways. i.e. Bombarding the nucleus with neutrons in nuclear reaction. Bombarding the nucleus with charged particles in nuclear acceleration Bombarding the nucleus with a stream of high energy rays in photodisintegration.Question 1. Explain why neutrons are preferred to charged particles in inducing nuclear fission or nuclear experiments.Neutrons are preferred because they carry no charge and are therefore not deflected by electrons or repelled by the nuclear charge. This enables them to penetrate deep into the nucleus. Question 2. Explain why radioactivity is referred to as a random process.Alpha particles. ( )This is a helium nucleus of 2 protons and 2 neutrons.Properties of Alpha particles.They are positively charged and therefore deflect in magnetic and electric fields.They have a low penetrating power because of their big mass and high charge.They have a very high ionizing power.They cause fluorescence on fluorescent materials.Beta particles. These are fast moving electrons.Properties of beta particles.They are negatively charged therefore they deflect in both magnetic and electric field.They have a high penetrating power because of their small mass.They have a high ionizing power.They cause fluorescence on fluorescent materials.Gamma rays.These are electromagnetic radiations travelling at the same speed as that of light.Properties of gamma rays.They carry no charge.They have a very high penetrating power.They travel in a straight line at the same speed as light.They have a low ionizing power.They cause photoelectric emission.They cause a heating effect when absorbed by matter.Activity (A)This is the number of integrations per second. Decay law.This states that the rate of decay(activity) is directly proportional to the number of radioactive atoms present. i.e. where N is the number of radioactive atoms present.dN/dt=-λN The negative shows that the numbers of particles reduce with increase in time.dN/N=-λ dt∫_(N_0)^(N_t)▒〖dN/N=-∫_0^t▒〖λ dt〗〗 In N_t-In N_0=-λtIn (N_t/N_0 )=-λtN_t/N_0 =e^(-λt)N_t=N_0 e^(-λt) Decay constant (λ) is the fractional number of atoms decaying per second.Half life (T_(1/2)) is the time taken for activity of a radioactive substance to reduce to half its original value. Or is the time taken for half of the radioactive atoms present to decay.Relationship between λ and T_(1/2)at t=0,N_t=N_0 and at t=T_(1/2),N_t=N_0/2from N_t=N_0 e^(-λt)at t=T_(1/2)N_0/2=N_0 e^(-λT_(1/2) )1/2=e^(-λT_(1/2) )2=e^(λT_(1/2) )In 2=In e^(λT_(1/2) )In 2=λT_(1/2)(In 2)/λ=T_(1/2)T_(1/2)=(In 2)/λ=0.6932/λAlso. Taking A (activity). From the Law of radioactivityA=λNat t=0,A_0=λN_0→N_0=A_0/λat t=t,A_t=λN_t→ N_t=A_t/λA_t/λ=A_0/λ e^(-λt)A_t=A_0 e^(-λt) Questions.25g of sodium decays to 4g. calculate Number of sodium atoms that decay. If the half-life of Na is 71 seconds, calculate its activity after the decay process.Answer. 23g→6.02×〖10〗^23 atoms from Avogadro’s law as one moles contains25g→(6.02×〖10〗^23)/23×25 atoms4g→(6.02×〖10〗^23)/23×4 atomsNumber of atoms that decayed =(6.02×〖10〗^23)/23×25-(6.02×〖10〗^23)/23×4 =5.5×〖10〗^23 atoms. from A_t=λN_t and T_(1/2)=(In 2)/λ →λ=(In 2)/T_(1/2) A_t=(In 2)/71×(6.02×〖10〗^23)/23×4〖 A〗_t= disintegrations Question.0.02kg of of half-life 5 years decays giving off alpha particles. Calculatei) the decay constantii) How much of the element is left after 3 years?iii). Calculate the activity after 3 years.Solution 〖 N〗_t=N_0 e^(-λt) =0.02×e^(-0.139×3)=0.013kg 0.054kg→6.02×〖10〗^23 atoms0.013kg→(6.02×〖10〗^23)/0.054×0.013 atomsfrom A_t=λN_t =0.139×(6.02×〖10〗^23)/0.054 0.013 =2.01×〖10〗^22 〖year〗^(-1)At a second time, an alpha particle detector registers a count rate of 32 per second. Exactly 10 days later, the count rate drops to 8 per second. Find the decay constant.A_t=A_0 e^(-λt) 8=32×e^(-λ×10×60×60×24)0.25=e^(-λ×36000) λ=1.6×〖10〗^(-6) s^(-1)If 1g of decays according to the equation Identify Q If the half-life of is 10 years, calculate its activity.Solution. 235+1=95+139+2+xx=092=42+57+yy=-7 from T_(1/2)=(In 2)/λλ=(In 2)/10λ=0.0693 〖year〗^(-1)A=λN235g contains 6.02×〖10〗^23 atoms117.5g conatins (6.02×〖10〗^23)/235×117.5 atomsA=0.0693×(6.02〖×10〗^23)/235×117.5A=2.08593×〖10〗^22 atoms per yearExperiment to determine the half-life.A radioactive substance is placed near the window of a Geiger Muller tube. (GM tube)The number of count rate is recorded in a given time interval.A graph of count rate against time is plotted.From the graph, the time taken for count rate to reduce to half the original value is determined which is the half-life.Radio-isotopes are nuclides which are unstable and undergo radioactive decay producing or emitting α,β,γ radiations to form a stable nuclide. E.g. A number of radio-isotopes is produced artificially by bombarding stable nuclides with high energy particles such as protons, alpha particles, neutrons and deuterium. E.g. bombarding with α particles the following equation Silver decays further forming a new deuterium β particle. Uses of radio-isotopes.Biological usesCarbon datingPlants take in carbon dioxide through photosynthesis which contains radioactive carbon 14. Upon death no more carbon dioxide is taken in and the proportion of carbon 14 atoms decreases as they disintegrate. Using the GM tube the activity (A) of the dead sample of plant measured and activity of a live sample plant is measured (A0). Given the half-life of carbon 14. The apparent age of the dead sample is determined from .Assessment of volume and concentration of blood in mammals.Radioactive sodium is mixed with a small volume of blood and then re-injected into the blood stream. A given sample of blood is then removed and the activity of sodium determined. A very high activity implies less volume of blood therefore high concentration.Example. A radioactive isotope has a half-life of 2 hours and activity 2.88×〖10〗^7 disintegrations per hour. Its infected into an animal and a sample of 2cm3 of blood taken from the animal. 16 hours later it was found to have an activity of 300 disintegrations per hour. Calculate the volume of blood in the animal.