advertisement

QUANTUM PHYSICS AND NUCLEAR PHYSICS 13.1 (SL Option B1) Quantum physics 13.2 (SL Option B2) Nuclear physics 13 AHL QUANTUM PHYSICS AND NUCLEAR PHYSICS 13.1 (SL oPTIoN B.1) QUANTUM PHYSICS The quantum nature of radiation charge involved in the photo-electric efect consisted of particles identical in every respect to those isolated by J. J. homson two years previously, namely, electrons. 13.1.1 Describe the photoelectric effect. 13.1.2 Describe the concept of the photon, and use it to explain the photoelectric effect. 13.1.3 Describe and explain an experiment to test the Einstein model. 13.1.4 Solve problems involving the photoelectric effect. © IBO 2007 13.1.1 (B.1.1) THE PHoToELECTRIC EffECT I n 1888, Hertz carried out an experiment to verify Maxwell’s electromagnetic theory of radiation. Whilst performing the experiment Hertz noted that a spark was more easily produced if the electrodes of the spark gap were illuminated with ultra-violet light. Hertz paid this fact little heed, and it was let to one of his pupils, Hallswach, to investigate the efect more thoroughly. Hallswach noticed that metal surfaces became charged when illuminated with ultra-violet light and that the surface was always positively charged. He concluded therefore that the ultra-violet light caused negative charge to be ejected from the surface in some manner. In 1899 Lenard showed that the negative Figure 1301 (a) shows schematically the sort of arrangement that might be used to investigate the photo-electric efect in more detail. he tube B is highly evacuated, and a potential diference of about 10 V is applied between anode and cathode. he cathode consists of a small zinc plate, and a quartz window is arranged in the side of the tube such that the cathode may be illuminated with ultraviolet light. he current measured by the micro-ammeter gives a direct measure of the number of electrons emitted at the cathode. When the tube is dark no electrons are emitted at the cathode and therefore no current is recorded. When ultraviolet light is allowed to fall on the cathode electrons are ejected and traverse the tube to the anode, under the inluence of the anode-cathode potential. A small current is recorded by the micro-ammeter. Figure 1301 (b) shows a plot of photoelectric current against light intensity for a constant anode-cathode potential. As you would expect, the graph is a a straight line and doubling the light intensity doubles the number of electrons ejected at the cathode. he graph of photoelectric current against light frequency, Figure 1301 (c) is not quite so obvious. he graph shows clearly that there is a frequency of light below which no electrons are emitted. his frequency is called the threshold frequency. Further experiment shows that the value of the threshold frequency is independent of the 333 CHAPTER 13 intensity of the light and also that its value depends on the nature of the material of the cathode. U.V. source μA P. E. current Cathode B 13.1.2 (B.1.2) EINSTEIN AND THE Anode 10 V P.E. current AHL Figure 1301 (a) Apparatus to investigate the photo-electric effect light intensity P.E. current Figure 1301 (b) f0 frequency Figure 1301 (c) In terms of wave theory we would expect photo-emission to occur for light of any frequency. For example: consider a very small portion of the cathode, so small in fact that it contains only one electron for photo-emission. If the incident light is a wave motion, the energy absorbed by this small portion of the cathode and consequently by the 334 electron will increase uniformly with time. he amount of energy absorbed in a given time will depend on the intensity of the incident light and not on the frequency. If the light of a given frequency is made very very feeble there should be an appreciable time lag during which the electron absorbs suicient energy to escape from within the metal. No time lag is ever observed. PHoToN he existence of a threshold frequency and spontaneous emission even for light of a very low intensity cannot be explained in terms of a wave theory of light. In 1905 Einstein proposed a daring solution to the problem. Planck had shown that radiation is emitted in pulses, each pulse having an energy hf where h is a constant known as the Planck constant and f is the frequency of the radiation. Why, argued Einstein, should these pulses spread out as waves? Perhaps each pulse of radiation maintains a separate identity throughout the time of propagation of the radiation. Instead of light consisting