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L9 Introduction to Atomic Physics L9 Intro to Atomic Physics In the last lecture we were introduced to some of the ideas and concepts associated with quantum mechanics In particular we saw that for a microscopic system con ned to some region of space the quantum formalism of the Schrodinger equation results in one or more quantum numbers Whose integer values 12 3 result in the quantity of interest such as energy or angular momentum being quantized which means that its values only come in certain bundles or packets In this lecture we will extend these ideas to the realm of atomic physics This study of the solution of the Schrodinger equation as applied to atoms explains the fundamentals of chemisnfy You can basically think of chemistry as the study of electronic states in atoms and groups of atoms either isolated or interacting with other atomic systems We will not go into such depth here however We will spend the overwhelming majority of our time in this lecture talking mostly qualitatively about the results of the quantum fomialism applied to the simplest of atoms lzydr0gen httpmtsn dlnhv 707nnPr fnrpcparf 7 I AJ 1 11 mmm mum 1 Page 1 of l A If Innnn nyorogemcqms Hydrogenic Quantum Numbers The hydrogen atom consists of one electron orbiting a single proton It is the simplest atom in the universe Unless you consider the equally simple case of anti hydrogen which consists of an anti electron also called a positron orbiting an anti proton Such atoms of anti matter have been produced in the laboratory If you have studied chemistry in high school or college then you have most likely learned some quantum mechanical results for this important atom In particular you may have learned that the quantum mechanical states or energy levels of the electron in the hydrogen atom depend on quantum numbers which result from the solution of the Schrodinger equation as discussed in the previous lecture The quantum numbers associated with the electronic states in the hydrogen atom are each given a special name however due to the importance of this fundamental atom more on this later The quantum numbers for the electronic states in the hydrogen atom are as follows The rst quantum number is called the principal quantum number and is denoted you guessed it n It can take on values n 1 2 3 Note that we cannot have n 0 The next quantum number is called the orbital angular momentum quantum number and is denoted I this is the small letter L not the number 1 For a given value of the principal quantum number n the orbital angular momentum quantum number can take on values I 0 I 2 up to n I For example for the quantum state with n 4 the orbital angular momentum quantum number I can take on the values I 0 I 2 or 3 It is traditional to use letters to represent quantum states with certain values of 1 For example an I 0 state is called an S state an I 1 state is called ap state l 2 is called a dstate l 3 is called a f state etc These names originally came from a study of atomic spectra The value of I gives us information about the value of the orbital angular momentum of the electron in its orbit around the proton The third quantum number is called the magnetic quantum number This is because the value of this quantum number speci es how the energy levels of the electron will be affected by an interaction with an external magnetic eld Of cially this quantum number specifies the z component of the orbital angular momentum of the electron in this state where the za irection is the direction of the external magnetic eld This quantum number is denoted m and for a given value of the angular momentum quantum number I can take on values m l l l I 1 2 0 I l 1 I For example ifthe electron is in an I 3 state then the possible values for m are ml 3 2 1 0 I 2 or 3 The last quantum number is the spin quantum number ms Of cially this is the zcomponent of spin angular momentum quantum number This quantum number can take on one of only two possible values ms or 12 These two values are often referred to as spin down and spin up states respectively referring to the negative and positive values of the zcomponent of the spin angular momentum For convenience we here summarize the possible values of the quantum http mtsueduphys2020LecturesPart2L6 Ll 1Lecture9HydrogenicQNsbodyhyd 262008 nyurogemcqms rage 4 OI A numbers for the electron in atomic hydrogen Principal Quantum Number 12 This quantum number can have any integer value from 1 up to in nity Orbital Angular Momentum Quantum Number I For a given value of n this quantum number can have any integer value from 0 up to n l Magnetic Quantum Number m I For a given value of I this quantum number can be any integer value starting at l and going up to 1 Spin Quantum Number m S This quantum number can only have one of two possible values 2 spin up and 2 Spin down httpmtsueduphy52020LecturesPart2L6 L1 1Lecture9HydrogenicQNsbodynhyd 262008 RUldUVJSUU EllCU L5 Special Relativistic Effects If you re really on the ball you should be wondering about something very important at this point A hydrogen atom exists in threedimensi0nal space That is the electron is held in some region of 3D space around the proton We would therefore expect that there should only be three quantum numbers to describe the quantum states of the electron in the hydrogen atom Indeed this is exactly how many quantum numbers result from the Schrodinger equation the rst three listed in the previous sectionkand yet we list four What 5 up In the