λ=(In 2)/T_(1/2) =(In 2)/2=3.47×〖10〗^(-1) 〖hr〗^(-1)A_0=2.88×〖10〗^7 disintegration 〖hr〗^(-1)A=A_0 e^(-λt)A=2.88×〖10〗^7×e^(-3.47×〖10〗^(-1)×16)A=112,500 disintegrations 〖hr〗^(-1)300 disintegrations 〖hr〗^(-1) are contained in 2 〖cm〗^3112,500 disintegrations 〖hr〗^(-1)are contained in 2/300×112,500=750〖cm〗^3Gamma rays from Cobalt 60 are used in treatment of cancer by inhibition of cancerous cells.Can be used in sterilization of equipment and bandages. Medical instruments are sterilized by exposing to gamma rays. The gamma rays are also being used to sterilize some food products.Gamma rays from Cobol also can be used an alternative to X-rays to produce radiographs.Industrial uses.Used for sterilization of equipment.Used as tracers of leakages in underground pipes.Used to check/trace imperfections around welded joints.They are used in automatic control of thickness of paper in paper industries or iron sheets by using β-particles and γ-raysUsed to determine the rate of wear of piston rings or car tyres.Agricultural uses.Used as tracers to determine how fertilizers, pesticides or nutrients are taken up by plants and how they perform their functions. Radio-isotopes have been used to induce plant mutations. This had led to improved varieties of seeds e.g. peas, beans, wheat etc. with high yields and high resistance to crop diseases.Exercise. 1.0μg of nuclide (_11^25)Na of half –life 1 minute decays by emission of Beta particles. Find The number of Na atoms initially present. The initial activity of the sample. The number of Na atoms present after 10 minutes. A rock contains and in the ratio 1:4 by weight respectively. Calculate the initial activity of in 6g of the rock sample if it has a half-life of 10 years. Element P exists as 1% of . If 20μg of Q is initially present. Calculate the activity of P after 25 seconds if it has a half-life of 10 seconds. 725μg of (_55^124)Cs decays with a half-life of 30.8 seconds. Calculate The activity after 2 minutes. After how long will the activity fall to less than 1 s-1Nuclear radiation detectors.Ionization chamber. It contains a gas at low pressure whose atoms or molecules are ionized to form electrons and positive ions by radiation from the source.Ionizing radiations enter the chamber through the wire gauze causing ionization of gas molecules by collision.The positive and negative ions formed are driven to opposite directions by the electric field between the anode and cathode.Electrons move to the anode which is usually pointed from where their detection is got by deflection on the electrometer. (E).The current depends on the nature of radiation and volume of the chamber. The magnitude of ionization current is then used as the measure of radioactivity of the sample.Note: Ionization current, I=ne where n is the number of electron pairs and e is the electronic chargeElectron Volt. (eV)Is the work done I moving an electron through a p.d of one volt. Example 1. The source of α particles has activity of 2.0×〖10〗^3 s^(-1). When α particles enter the ionization chamber, a saturation current of 2×〖10〗^(-9) A is obtained. If the energy required to produce an ion pair is 32 electron volts, determine the energy of an alpha particles. (e=1.6×〖10〗^(-19) C) from I=ne→n=I/en=(2×〖10〗^(-9))/(1.6×〖10〗^(-19) )=1.25×〖10〗^10 pairs32eV=32×1.6×〖10〗^(-19)=5.12×〖10〗^(-18) JTotal enery of ion pairs=5.12×〖10〗^(-18)×1.25×〖10〗^10Total energy of α-particles=6.4×〖10〗^(-8) JEnergy of an α-particle=(6.4×〖10〗^(-8))/(2×〖10〗^3 ) =3.2×〖10〗^(-11) JExample 2. A radioactive source produces alpha particles each of energy 60Mev. If 20% of the alpha particles enter the ionization chamber, a current of 0.2μA flows. Find the activity of the alpha source if the energy needed to make an ion pair in the chamber is 32MeV.n=I/e=(0.2×〖10〗^(-6))/(1.6×〖10〗^(-19) )=1.25×〖10〗^12 pairsTotal energy=32×1.25×〖10〗^12 =4×〖10〗^13 MeVTotal energy of one alpha particle=(4×〖10〗^13 Mev)/(60 Mev)=6.67×〖10〗^11 disintegrations s^(-1)20%→6.67×〖10〗^11 s^(-1)100%→(6.67×〖10〗^11)/(100%)×20%=1.334×〖10〗^11 s^(-1)Exercise. A radioactive source emits 2×〖10〗^5 alpha particles per second. The particle produces a saturation current of 1.1×〖10〗^(-8) A in an ionisation chamber. If energy ion pairs is 32eV. Determine the energy in Mev of alpha particles emitted by the source. (Ans 11MeV) A radioactive source emits alpha particles each of energy 2MeV. On entering an ionisation chamber 80% of the alpha particles produces ion pairs giving a saturation current of 1.2×〖10〗^(-8) A. Calculate the activity of the source if the energy required to produce one ion pair is 32eV. Alpha particles of energy 1MeV enters into an ionisation chamber. 75% of the alpha particles produce ion pairs giving a current of 2A. if the energy required to produce an ion pair is 32eV. Find the activity of the source.A graph of ionization current against applied voltage Between A and B current detected increases gradually but the p.d is not large enough to prevent recombination of the ions.Along BC, ionization current is almost constant. All ions reach the electrometer before recombination but there is no secondary ionization.Beyond C, current increases rapidly for a small change in p.d. secondary ionization takes place leading to many ion pairs being produced hence a large current detected.Geiger Muller Tube. (GMT) When ionising particles enter the chamber through the mica window, ion pairs are produced through collision with argon atoms.The electrons produced are accelerated towards the anode and more ions are produced by repeated collisions. This results into gas amplification (an avalanche).On reaching the anode, discharge occurs and causes a current pulse through resistor R.The voltage which depends is amplified and operates a counter.The tube is made ready (quenched) for the next count.The positive ion which would have caused secondary discharge on reaching the cathode re slowed down by collision with bromine molecules.Qn. Explain why the anode is in form of a thin wire gauze.The anode is in form of a thin wire such that the charge on it creates a strong electric field close to its surface which drives the ions at a very high speed producing an avalanche of electrons.Question: What is meant by an avalanche?An avalanche is a large number of moving ionized particles created as a result of secondary ionization due to collision between ions and gas atoms.When the ions are accelerated by high p.d, each ionization leads to formation of more ion pairs which in turn cause further ionisations. Dead time is the time taken by positive ions to travel towards the cathode before the electric field at the anode returns to the level large enough for another avalanche to start.Ionising particles arriving within this time will not be detected.Explain what is meant by a quenching agent.Quenching agent is a gas (vapour) inside the GM tube to ensure that only one pulse is produced by each ionising