of a train of waves we should think of it as consisting of a hail of discrete energy bundles. On this basis the signiicance of light frequency is not so much the frequency of a pulsating electromagnetic ield, but a measure of the energy of each ‘bundle’ or ‘particle of light’. he name given to these tiny bundles of energy is photon or quantum of radiation. Einstein’s interpretation of a threshold frequency is that a photon below this frequency has insuicient energy to remove an electron from the metal. he minimum energy required to remove an electron from the surface of a metal is called the work-function of the metal. he electrons in the metal surface will have varying kinetic energy and so at a frequency above the threshold frequency the ejected electrons will also have widely varying energies. However, according to Einstein’s theory there will be a deinite upper limit to the energy that a photo-electron can have. Suppose we have an electron in the metal surface that needs just φ units of energy to be ejected, where φ is the work function of the metal. A photon of energy hf strikes this electron and so the electron absorbs hf units of energy. If hf ≥ φ the electron will be ejected from the metal, and if energy is to be conserved it will gain an amount of kinetic energy EK given by EK = hf - φ Since φ is the minimum amount of energy required to eject an electron from the surface, it follows that the above QUANTUM PHYSICS AND NUCLEAR PHYSICS electron will have the maximum possible kinetic energy. We can therefore write that EK max = hf - φ Planck constant is measured. he modern accepted value is 6.2660693 × 10-34 J s. he intercept on the frequency axis is the threshold frequency and intercept of the Vs axis is numerically equal to the work function measured in electron-volt. Or EK max = hf – hf0 Where f0 is the threshold frequency. Either of the above equations is referred to as the Einstein Photoelectric Equation. It is worth noting that Einstein received the Nobel prize for Physics in 1921 for “his contributions to mathematical physics and especially for his discovery of the law of the photoelectric efect”. Figure 1302 shows the typical results of Millikan’s experiment which shows the variation with frequency f of the maximum kinetic energy Ek . It is let to you as an exercise, using data from this graph, to determine the Planck constant, and the threshold frequency and work function of the metal used for the cathode. (Ans 6.6 × 10-34 J s, 4.5 eV) 6 PHoToELECTRIC EQUATIoN In 1916 Millikan veriied the Einstein photoelectric equation using apparatus similar to that shown in Figure 1301 (a). Millikan reversed the potential diference between the anode and cathode such that the anode was now negatively charged. Electrons emiited by light shone onto the cathode now face a ‘potential barrier’ and will only reach the anode if they have a certain amount of energy. he situation is analogous to a car freewheeling along the lat and meeting a hill. he car will reach the top of the hill only if its kinetic energy is greater than or equal to its potential energy at the top of the hill. For the electron, if the potential diference between cathode and anode is Vs (‘stopping potential’) then it will reach the anode only if its kinetic energy is equal to or greater than Vse where e is the electron charge. In this situation, the Einstein equation becomes Vse = hf – hf0 Millikan recorded values of the stopping potential for diferent frequencies of the light incident on the cathode. For Einstein’s theory of the photoelectric efect to be correct, a plot of stopping potential against frequency should produce a straight line whose gradient equals eh . A value for the Planck constant had been previously determined using measurements from the spectra associated with hot objects. he results of Millikan’s experiment yielded the same value and the photoelectric efect is regarded as the method by which the value of the 5 AHL 13.1.3 (B.1.3) MILLIkAN’S vERIfICATIoN of EINSTEIN’S 4 Ek /eV 3 2 1 0 0 0.5 1 1.5 2 2.5 15 f /10 Hz Figure 1302 The results of Millikan’s experiment In summary, Figure 1303 shows the observations associated with the photoelectric efect and why the Classical theory, that is the wave theory, of electromagnetic radiation is unable to explain the observations i.e. makes predictions inconsistent with the observations. Observation Emission of electrons is instantaneous no matter what the intensity of the incident radiation. he existence of a threshold