particle that enters the tube. It slows down the positive ions and prevents further ionisation.Characteristic curve of a GM tube. Threshold voltage is the minimum p.d below which no pulse is detected.Up to the threshold voltage, no counts are recorded at all. The amount of gas amplification is not enough to give pulses of sufficient magnitude to detect.Between A and B, the magnitude of the pulse developed in the tube depends on the initial ionisation which in turn depends on the energy of incident particles.Only energetic electrons give pulses of sufficient magnitude to be recorded. But their number increases with applied voltage.Between B and C, the count rate is almost constant, a full avalanche is obtained along the entire length of the anode and all particles regardless of their energy produce detectable pulses.Beyond C, count rate increases rapidly with voltage due to incomplete quenching.Note. The GM tube must be operated in the plateau region (BC) preferably in the middle of the region because its sensitivity is greatest in this region and is independent of the applied voltage.Cloud chambers. These are used to differentiate the particles given off.Wilson Cloud Chamber. (Expansion type) The liquid alcohol or methylated spirit is placed on a dark pad of the piston.The chamber suddenly expands adiabatically when the piston is moved down suddenly and the air cools.The dust nuclei are all carried away after a few expansions by drops forming on them after condensation.The dust free air is now subjected to a controlled expansion making the air super saturated.Simultaneously the air is exposed to ionising radiations from the source. The water drops collect around the ions produced. These droplets of water around the ions reflects light when illuminated.A photograph of the chamber taken at a right time reveals the tracts of the ionising particles. Background radiations.These are radiations from the surrounding. These are obtained from;- luminous paints- cosmic raysWeathering rocksNote. Background radiations are detected by a reading on the G-M tube without any radioactive source in front of its window. Diffusion Cloud Chamber. Air in the chamber is saturated with alcohol vapour. The metal base is maintained at very low temperatures creating a temperature gradient from the top to the bottom.This enables the alcohol vapour to steadily diffuse from the top to the bottom where it becomes super saturated close to the metal base.When the shield from the radioactive source is removed, ionising radiations ionize the air molecules and the ions formed diffuse to the bottom where they act as condensation centers on the ions forming a string of liquid droplets known as chamber tracks.These tracks appear white on a black background when viewed the microscope.The length, thickness and nature of the tracks shows the intensity of ionisation. Nuclear EnergyEinstein’s Mass-Energy relation.It states that E=mC^2 where m is mass, E is energy and C is the speed of light in a vacuum.Unified atomic mass unit.(1U)This is a 〖1/12〗^th of the mass of the carbon atom .from Avogardo^' scantant12g contains 6.02×〖10〗^23 atoms1 atom weighs (12×〖10〗^(-3))/(6.02×〖10〗^23 ) kg〖1/12〗^th of the mass of one atom=(1/12 ((12×〖10〗^(-3))/(6.02×〖10〗^23 ) ))kg 1U =1.66×〖10〗^(-27) kgfrom E=mC^2E=(1.66×〖10〗^(-27) )×(3.0×〖10〗^8 )^2=1.49×〖10〗^(-10) Jfrom 1eV=1.6×〖10〗^(-19) J1.49×〖10〗^(-10) J=((1.49×〖10〗^(-10) J)/(1.6×〖10〗^(-19) ))eV=9.31×〖10〗^8 eV=931Mev bydividing by 〖10〗^61U=1.66×〖10〗^(-27) kg=1.49×〖10〗^(-10) J=9.31×〖10〗^8 eV=931MevBinding energyThis is the energy required in forming a nucleus from its constituent protons and neutrons Or This is the energy required to split the nucleus into its constituent protons and neutrons.Binding energy per nucleonIs the ratio of the binding energy to the mass number.Mass defect.Is the mass equivalent of the energy required in forming a nucleus from its constituent protons and neutrons.NB: The energy given off during nuclear fusion is as a result of mass defect converted into energy. i.e. ΔE=(Δm).C^2 from Einstein’s mass energy relationExamples 1Calculate the binding energy for a deuterium nucleus given the following (_1^1)H=1.0076U (_1^2)H=2.0142U (_0^1)∩=1.0090U 1U=934MeVSolution(_1^2)H→(_1^1)H+(_0^1)∩+energy Or (_1^1)H+(_0^1)∩→(_1^2)H+energy Total mass on R.H.S =[1.0076+1.0090]U=2.0166UMass defect ∆m =[2.0166-2.0142]U=0.0024UBinding Energy =0.0024×934MeV =2.2416MeV =2.2416×〖10〗^6×1.6×〖10〗^(-19) J=3.587×〖10〗^(-18) JExample 2. Calculate the binding energy per nucleon in MeV of Boron given the following (_5^10)Bo=10.129U (_1^1)H=1.0080U (_0^1)∩=1.0087U 1U=1.66×〖10〗^(-27) kgSolution. (_5^10)Bo→5(_1^1)H+5(_0^1)∩+energy Total mass on R.H.S =[5×1.0080+5×1.0087]U=10.087UMass defect, m =[10.129-10.087]U=0.0455U=0.0455×1.66×〖10〗^(-27) kg =7.553×〖10〗^(-29) kgBinding energy, E =mC^2=7.553×〖10〗^(-29)×(3×〖10〗^8 )^2=6.798×〖10〗^(-12) JBinding energy per nucleon =(Binding energy)/(mass number) =(6.798×〖10〗^(-12))/10=6.798×〖10〗^(-13) J =(6.798×〖10〗^(-13))/(〖10〗^6×1.6×〖10〗^(-19) ) MeV =4.249MeVNuclear fusionThis is the joining of two light nuclear to form one heavy nucleus with the release of energy.Conditions required The fusing nucleus must approach each other at very high speeds and high temperature so as to overcome the mutual electrostatic repulsion. Fusing nuclei must be lightNote: nuclear fusion is the principle source of the sun’s energy as Hydrogen is fused into helium at temperatures between 〖10〗^8-〖10〗^9 K Nuclear fissionIs the splitting of a heavy unstable nucleus into two light nuclei with the release of energy.Conditions required The nucleus must be heavy and unstable.Neutrons are preferred to charged particles in inducing nuclear fission because being neutral they are not deflected by atoms electrons or repelled by nuclear charge and can therefore penetrate deep into the nucleus. A graph of binding energy per nucleon against mass number. Binding energy per nucleon of very small and very large nuclides is small.There exists a maximum binding energy per nucleon at a mass number approximately equal to 56.There are peaks for smaller nuclides whose number of protons equals the number of neutrons.Qn. Using the graph above, explain why energy is released during nuclear fusion and fission.During fusion, two light nuclei of low binding energy per nucleon combine to form a heavier nucleus of greater binding energy per nucleon. The total mass of light nuclei is greater than the mass of bigger nucleus. This difference in mass accounts for the energy released during fusion. During fission, a heavy unstable nucleus (A>56) of low binding energy per nucleon splits to give two or more light nuclei of greater binding energy per nucleon. The total mass of the two or more light nuclei is less than the mass of the heavy nucleus. This difference in mass accounts for the energy released during fission.X-RaysThese are electromagnetic radiations of very short wave length produced when fast moving electrons are stopped by a hard metal target.Production of X-rays. The filament is heated by a low voltage supply emitting electrons by thermionic emission.The electrons are accelerated to the anode by a high p.d between the anode and cathode (E.H.T) and focused on the target by the focusing cap. The electrons strike the metal target embedded in the anode.A small percentage of the kinetic energy of the electrons is transformed into X-rays while the rest is changed into heat.The anode heats up and is cooled by the cooling fins. Energy changes during the working of the X-ray tube.elctrical energy→heat energy→kinetic energy→heat energyNote: Tungsten or molybdenum metals of high melting point can therefore withstand the heat generated at the target. The anode is made of copper which is a good conductor of heat therefore conducts heat away from the target to the cooling fins. The tube is evacuated to allow free acceleration of electrons with no loss of kinetic energy. The lead shield is used to protect workers from the X-rays.Intensity of X-rays.Intensity increases by the number of electrons striking the target per second.The number of electrons striking the target per second depends on the filament temperature which temperature is controlled by the filament current.Types of X-rays.Hard X-raysThese are produced using very high accelerating p.d. (E.H.T) giving out electrons of very high kinetic energy. The resulting X-rays have a very short wave length (high frequency) and very high penetrating power. Such X-rays are very dangerous.Soft X-raysThese are produced using low accelerating p.d which results into electrons of low kinetic energy giving X-rays of long wave lengths and low penetrating power.Properties of X-raysTravel in a straight line at the same speed as light.They carry no charge.They are not deflected in both magnetic and electric fields.Easily penetrate matter of low density / small atomic mass.Eject electrons from the metal surface by the photoelectric effect.Cause fluorescence when sprayed on fluorescent materials.Darken photographic plates. They ionize gases through which they pass therefore a charged gold leaf loses its charge when surrounding air is irradiated with X-rays.They are diffracted by crystals showing that they are waves.Uses of X-raysUsed in the production of radiographs. (X-ray photographs)Used in the treatment of cancer by destroying cancer cells.Used to inspect welded joints for internal imperfections.Used in the study of the structure of crystals.Dangers of X-rays.They damage body cells.They cause mutations leading to abnormalities and the growth of cancer cells.They cause barrenness / deformed offspring when sprayed on the reproductive areas.They encourage the growth of skin and blood cancer cells.Precautions taken when handling X-rays.Too much and unnecessary exposure must be avoided. Personnel handling X-rays must wear protective clothing coated with a layer of lead.X-ray equipment must be stored in thick lead containers.X-ray spectra.This consists of continuous spectrum and line spectrum.A graph of intensity versus wave length of an X-ray spectrum. The lines are made of series denoted by the letters KLMN in order of increasing wave length and the lines are denoted by the letters α,β and γ in order of decreasing wave length. Line Spectrum.A line spectrum is formed when a highly energetic electron ejects an electron out of the inner most shells of the target atom. This puts the atom in an excited state because of the vacancy left. To restore stability, an electron from the outer shells moves to fill the vacancy left. Electron transition to the vacancies left results into emission of X-rays of definite wave length hence a line spectrum.Increasing the tube voltage increases the kinetic energy of the bombarding electrons which then increases the number of lines produced.ExampleThe figure below shows energy levels involved in the X-ray spectra. Identify series A and B Identify line Q and P account for it.Continuous spectrumThis is as result of multiple collisions of energetic electrons with target atoms. Different amounts of energy are lost during this collision giving off X-rays of different wavelengths ranging from definite cut off wave length.Definite cut off wave lengthThis is a wavelength of an X-ray produced as a result of an energetic electron losing all its kinetic energy in a single collision with the target.The X-ray photon has maximum energy with the shortest possible wavelength.NB: The energy of X-rays (photons), E=hf where h is Planck’s constant and f is frequency of X-rays.At minimum wavelength, λ_minenergy of photon=kinetic energy of electron hf_max=eVf_max=eV/hC/λ_min =eV/hλ_min=hC/eVExamples. An X-ray tube is operated at 10kV. Calculate the minimum wavelength of X-rays that can be produced.(h=6.63×〖10〗^(-34) J,e=1.6×〖10〗^(-19) C,C=3×〖10〗^8 ms^(-1))V=10×〖10〗^3 Vλ_min=hC/eV =(6.63×〖10〗^(-34)×3×〖10〗^8)/(1.6×〖10〗^(-19)×10×〖10〗^3 ) =1.243×〖10〗^(-10) m In a water cool X-ray tube operating at 80kV has a filament current of 12mA. Calculate Number of electrons striking anode per second. Assuming 98% of the electrons have their kinetic energy converted into heat, calculate the rate of flow of water required to maintain the target at 50% assuming the incoming water is at 370C. (S.H.C of water is 4200Jkg-1K-1)Solution I=neI=12mA=12×〖10〗^(-3) A,e=1.6×〖10〗^(-19) Cn=I/e =(12×〖10〗^(-3))/(1.6×〖10〗^(-19) ) =7.5×〖10〗^16 electrons rate of heat produced=rate of heat gained by water98/100×IV =mcθ98/100×12×〖10〗^(-3)×80000=m×4200×(50-37)m=1.72×〖10〗^(-2) kgs^(-1) An X-ray tube has an electron beam of 10mA and accelerating p.d 50kV. Find the number of electrons striking the target per second. Speed of electrons when they hit the target. Minimum wavelength of X-rays produced.Solution I=nen=I/e=(10×〖10〗^(-3))/(1.6×〖10〗^(-19) )=6.25×〖10〗^16 electrons. 