frequency. Classical theory predictions Energy should be absorbed by the electron continuously until it has suicient energy to break free from the metal surface. he less the intensity of the incident radiation, the less energy incident of the surface per unit time, so the longer it takes the electron to be ejected. he intensity of the radiation is independent of frequency. Emission of electrons should occur for all frequencies. Figure 1303 Observations associated with the photoelectric effect 335 CHAPTER 13 he fact that the photoelectric efect gives convincing evidence for the particle nature of light, raises the question as to whether light consists of waves or particles. If particulate in nature, how do we explain such phenomena as interference and difraction?. his is an interesting area of discussion for TOK and it is worth bearing in mind that Newton wrote in his introduction to his book Optics ‘It seems to me that the nature of light be particulate’. 13.1.4 (B.1.4) SoLvE PRoBLEMS oN THE PHoToELECTRIC EffECT AHL Example 1 Calculate the energy of a photon in light of wavelength 120 nm. State and explain two observations associated with the photoelectric efect that cannot be explained by the Classical theory of electromagnetic radiation. 3. In an experiment to measure the Planck constant, light of diferent frequencies f was shone on to the surface of silver and the stopping potential Vs for the emitted electrons was measured. he results are shown below. Uncertainties in the data are not shown. Vs / V 1.2 1.6 2.0 2.5 3.0 3.2 f / 1015 Hz 0.25 1.7 3.3 5.6 7.7 8.4 Plot a graph to show the variation of Vs with f. Draw a line of best-it for the data points. Solution f = 2. c 3.0 ×10 8 = = 2.5 × 1015 Hz λ 1.2 ×10 −7 Use the graph to determine E = hf = 6.6 × 10 × 2.5 × 10 = 1.7 × 10 J -34 15 -18 Example 2 he photoelectric work function of potassium is 2.0 eV. Calculate the threshold frequency of potassium. (i) (ii) a value of the Planck constant the work function of silver in electron-volt. The wave nature of matter 13.1.5 Describe the de Broglie hypothesis and the concept of matter waves. 13.1.6 Outline an experiment to verify the de Broglie hypothesis. Solution 13.1.7 Solve problems involving matter waves. φ 2.0 ×1.6 ×10 −19 f0 = = = 4.8 × 1014 Hz h 6.6 ×10 −34 Exercises 1. 336 Use data from example 2 to calculate the maximum kinetic energy in electron-volts of electrons emitted from the surface of potassium when illuminated with light of wavelength 120 nm. © IBO 2007 13.1.5 (B.1.5) THE DE BRogLIE HYPoTHESIS AND MATTER WAvES In 1923 Louis de Broglie suggested that since Nature should be symmetrical, that just as waves could exhibit particle properties, then what are considered to be particles should exhibit wave properties. QUANTUM PHYSICS AND NUCLEAR PHYSICS We have seen (Section 7.3.4) that as a consequence of Special relativity the total mass-energy of a particle is given by nickel crystal accelerated electron beam E = mc2 We can use this expression to ind the momentum of a photon by combining it with the Planck equation E = hf such that E = hf = hc = mc 2 λ scattered electrons electron detector Figure 1305 The scattering of electrons by a nickel crystal h mc = λ But mc is the momentum p of the photon, so that p= h λ Based on this result, the de Broglie hypothesis is that any h particle will have an associated wavelength given by p = λ he waves to which the wavelength relates are called matter waves. For a person of 70 kg running with a speed of 5 m s-1, the wavelength λ associated with the person is given by λ= h 6.6 ×10 −34 = ≈ 2 × 10-36 m p 70 × 5 his wavelength is minute to say the least. However, consider an electron moving with speed of 107 m s-1, then its associated wavelength is λ= h 6.6 ×10 −34 = ≈ 7 × 10-11 m p 9.1 ×10 −31 ×10 7 Although small this is measurable. 13.1.6 (B.1.6) ExPERIMENTAL vERIfICATIoN of THE DE BRogLIE HYPoTHESIS In 1927 Clinton Davisson and Lester Germer who both worked at the Bell Laboratory in New Jersey, USA, were studying the scattering of electrons by a nickel crystal. A schematic diagram of their apparatus is shown in Figure 1305. heir vacuum system broke down and the crystal oxidized. To remove the oxidization, Davisson and Germer heated the crystal to a high temperature. On continuing the experiment they found that the intensity of the scattered electrons went through a series of maxima and minimathe electrons were being difracted. he heating of the nickel crystal had changed it into a single crystal and the electrons were now behaving just as scattered X-rays do. (See Chapter 18 Topic G.6). Efectively, that lattice ions of the crystal act as a difraction grating whose slit width is equal to the spacing of the lattice ions. Davisson and Germer were able to calculate the de Broglie wavelength λ of the electrons from the potential diference V through which they had been accelerated. Using the relationship between kinetic energy and momentum, we have E k = Ve = p2 2m herefore p = 2mVe Using the de Broglie hypothesis p = λ= h 2 mVe h , we have λ hey knew the spacing of the lattice ions from X-ray measurements and so were able to calculate the predicted difraction angles for a wavelength equal to the de Broglie wavelength of the electrons. he predicted angles were in close agreement with the measured angles and the de Broglie hypothesis was veriied – particles behave as waves. Of course we now have a real dilemma; waves behave like particles and particles behave like waves. How can this be? his so-called wave-particle duality paradox was not resolved until the advent of Quantum Mechanics in 337 AHL from which CHAPTER 13 1926-27. here are plenty of physicists today who argue that it has still not really been resolved. To paraphrase what the late Richard Feynman once said, ‘If someone tells you that they understand Quantum Mechanics, they are fooling themselves’. 13.1.7 (B.1.7) SoLvE PRoBLEMS INvoLvINg MATTER WAvES Example AHL Solution λ= energy states 13.1.8 Outline a laboratory procedure for producing and observing atomic spectra. 13.1.9 Explain how atomic spectra provide evidence for the quantization of energy in atoms. 13.1.10 Calculate wavelengths of spectral lines from energy level differences and vice versa. Calculate the de Broglie wavelength of an electron ater acceleration through a potential diference of 75 V. Use λ = Atomic spectra and atomic 13.1.11 Explain the origin of atomic energy levels in terms of the “electron in a box” model. 13.1.12 Outline the Schrödinger model of the hydrogen atom. h 2 mVe 6.6 ×10 −34 2 × 9.1 ×10 −31 ×75 ×1.6 ×10 −19 13.1.13 Outline the Heisenberg uncertainty principle with regard to position– momentum and time–energy. © IBO 2007 = 1.4 nm. Exercises 1. Repeat the example above but for a proton. 2. Determine the ratio of the de Broglie wavelength of an electron to that of a proton accelerated through the same magnitude of potential diference. 13.1.8-10 (B.1.8-10) oBSERvINg AToMIC SPECTRA hese topics are discussed in Topic 7.1.4. he following example will serve as a reminder and also reinforce the concept of the photon. Example he diagram below shows some of the energy levels of the hydrogen atom. -0.85 eV B -1.50 eV A -3.4 eV Calculate the frequency associated with the photon emitted in each of the electron transitions A and B. 338 QUANTUM PHYSICS AND NUCLEAR PHYSICS p2 Using E k = 2 m we have therefore that the energy of the electron En with wavelength λn Solution En = A ∆E = 3.4 – 0.85 = 2.6 eV = 2.6 × 1.6 × 10-19 J = 4.2 × 10-19 J = hf to give f = 4.2 ×10 −19 = 6.4 × 1014 Hz 6.6 ×10 −34 13.3.12 (B.1.12) THE SCHRöDINgER Transition A gives rise the blue line in the visible spectrum of atomic hydrogen and B to a line in the infrared region of the spectrum. 13.3.11 (B.1.11) THE oRIgIN of THE ENERgY LEvELS he electron is bound to the nucleus by the Coulomb force and this force will essentially determine the energy of the electron. If we were to regard the hydrogen atom for instance as a miniature Earth-Moon system, the electron’s energy would fall of with inverse of distance from the nucleus and could take any value. However we can the origin of dither existence of discrete energy levels within the atom if we consider the wave nature of the electron. To simplify matters we shall consider the electron to be conined by a one dimensional box rather than a three dimensional “box” whose ends follow a 1 shape. r In classical wave theory, a wave that is conined is a standing wave. If our electron box is of length L then the allowed wavelengths λn are give by (see Topic 11.1) 2L where n = 1, 2 , 3 …) n However from the de Broglie hypothesis we have that pn = h nh = λn 2 L In 1926 Erwin Schrödinger proposed a model of the hydrogen atom based on the wave nature of the electron and hence the de Broglie hypothesis. his was actually the birth of Quantum Mechanics. Quantum mechanics and General Relativity are now regarded as the two principal theories of physics he mathematics of Schrödinger’s so-called wavemechanics