1/2 mv^2=eVv^2=2eV/mv=√(2eV/m)v=√((2×1.6×〖10〗^(-19)×50000)/(9.11×〖10〗^(-31) ))=1.33×〖10〗^8 ms^(-1) λ_min=hc/eV =(6.63×〖10〗^(-34)×3×〖10〗^8)/(1.6×〖10〗^(-19)×50000)=2.49×〖10〗^(-11) mDiffractionThis is the spreading of waves after passing through a narrow aperture.An experiment to show that X-rays are waves in nature. A narrow beam of monochromatic X-rays is incident on the crystal behind which is a photographic plate.On developing the photographic film, a central dark spot is obtained surrounded by faint spots in concentric circles.The central dark spot shows that most of the X-rays were able to pass through the crystal.The faint spots surrounding the central dark spot is due diffraction of X-rays by interaction with electrons in the atom of the crystal hence confirming the wave nature of X-rays.Note. Most of the X-rays are able to pass through the crystal giving a central dark spot. The regularity of the faint spots shows that atoms in the crystal are arranged in a regular pattern.Bragg’s law.This states that 2dsin θ=nλ where d-atomic spacing/ distance between two atoms.θ-glancing angle.n-order of diffractionλ-wave length of the X-rays.Derivation Path difference =AB+BCConsider ABN, sin⁡〖θ=AB/d〗BC=AB=d sin⁡θPath difference =d sin⁡〖θ+d sin⁡〖θ=2d sin⁡θ 〗 〗For two waves in phase i.e. in constructive interference, path difference is an integral multiple of the wave length. Therefore path difference =nλ2d sin⁡θ=nλWhere n is an integer n=1,2,3…For maximum, n, sin⁡〖θ=1, or θ=90°〗n_max=2d/λExplain X-ray diffraction by a crystal and state Bragg’s law.When a parallel beam of X-rays is incident on crystals, atoms in the plane scatter a fraction of X-rays. Constructive interference occurs for those rays that are scattered parallel otherwise destructive interference occurs.Examples Calculate the Bragg’s angle for X-rays of wave length 8.42×〖10〗^(-9) cm incident on the set of cubic planes in a crystal of spacing 2.82×〖10〗^(-10) m if first order diffraction maximum is obtained.n=1,d=2.82×〖10〗^(-10) m λ=8.42×〖10〗^(-9) m from 2d sin⁡〖θ=nλ〗sin⁡〖θ=nλ/2d=(1×8.42×〖10〗^(-9)×〖10〗^(-2))/(2×2.82×〖10〗^(-10) )〗=0.149θ= Calculate the atomic spacing if X-rays of wave length 2×〖10〗^(-10) m are incident on a crystal and second order of diffraction maximum occurs for 11°40'. X-rays of wavelength 0.7×〖10〗^(-10) m are incident on a set of cubic planes in a crystal. If the first order diffraction maximum occurs for 10.4°, calculate the volume of the unit cell in a crystal.2d sin⁡〖θ=nλ〗d=nλ/(2 sin⁡θ ) =(1×0.7×〖10〗^(-10))/(2×sin⁡〖10.4°〗 ) =1.94×〖10〗^(-10) mvolume=d^3 =(1.94×〖10〗^(-10) )^3 =7.3×〖10〗^(-30) m^3 Monochromatic beam of X-rays of wavelength 4×〖10〗^(-10) m are incident on cubic planes of . If the first order diffraction maximum occurs at Bragg’s angle 9°, calculate the density of the crystal.2d sin⁡〖θ=nλ〗d=nλ/(2 sin⁡θ )=(1× 4×〖10〗^(-10))/(2×sin⁡〖9°〗 )=1×〖10〗^(-9) mvolume=d^3=〖(1×〖10〗^(-9))〗^3=2.0×〖10〗^(-27) m^36.02×〖10〗^23 atoms weighs 23g1 atom weighs 23/(6.02×〖10〗^23 ) gdensity=(23/(6.02×〖10〗^23 ))/(2.0×〖10〗^(-27) )=19.1g m^3 A beam of X-rays of wavelength 1×〖10〗^(-10) m is incident on a set of cubic planes in a sodium chloride crystal. The first order diffraction maximum is obtained at a glancing angle of 10.2°, find Spacing between the consecutive planes. Density of the sodium chloride crystal. (molecular mass of NaCl =58.3g) Calculate the atomic spacing of sodium Chloride crystal if the relative atomic mass of sodium is 23 and Cl is 35.5. ( density of NaCl 2.18×〖10〗^3 kgm^(-3))ATOMIC STRUCTURERutherford’s model of an atom The positive charges of an atom and nearly all the mass are concentrated in a small volume of the center. Electrons are moving very fast and hence their effect on alpha particles is negligible. Electrons are in motion in spheres around the nucleus and the electron cloud accounts for the volume of the atom.Rutherford’s scattering experiment. A source of alpha particles is placed behind a thin gold foil. A fluorescent screen is then placed in front of it. In analyzing the following observations were obtained;Observations Most alpha particles passed through the thin gold foil undeflected. Few alpha particles were scattered through small angles.(angles less than 90°) Very few alpha particles were scattered through angles more than 90°)ExplanationSince most alpha particles passed through undeflected, most space of the atom is empty.The small deflection of the alpha particles was due to the repulsion by the positive charge in the nucleus.The nucleus occupies a very small space of the atom therefore alpha particles direct to the nucleus are strongly repelled almost backward.Conclusion The positive charge of the atom and nearly all the mass are concentrated at a very small volume at the centre.Electrons are in motion in spheres around the nucleus therefore the volume of the atom is accounted for by this electron cloud.Electrons move very fast hence their effect on the alpha particles is negligible. Distance of closest approach. An alpha particle directly approaching the nucleus is slowed down and momentarily comes to rest at a distance x from the nucleus before it is repelled back. At the point of repulsion, the initial kinetic energy of the alpha particle is converted into electrical potential energy of the alpha particle nucleus system.kinetic energy=1/2 mV^2 electrical potential energy=(Q_1 Q_2)/(4πε_0 x) charge of α particles,Q_1=2echarge of nucleus,Q_2=Ze where Z is the atomic number1/2 mV^2=(2Ze^2)/(4πε_0 x)x=(Ze^2)/(πε_0 mV^2 )Example An alpha particle of mass 4U moving with a velocity of 2×〖10〗^7 ms^(-1) is accelerated towards a gold nucleus of atomic number 79. Find the distance of closet approach of the alpha particle towards the gold nucleus.x=(Ze^2)/(πε_0 mV^2 )=(79×〖(1.6×〖10〗^(-19))〗^2×9×〖10〗^9)/(1.66×〖10〗^(-27)×〖(2×〖10〗^7)〗^2 )=2.741×〖10〗^(-14) mExercise A beam of alpha particles of energy 4.2MeV is incident normally on a thin gold foil of atomic number 79. What is the closest distance of approach by the alpha particle to the nucleus? Calculate the least distance of approach of a 3.5MeV alpha particle to the nucleus of a gold atom. (Atomic number of gold=79) In a head on collision between an alpha particle and a gold nucleus the minimum distance of approach is 5×〖10〗^(-14) m. Calculate the energy of the alpha particle in MeV. (Atomic number of gold=79) Alpha particles of mass 4U moving with velocity 2×〖10〗^7 m are accelerated towards the gold nucleus of atomic number 79. Find the distance of closest approach towards the nucleus. (1/(4πε_0 )=9×〖10〗^9 Fm^(-1))Rutherford’s failures Orbiting electrons are accelerated