is somewhat complicated so at this level, the best that can be done is to outline his theory. Essentially, he proposed that the electron in the hydrogen atom is described by a wave function. his wave function is described by an equation known as the Schrödinger wave equation, the solution of which give the values that the wave function can have. If the equation is set up for the electron in the hydrogen atom, it is found that the equation will only have solutions for which the energy E of the electron is given by E = (n + ½ )hf. Hence the concept of quantization of energy is built into the equation. Of course we do need to know what the wave function is actually describing. In fact the square of the amplitude of the wave function measures the probability of locating the electron in a speciied region of space. he solution of the equation predicts exactly the line spectra of the hydrogen atom. If the relativistic motion of the electron is taken into account, the solution even predicts the ine structure of some of the spectral lines. (For example, the red line on closer examination, is found to consist of seven lines close together.) he Schrödinger equation is not an easy equation to solve and to get exact solutions for atoms other than hydrogen or singly ionised helium, is well-nigh impossible. Nonetheless, Schrödinger’s theory changed completely the direction of physics and opened whole new vistas- and posed a whole load of new philosophical problems. 339 AHL 0.65 ×1.6 ×10 −19 = 1.6 × 1014 Hz 6.6 ×10 −34 he corresponding wavelengths are A = 470 nm and B = 190 nm λn = Hence we see that the energy of the electron is quantized. MoDEL of THE HYDRogEN AToM B f = n2 h2 8me L CHAPTER 13 13.3.13 (B.1.13) THE HEISENBERg UNCERTAINTY PRINCIPLE means is that if its momentum is deined precisely, then its associated probability wave is ininite in extent and we have no idea where it is. Efectively, the more precisely we know the momentum of a particle, the less precisely we know its position and vice versa. In 1927 Werner Heisenberg proposed a principle that went along way to understanding the interpretation of the Schrödinger wave function. Suppose the uncertainty in our knowledge of the position of a particle is ∆x and the uncertainty in the momentum is ∆p, then the Uncertainty Principle states that the product ∆x∆p is at least the order of h, the Planck constant. A more rigorous analysis shows that From the argument above, we see that in the real world waves are always made up of a range of wavelengths and form what is called a wave group. If this group is of length ∆x, then classical wave theory predicts that the wavelength spread ∆λ in the group is given by AHL ∆x∆p ≥ h 4π To understand how this links in with the de Broglie hypothesis and wave functions, consider a situation in which the momentum of a particle is known precisely. h In this situation, the wavelength is given by λ = p and is completely deined. But for a wave to have a single wavelength it must be ininite in time and space. For example if you switch on a sine-wave signal generator and observe the waveform produced on an oscilloscope, you will indeed see a single frequency/wavelength looking wave. However, when you switch of the generator, the wave amplitude decays to zero and in this decay there will be lots of other wavelengths present. So if you want a pure sine wave, don’t ever switch the generator of and, conversely, never switch it on. For our particle what this 1 ∆ ∆x ≈ 1 λ If we consider the wave group to be associated with a particle i.e. a measure of the particle momentum, then 1 ∆p ∆ = λ h that is ∆p ∆x ≈ 1 h or ∆p ∆x ≈ h We have in fact used the classical idea of a wave but we have interpreted the term ∆ 1 as a measure of the λ Tok What is quantum mechanics? he advent of quantum mechanics meant that the determinism of classical physics was a thing of the past. In classical physics, it was believed that if the initial state of a system is known precisely, then the future behaviour of the system could be predicted for all time. However, according to quantum mechanics, because of the inability to deine the initial data with absolute precision, such predictions can no longer be made. In classical physics it was thought that the only thing that limited knowing the initial state of the system with suicient precision was determined basically by precision of the measuring tool. he Uncertainty principle put paid to this idea – uncertainty is an inherent part of Nature. here have been