electric charges and should therefore emit electromagnetic radiations losing energy. This implies that the electrons would spiral (move) into the nucleus and the atom would collapse within a short time. The atom is in fact a stable structure and therefore Rutherford cannot explain the stability of the atom. Since electrons are constantly accelerating around the nucleus emitting radiations, a continuous emission spectrum should be emitted by the orbiting electrons as the atomic spectrum. However experimental observations show that a line emission spectrum is obtained as the atomic emission spectrum.Bohr atomSuggestions madeElectrons can revolve around the nucleus only in certain allowed orbits and while in these orbits, they don’t emit radiations.An electron can jump from a higher orbit to a lower orbit and the difference in the levels emitted as electromagnetic radiations of frequency f and ∆E=E_2-E_1=hfWhen an electron is in a stable orbit, its angular momentum varies as an integral multiple of h/2π i.e. mvr=n(h/2π)Where m-mass of an electron. v-velocity of electrons h-Planck^' s constantn-principal quantum numberNote: Angular momentum is the moment of linear momentum.DefinitionA Bohr atom is an atom with a small central positive nucleus consisting of electrons revolving around the nucleus only in certain allowed circular orbits and while in these orbits, they don’t emit radiations but when an electron makes a transition from a higher energy orbit to a lower energy orbit an electromagnetic radiation of frequency f given by ∆E=hf is emitted.Applications of Bohr’s suggestions to a hydrogen atomEnergy of an electron Assumptions Each electron moves in a circular orbit centered by the nucleus. The necessary centripetal force is provided by the electrostatic force of attraction between the positively charged nucleus and negatively charged electron. Kinetic energycentripetal force=electrostatic force(mV^2)/r=e^2/(4πε_0 r^2 )mV^2=e^2/(4πε_0 r)1/2 mV^2=e^2/(8πε_0 r) Potential energyIs equivalent to work done to move an electron from infinity to an orbit at a distance r from the nucleus. ∆w=F drw=∫_∞^r▒〖F dr〗w=∫_∞^r▒〖e^2/(4πε_0 r^2 ) dr〗=e^2/(4πε_0 ) ∫_∞^r▒〖1/r^2 dr〗=e^2/(4πε_0 ) [-1/r]_∞^rP.E=-e^2/(4πε_0 r) Total energy (mechanical energy)E=K.E+P.EE=e^2/(8πε_0 r)+-e^2/(4πε_0 r)E=-e^2/(8πε_0 r)from mvr=n(h/2π)V=nh/2πmr V^2=(n^2 h^2)/(4πm^2 r^2 )from (mV^2)/r=e^2/(4πε_0 r^2 )V^2=e^2/(4πε_0 mr)e^2/(4πε_0 mr)=(n^2 h^2)/(4πm^2 r^2 )r=(n^2 h^2 ε_0)/(e^2 πm)E=-e^2/(8πε_0 ((n^2 h^2 ε_0)/(e^2 πm)) )E=-(me^4)/(8(ε_0 )^2 n^2 h^2 ) where n=1,2,3,…Calculate the wave length of radiation that will be emitted when an electron makes a transition from n=4 to n=3.∆E=hf∆E=h C/λE_4=-(me^4)/(128(ε_0 )^2 h^2 ) and E_3=-(me^4)/(72(ε_0 )^2 h^2 )ΔE=E_4-E_3h C/λ=(me^4)/((ε_0 )^2 h^2 ) (1/72-1/128)1/λ=(9.11×〖10〗^(-31)×(1.6×〖10〗^(-19) )^4)/((8.85×〖10〗^(-12) )^2×(6.63×〖10〗^(-34) )^3×3×〖10〗^8 ) (1/72-1/128)λ=1.888×〖10〗^(-6) mWhy is the energy negative?It is because electrons are bound to the nucleus of the atom. Work must be done to remove the electron from the atom to infinity where energy is zero and this work is done against the nucleus binding the electrons to the atom.Electrons in an atom have a number of separate energy values. These energy values are called the energy levels of atom. An electron is allowed to move from one level to another by gaining or losing energy but it is not allowed to have an amount of energy that would put it in between two levels thus an atom can accept packets of energy of different size called photons / quanta.The most stable state of an atom is called the ground state where electrons have the lowest energy. An atom is in an excited state when electrons have energy more than that of the ground state.Spectral lines of hydrogen. Lyman seriesThese correspond to series of lines with different frequency/ wavelength emitted by an electron jumping from the exited states to the ground states. They correspond to the ultraviolet part of the spectrum.Balmer series These correspond to the series of lines with different frequency/ wavelength emitted by an electron jumping from exited states to the first exited state. (n=2). They correspond to the visible part of the spectrum.Paschen SeriesThese correspond to a series of lines with different frequency/ wavelength emitted by electron jumping from exited states to the second exited state. (n=3) They correspond to the infrared part of the spectrum.Ionisation energyIs the energy required to remove an electron from an atom from its ground state to infinity so that it is completely lost.ionisation energy=E_∞-E_1=0-(-E_1 )=E_1Excitation energy is the energy required to excite an atom.Note. Energy and potential are numerically equal but of different units. QuestionThe energy levels in a mercury atom are -10.4eV, -5.5eV, -3.7eV and -1.6eV Find the ionisation energy of mercury in joules. What is likely to happen if a mercury atom in an unexcited state is bombarded with an electron of 4.0eV, 6.7eV or 11.0eV?Line emission spectrumIt is emitted when atoms of elements are excited in some form of heat from electricity.When electrons make a transition to higher energy levels, the atom becomes unstable since the energy is increased.Electron transition may occur to a vacancy left in the lower energy and electromagnetic radiation is emitted and a line is formed on the spectrum. These lines appear bright against dark background.These lines are separated and discontinuous and this gives evidence that the energy levels of an atom are separate.What is line spectrum?These are discontinuous lines produced by excited atoms and ions as they fall back to the lower energy levels. Line absorption spectrumAtom’s energy can change only by discrete amounts. If a photon of energy hf is just sufficient to excite the atom when an energetic radiation is passed through the atoms, the photon is absorbed. As a result, the intensity of the incident radiation is reduced since it has lost a photon.A dark line on a bright background is observed whose wavelength or frequency is that of the absorbed photon. These lines constitute the line absorption spectrum.Bohr’s failuresBohr’s theory cannot explain the fine structure of the spectral lines of hydrogen. Bohr said that electron orbits are circular but later it was discovered by Summerfield that they are elliptical.Bohr’s theory is only successful in explaining spectra for simpler atoms which contain one electron like hydrogen but for complicated