many attempts to understand what quantum mechanics is really all about. On a pragmatic level, many physicists accept that it works and get on with their job. Others worry about the many paradoxes to which it leads. One of the true mysteries (apart from the ever famous Schrödinger cat) is the double slit experiment and its interpretation. Fire electrons at a double slit and just like light waves, an observable interference pattern can be obtained. However, if you observe through which slit each electron passes, the interference pattern disappears and the electrons behave like particles. If you can interpret this then you can “understand” what quantum mechanics is all about. However, remember what Richard Feynman had to say on this topic. Finally, if quantum mechanics is the “correct” physics, why do we in this course spend so much time learning classical physics? An example might suice to answer this question. If you apply Newtonian mechanics to the motion of a projectile, you get the “right” answer, if you apply it to the motion of electrons, you get the “wrong” answer. However, if you apply quantum mechanics to each of these motions, in each case you will get the right answer. he only problem is, that using quantum mechanics to solve a projectile problem is like using the proverbial sledge-hammer to crack a walnut. We must move on. 340 QUANTUM PHYSICS AND NUCLEAR PHYSICS uncertainty in the momentum of a particle and this leads to the idea of the Uncertainty Principle. You will not be expected to produce this argument in an IB examination and is presented here only as a matter of interest. he Principle also applies to energy and time. If ∆E is the uncertainty in a particle’s energy and ∆t is the uncertainty in the time for which the particle is observed is ∆t, then h 4π path of deflected α-particle region occupied by gold nucleus path of Figure 1307 An α-particle colliding with a gold nucleus his is the reason why spectral lines have inite width. For a spectral line to have a single wavelength, there must be no uncertainty in the diference of energy between the associated energy levels. his would imply that the electron must make the transition between the levels in zero time. 13.2 (SL oPTIoN B.2) NUCLEAR PHYSICS he kinetic energy of the α-particle when it is a long way from the nucleus is Ek. As it approaches the nucleus, due to the Coulomb force, its kinetic energy is converted into electrostatic potential energy. At the distance of closest approach all the kinetic energy will have become potential energy and the α-particle will be momentarily at rest. Hence we have that Ze × 2e Ze 2 = Ek = 4πε 0 d 2πε 0 d AHL ∆E ∆t = d = distance of closest approach Where Z is the proton number of gold such that the charge of the nucleus is Ze . he charge of the α-particle is 2e. 13.2.1 Explain how the radii of nuclei may be estimated from charged particle scattering experiments. For an α-particle with kinetic energy 4.0 MeV we have that 13.2.2 Describe how the masses of nuclei may be determined using a Bainbridge mass spectrometer. =1.2 × 10-13 m 13.2.3 Describe one piece of evidence for the existence of nuclear energy levels. © IBO 2007 13.2.1 (B.2.1) NUCLEI RADII 2πε 0 E k 2 × 3.14 × 8.85 ×10 −12× 4.0 ×10 d= = Ze 2 79 ×1.6 ×10 −19 6 he distance of closest approach will of course depend on the initial kinetic energy of the α-particle. However, as the energy is increased a point is reached where Coulomb scattering no longer take place. he above calculation is therefore only an estimate. It is has been demonstrated at separations of the order of 10-15 m, the Coulomb force is overtaken by the strong nuclear force. In 7.1.2 we outlined how the Geiger-Marsden experiment, in which α-particles were scattered by gold atoms, provided evidence for the nuclear model of the atom. he experiment also enabled an estimate of the nuclear diameter to be made. 13.2.2 (B.2.2) MEASURINg NUCLEAR Figure 1307 shows an α-particle that is on a collision course with a gold nucleus and its subsequent path. Since the gold nucleus is much more massive than the α-particle we can ignore any recoil of the gold nucleus. he measurement of nuclear (isotope) masses is achieved using a mass spectrometer. A form of mass spectrometer is shown in Figure 1308. MASSES Positive ions of the element under study are produced in a high voltage discharge tube (not shown) and pass through a slit (S1) in the cathode of the discharge tube. he beam of ions is further collimated by passing through slit S2 which provides an entry to the spectrometer. In the region X, the ions move in crossed electric and magnetic ields. 