atoms with large numbers of electrons the theory fails to explain their spectra.Thermionic emissionThis is the ejection of electrons from metals heated to very high temperatures.Mechanism of thermionic emissionThe conduction (valence) electrons in a metal are loosely bound to their parent nuclei. These electrons move randomly throughout the metal lattice. When a metal is heated sufficiently to high temperatures, the surface electrons gain sufficient kinetic energy to overcome their attraction by the atomic nucleus within the surface of a small electric potential gradient leave the metal surface. This ejection of electrons from a heated metal is called thermionic emission.Cathode rays. This is beam of fast moving electrons.Properties of cathode rays.They are negatively charged.They travel in a straight line.They affect photographic plates.Can be deflected by both in electric and magnetic fields.They possess kinetic (mechanical) energy.They cause fluorescence on certain substances e.g. phosphorous and zinc sulphate.They produce X-rays when they strike some metalsCathode Rays Oscilloscope. Cathode.Produces electrons thermionically.AnodeFocus electrons onto the screen.Accelerates electrons along the tube.Control grid.Concentrates the electron beam.Controls the number of electrons striking the screen per second. (brightness)X-plates.Vertical in position and deflect the electron beam horizontally when p.d is applied between them.Y-plates.Horizontal in position and deflect the electron beam vertically when p.d is applied between them.Fluorescent screenEmits light when struck by the electron beam. i.e. kinetic energy of electrons changes to light when they strike the screen.Graphite coating Prevents (shields) the electron beam from external electric field’s influence.Collects electrons produced by secondary emission which provides a return path to the E.H.T/earth.Uses of a C.R.OIt’s used as a clockIt’s used in comparing frequencies and display of wave forms.It’s used in measurement of phase difference.Used as a voltmeter for measuring both a.c and d.c voltages.Advantages of using a C.R.O as a voltmeter over moving coil voltmeter.It measures both a.c and d.c voltages.It responds instantaneously.It’s very accurate since it doesn’t draw any current from the circuit. It can’t be damaged by over loading since it has no coils to burn out.Time baseThis is an electrical circuit internally connected to X-plates and provides a saw tooth p.d that sweeps the electron spot from left to right of the screen at a constant rate also it helps in studying a variation of a quantity with time.A graph of voltage against time. Explain what happens during linear sweep and fly back.During linear sweep, the electron spot moves from left to right of the screen at a constant rate taking a definite time which can be controlled.During fly back, the electron spot just jumps from right to left of the screen but during this time, the grid is automatically made more negative and this reduces the brightness of the spot to zero at the start of the fly back and is not seen.The time taken for the right to left sweep of this spot is called fly back time.Appearance of electron spot on the screen. When there is no voltage on both the X and Y plates. (when both are earthed). The electron spot is at the center of the screen. When D.C voltage is connected to Y-plates, X-plates earthed and time base off The electron spot is deflected vertically upwards.applied p.d ∝deflectionp.d=(Voltage gaain)×dVoltage gain=Y-sensitivityQn. Explain how D.C voltage is measured using a C.R.O.D.C voltage is connected to Y-plates, X-plates earth and time base is switched off. Voltage gain is set.Deflection (d) of the electron spot on the screen is measured.p.d=Voltage gain×d A.C voltage is connected to Y-plates, X-plates earthed and time off. 2V_0∝l2V_0=Voltage gain×lV_(r.m.s)=V_0/√2Qn. Describe how root mean square voltage is measured using the C.R.OA.C voltage is connected to Y-plates, X-plates are earthed and time base switched off.Voltage gain is set.Length l of vertical line formed on the screen is measured.2V_0=Voltage gain×lV_(r.m.s)=V_0/√2 When the Y-plates are earthed, time base on and p.d applied on X-plates. When Y-plates are connected to a D.C source, time base on and X-plates earthed. When Y-plates are connected to A.C source secondary of the transformer and time base on. Qn. A C.R.O has its voltage gain set at 10V per cm. A sinusoidal input voltage is suitably applied to give a steady trace. When the time base is switched on, the electron beam takes 0.01s to traverse the screen. If the trace seen has a peak to peak height of 4cm and contains 2 complete cycles. Find the (i) value of the root mean square input voltage. (ii) frequency of the input signal.Solution (i) 2V_0=(voltage gain)×l2V_0=10×4V_0=20VV_(r.m.s)=V_0/√2 =20/√2=14.4V(ii) 2 cycless→0.01s1 cycle→0.01/2=0.005sT=0.005sf=1/T=1/0.005=200H_zQn. A C.R.O has its y-sensitivity set at 12Vcm-1. An A.C voltage of V_(m.s.v)=25.7V is steadily applied. When the time base is on, the beam takes 4×〖10〗^(-3) s to transverse a screen with 2.5 cycles formed. Calculate the. Peak to peak height Frequency of the power source.Experiment to determine voltage gain. (y-sensitivity) Connect D.C supply to Y-plates, X-plates earthed and time base switched off. A deflection is seen.Applied voltage across the Y-plates is varied using a rheostat and measured using a voltmeter while X-plates are earthed and time base is switched off.The deflection (d) of the electron spot on the screen is measured for various values of p.d using a graduated scale on the screen, the results are tabulated.A graph of p.d against d is plotted.The slope is calculated and using this gives the voltage gain or y-sensitivity.Voltage gain is the p.d per unit deflection of the electron spot of the screen.Qn. In what way does a T.V tube differ from a C.R.O.In a T.V, deflection of the electron beam is by magnetic field while in C.R.O its by electric field.Qn. Give a reason why it is possible to have a wider screen in a T.V set than in the C.R.O. In a T.V set the electron beam is deflected by magnetic field which gives a wider deflection than an electric field in a C.R.O which gives a smaller deflection.Describe how cathode rays are produced by thermionic emission. When a cathode inside an evacuated glass tube is heated by low voltage supply, electros are produced thermionically.The electrons are accelerated by a positive high voltage applied between the cathode and the anode.The electrons travel undeviated across the vacuum past the anode and produce a glow when they collide with a fluorescent screen and give up their energy.Its this beam of fast moving electrons from the cathode that constitute the cathode rays.Experiment to show that cathode rays carry a negative charge.The cathode rays are made to pass through an electric field. They are deflected towards the positive phase and this implies that they carry a negative charge.Experiment to show that cathode rays travel in a straight line. The cathode is heated by low voltage supply and emits electrons by thermionic emission. The anode accelerates the emitted electrons towards a Maltese cross placed in the center of the glass tube.A sharp shadow of the Maltese cross appears on the screen of the tube.This shows that that cathode rays travel in s a straight line.Experiment to show that cathode rays possess kinetic energy.See the diagram above.The cathode is heated by low voltage supply and emits electrons by thermionic emission. The anode accelerates the emitted electrons to high velocity and cathode rays are formed. When the cathode rays strike the Maltese cross, it rotates. This means that cathode rays possess kinetic energy.Electron deflection in a magnetic field only. When an electron moving with a velocity, V perpendicular to the magnetic field of flux density, B, it experiences a magnetic force given by F=BeV.If the speed V remains constant, then the electron describes a circular path because of the deflection caused by the magnetic field. When in a circular path, centripetal force acting on the electron is given by F=m V^2/r . where r is the radius of the path and m is the mass of an electron.magnetic force=cenripetal forceBeV=m V^2/rB=mV/reAndV=Ber/mangular velocity,ω=V/r=Be/mPeriod,T=2π/ω=2πm/Befrequency,f=1/T=Be/2πmExample 1. A beam of positive ions is accelerated through a potential difference of 1×〖10〗^3 V into a region of uniform magnetic field/flux density 0.2T. while in the magnetic field it moves in circular path of radius 2.3cm. derive an expression for the charge to mass ratio of ions and calculate its value.magnetic forc=centripetal forceBqu=m u^2/ru=Bqr/mThe kinetic energy must balance the energy due to the electric filed.1/2 mu^2=qV1/2 m(Bqr/m)^2=qVq/m=2V/(Br)^2 q/m=(2×1×〖10〗^3)/(0.2×2.3×〖10〗^(-2) )^2 =9.5×〖10〗^7 C〖kg〗^(-1)Electron Deflection in electric field only.Case 1 Let u be the velocity of electrons as they enter the electric field.Let v be the velocity of electrons as they emerge out of the electric field.Let V be the p.d between the plates and d be the plate separation.Horizontal motion Vertical motionu_x=ua_x=0v_x=u_x+a_x t=us_x=u_x+1/2 a_x t^2l=ut →t=l/u u_y=0a_y=F/m=Ee/mv_y=u_y+a_y tv_y=Eel/mus_y=u_y t+1/2 a_y t^2y=1/2 (Ee/m) (l/u)^2y=Ee/(2mu^2 ) l^2in the form y=kl^2 where k=Ee/(2mu^2 )The motion of electrons in an electric field is parabolic.At A, final velocity as the electrons emerge from the electric field.v=√(〖v_x〗^2+〖v_y〗^2 )v=√(u^2+(Eel/mu)^2 )tan⁡〖θ=v_y/v_x 〗θ=tan^(-1)⁡(Eel/(mu^2 ))tan⁡〖θ=y/OB〗Eel/(mu^2 )=(Ee/(2mu^2 ) l^2)/OBOB=1/2 lExample 1. A beam of electrons is accelerated through a p.d of 2kV and directed middle way between two horizontal plates of length 5cm and having a separation of 2cm. the p.d between the plates is 80V. Calculate The speed of electrons as they enter the region between the plates. Speed of electrons as they emerge from the region between the plates. Angle between the final and initial direction.Solution(i) accelerating p.d,V_a=2000Vp.d between plates,V=80Vlength of plates=5cm=0.05mplate separtion,d=2cm=0.02mwork done=change in kinetic energyeV_a=1/2 mu^2u=√((2×1.6×〖10〗^(-19)×2000)/(9.11×〖10〗^(-31) ))=2.65×〖10〗^7 ms^(-1)(ii) v=√(u^2+(Eel/mu)^2 )=but E=V/dv=√(u^2+(Vel/mud)^2 )v=√(〖(2.65×〖10〗^7)〗^2+((80×1.6×〖10〗^(-19)×0.05)/(9.11×〖10〗^(-31)×2.65×〖10〗^7×0.02))^2 )=2.65×〖10〗^7 ms^(-1)(iii) tan⁡〖θ=Eel/(mu^2 )〗θ=tan^(-1)⁡(Eel/(mu^2 ))θ=tan^(-1)⁡〖(Vel/(mu^2 d))=tan^(-1)⁡〖((80×1.6×〖10〗^(-19)×0.05)/(9.11×〖10〗^(-31)×(2.65×〖10〗^7 )^2×0.02))=2.9°〗 〗Explain the motion of electrons between the plates.The electrons are deflected from their original path towards the positively charged plate. They move with a vertical uniform acceleration given by Ee/m and have a uniform horizontal velocity. The path followed describes a parabola.Case 2 D- deflection of electron spot on a fluorescent screen at a distance L from the end of the electric plates. from tan⁡〖θ=Eel/(mu^2 )〗 D/(1/2 l+L)=Eel/(mu^2 )D=Eel(1/2 l+L)/(mu^2 )Example. In a cathode ray oscilloscope, an electron beam passes between the deflector plates each of 5cm long and 0.5m apart. The distance between the center of the Y-plates and the screen is 20cm and the potential difference between the anode and the electron gun is 250V. Determine the deflection of the electron beam of the electron beam on the screen of the C.R.O.D=Eel(1/2 l+L)/(mu^2 )E=V/d=250/0.5=500Vm^(-1)change in kinetic energy=work done1/2 mu^2=eVu=√((2×1.6×〖10〗^(-19)×250)/(9.11×〖10〗^(-31) )) =9.4×〖10〗^6 ms^(-1)D=(500×1.6×〖10〗^(-19)×0.05(0.025+0.2))/(9.11×〖10〗^(-31)×(9.4×〖10〗^6 )^2 )=Example. The figure above shows two metal plates 2cm long and placed 5mm apart. A fluorescent screen is placed 20cm away. An electron of energy 3.2×〖10〗^(-16) J is incident midway between the plates.Calculate the voltage which must be applied across the plates to deflect the electron 2.1cm along the screen, assuming that the space through which the electron moves is evacuated.work done=gain in kinetic energyeV_a=1/2 mu^21/2 mu^2=3.2×〖10〗^(-16)u=√((2×3.2×〖10〗^(-16))/(9.11×〖10〗^(-31) ))=2.7〖×10〗^7 ms^(-1)D=Vel(1/2 l+L)/(dmu^2 ) where V is calculatedSpecific charge of an electron. This is the ratio of charge to mass of an electron.Measurement of specific charge of an electron using J.J Thomson’s experiment. Electrons are emitted from the cathode filament thermionically and accelerated to the anode.Without electric and magnetic fields applied at the plates, the electron beam strikes the screen at point O which is noted.A magnetic field of known flux (B) is applied to deflect the electron beam to position C.An electric field of intensity (E) is simultaneously applied and adjusted until the beam goes back to O.The p.d across the plates (V) and plate separation (d) and accelerating voltage (V_a) are noted.Let v be the velocity of the electrons as they enter the field.electric force=magnetic forceEe=Bevv=E/B but E=V/d v=V