341 CHAPTER 13 S1 Plate P1 X S2 P2 S3 B’ Y electron transitions in the hydrogen atom which are only of the order of several eV. he existence of nuclear energy levels receives complete experimental veriication from the fact that γ-rays from radioactive decay have discrete energies consistent with the energies of the α-particles emitted by the parent nucleus. Not all radioactive transformations give rise to γ-emission and in this case the emitted α-particles all have the same energies. AHL Figure 1308 Measuring nuclear masses he electric ield is produced by the plates P 1 and P 2 and the magnetic ield by a Helmholtz coil arrangement. he region X acts as a velocity selector. If the magnitude of the electric ield strength in this region is E and that of the magnetic ield strength is B (and the magnitude of the charge on an ion is e) only those ions which have a v velocity given the expression Ee = Bev will pass through the slit S3 and so enter the main body, Y, of the instrument. A uniform magnetic ield, B´, exists in this region and in such a direction as to make the ions describe circular orbits. From Sections 2.4 and 6.3 we see that for a particular ion the radius r of the orbit is given by, mv2 mv = B´ev that is r = r B´e Since all the ions have very nearly the same velocity, ions of diferent masses will describe orbits of diferent radii, the variations in value depending only on the mass of the ion. A number of lines will therefore be obtained on the photographic plate P, each line corresponding to a diferent isotopic mass of the element. he position of a line on the plate will enable r to be determined and as B´, e and v are known, m can be determined. Radioactive decay 13.2.4 Describe β+ decay, including the existence of the neutrino. 13.2.5 State the radioactive decay law as an exponential function and define the decay constant. 13.2.6 Derive the relationship between decay constant and half-life. 13.2.7 Outline methods for measuring the half-life of an isotope. 13.2.8 Solve problems involving radioactive halflife. © IBO 2007 13.2.4 (B.2.4) β- DECAY We say in 7.2.2 that β- decay results from the decay of a neutron into a proton and that β+ decay results from the decay of a proton in a nucleus into a neutron viz, 13.2.3 (B.2.3) NUCLEAR ENERgY LEvELS he α-particles emitted in the radioactive decay of a particular nuclide do not necessarily have the same energy. For example, the energies of the α-particles emitted in the decay of nuclei of the isotope thorium-C have six distinct energies, 6.086 MeV being the greatest value and 5.481 MeV being the smallest value. To understand this we introduce the idea of nuclear energy levels. For example, if a nucleus of thorium emits an α-particle with energy 6.086 MeV, the resultant daughter nucleus will be in its ground state. However, if the emitted α-particle has energy 5.481 MeV, the daughter will be in an excited energy state and will reach its ground state by emitting a gamma photon of energy 0.605 MeV. Remember that the energy of photons emitted by 342 1 0 n → 11 p + −01 e + – ν 1 1 p → 01 n + −01 e + ν and It is found that the energy spectrum of the β-particles is continuous whereas that of any γ-rays involved is discrete. his was one of the reasons that the existence of the neutrino was postulated otherwise there is a problem with the conservation of energy. α-decay clearly indicates the existence of nuclear energy levels so something in β-decay has to account for any energy diference between the maximum β-particle energy and the sum of the γ-ray plus intermediate β-particle energies. We can illustrate how the neutrino accounts for this discrepancy by referring QUANTUM PHYSICS AND NUCLEAR PHYSICS to Figure 1309 that show the energy levels of a ictitious daughter nucleus and possible decay routes of the parent nucleus undergoing β+ decay. therefore ln N − ln N0 = −λ t or N = N0 e− λt parent nucleus β+ ν γ Figure 1309 excited level of daughter nucleus his is the radioactive decay law and veriies mathematically the exponential nature of radioactive decay that we introduced in 7.2.6. ground state of daughter nucleus 13.2.6 (B.2.6) HALf-LIfE β+ γ ν γ Neutrinos and the conservation of energy he igure shows how the neutrino accounts for the continuous β spectrum without sacriicing the conservation of energy. An equivalent diagram can of course be drawn for β- decay with the neutrino being replaced by an anti-neutrino. he radioactive decay law enables us to determine a relation between the half-life of a radioactive element and the decay constant. If a sample of a radioactive element initially contains N0 atoms, ater an interval of one half-life the sample will contain N2 atoms. If the half-life of the element is T½ from the decay law we can write that 0 − λT1 N0 = N0 e 2 2 13.2.5 (B.2.5) THE RADIoACTIvE DECAY LAW We have seen in 7.2.6 that radioactive decay is a random process. However, we are able to say that the activity of a sample element at a particular instant is proportional to the number of atoms of the element in the sample at that instant. If this number is N we can write that ∆N = −λ N ∆t where λ is the constant of proportionality called the decay constant and is deined as ‘the probability of decay of a nucleus per unit time’. he above equation should be written as a diferential equation i.e. dN = −λ N dt his equation is solved by separating the variables and integrating viz, such that dN = −λ dt N ln N = − t + constant It is then said that at time t = 0 the number of atoms is N0 such that ln N0 = constant or e-λ T½ = 1 2 that is T½ = ln 2 λ 13.2.7 (B.2.7) MEASURINg RADIoACTIvE HALf-LIfE he method used to measure the half-life of an element depends on whether the half-life is relatively long or relatively short. If the activity of a sample stays constant over a few hours it is safe to conclude that it has a relatively long half-life. On the other had if its activity drops rapidly to zero it is clear it has a very short half-life. Elements with long half-lives Essentially the method is to measure the activity of a known mass of a sample of the element. he activity can be measured by a Geiger counter and the decay equation in its diferential form is used to ind the decay constant. An example will help understand the method. 343 AHL β+ CHAPTER 13 A sample of the isotope uranium-234 has a mass of 2.0 µg. Its activity is measured as 3.0 × 103 Bq. he number of atoms in the sample is 13.2.8 (B.2.8) SoLvE PRoBLEMS INvoLvINg RADIoACTIvE 2.0 ×10 −6 2.0 ×10 −6 = = 3.3 × 1016 NA 6.0 ×10 23 ∆N Using = −λ N we have ∆t ∆N 3.0 ×10 3 ∆ t = 9.0 × 10-14 s-1 λ= = 3.3 ×10 16 N HALf-LIfE Example 1. Using the relation between half-life and decay constant we have (i) (ii) 0.69 ln 2 = = 7.6 × 1012 s ≈ 2.4 × 105 years T½ = λ 9.0 ×10 −14 AHL Elements with short half-lives he above is just an outline of the methods available for measuring half-lives and is suicient for the HL course. Clearly in some cases the actual measurement can be very tricky. For example, many radioactive isotopes decay into isotopes that themselves are radioactive and these in turn decay into other radioactive isotopes. So, although one may start with a sample that contains only one radioactive isotope, some time later the sample could contain several radioactive isotopes. 344 the decay constant for radium-223 the fraction of a given sample that will have decayed ater 3 days. Solution For elements that have half-lives of the order of hours, the activity can be measured by measuring the number of decays over a short period of time (minutes) at diferent time intervals. A graph of activity against time is plotted and the half-life read straight from the graph. Better is to plot the logarithm of activity against time to yield a straight line graph whose gradient is equal to the negative value of the decay constant. For elements with half-lives of the order of seconds, the ionisation properties of the radiations can be used. If the sample is placed in a tube across which an electric ield is applied, the radiation from the source will ionise the air in the tube and thereby give rise to an ionisation current. With a suitable arrangement, the decay of the ionisation current can be displayed on an oscilloscope. he isotope radium-223 has a half-life of 11.2 days. Determine (i) 0.0616 day-1 (ii) 0.169 Exercises 1. he isotope technetium-99 has a half-life of 6.02 hours. A freshly prepared sample of the isotope has an activity of 640 Bq. Calculate the activity of the sample ater 8.00 hours. 2. A radioactive isotope has a half-life of 18 days. 2 Calculate the time it takes for of the atoms in a 5 sample of the isotope to decay. 3. A nucleus of potassium-40 decays to a stable nucleus of argon-40. he half-life of potassium-40 is 1.3 × 109 yr. In a certain lump of rock, the amount of potassium-40 is 2.1 µg and the amount of trapped argon-40 is 1.7 µg. Estimate the age of the rocks.