early 190039s experimental measurements of the radiation given off by hydrogen atoms showed some minor but nevertheless unexplainable discrepancies with the Schrodinger equation results the emission spectrum of atomic hydrogen will be discussed shortly This implied that something was wrong with the theory In 1929 Paul Dirac incorporated Einstein s special theory of relativity the physics of things moving at speeds close to the speed of light that was published by Einstein in 1905 into the formalism of quantum mechanics to form a relativistic version of quantum mechanics There were two very signi cant results that came out of Dirac s theory The first was that there should exist negative energl electrons which had the same mass as the electron but opposite its charge This was the prediction of antielectrons or positrons The positron was discovered in 1932 the same year as the discovery of the neutron The second conclusion Dirac drew from his calculations was that the electron should have a new type of angular momentum which was called spin angular momentum Despite this classically descriptive name for this new angular momentum it is very wrong to think of the electron as a spinning ball despite the fact that this description is often used to explain some characteristics involving spin In his solution to the hydrogen atom Dirac found the fourth quantum number m S As already mentioned 39Dirac s quantum theory incorporated special relativity into its formalism One of the characteristics of special relativity is that it places the temporal coordinate time on the same mathematical footing as spacial coordinates the three dimensions of space It thus deals in a 4D spacetime The mathematical description of the electron in the hydrogen atom in a 4D region of space time around the proton in Dirac s relativistic version of quantum mechanics is what results in the fourth quantum number The spinangular momentum that came naturally out of Dirac s theory is a purely relativistic effect and cannot be correctly described in classical terms httpmtsueduphysZOZOLecturesPart2L6L1 1Lecture9RelativisticEffectsbodyr rage I or 1 262008 n W aveiuncuon Page 1 0i 2 The Hydrogen Wavefunctions The solution of the Schrodinger equation to nd the wavefunctions for the electron in the hydrogen atom results in wavefunctions that in general depend on the three quantum numbers n l and m l as well as on the position in space and the time Dependence on the spin quantum number m S can be added in as a result of Dirac s formalism but it does not enter directly into the solution from the Schrodinger equation As a result we can symbolically represent the resulting wavefunction as 1 1m position time In this notation the quantum numbers specifying the non relativistic quantum state of the electron in the hydrogen atom are written as subscripts to label which state the speci c wavefunction under consideration corresponds to and the position and time in parentheses simply denote that the wavefunction is a metion of the position and time at which we try to detect the electron around the nucleus The electron traveling around the nucleus generally has a probability of being found in any direction from the nucleus so it tends to form a spherically shaped distribution of possible positions around the nucleus For this reason 3km from spherical coordinates tend to be used when describing the V position of the electron relative to the nucleus instead of the paging rectangular coordinates ryz While we won t need to know any speci cs about spherical coordinates here it will be important for us to understand that the spherical coordinates for a point in 3 D space consist of three Distance 3 4 r coordinates two of which are angles and the third of which is the radial distance from the nucleus to the electron s position This radial position is denoted r A result of all of this is that the hydrogen wavefunction can n mucus be broken up into three parts all multiplied together The rst part is the radial part of the wavefunction which we will write as Rr showing that it is only a function of the radial distance r from the nucleus to the electron position It turns out that this part of the wavefunction only depends on the principal and orbital angular momentum quantum numbers n and 1 so we will label this wavefunction as follows R quot10 The second part of the hydrogen wavefunction is the angular part so we will write it as Aangles showing that it is a function of the spherical coordinate angles which we are not going to describe any more here ask your instructor if you re interested in how they are de nedl This part of the wavefunction only depends on the orbital angular momentum quantum numbers I and m I so we write it as A m angles The last part of the wavefunction is the part depending on time so we will simply write it as T nt We thus have that the total hydrogen wavefunction for the Schrodinger equation can be written in the form Pnfm positiontime Rm 1AgmI anglesTn 91 http mtsueduphy52020LecturesPart2L6Ll 1LectureW9HWavchtnbodyhwav 262008 H W 39aVCLLUlULlUll rage 1 U1 A It is the rst radial part of the wavefunction R1110 that describes how far from the nucleus the electron might be found In particular remember from the last lecture that the absolutesquare of the wavefunction gives us information about a probability function In the case of the wavefunction of Eq 91 its absolutesquare gives us information about the probability of finding the electron around the proton in a hydrogen atom as a function of the 3 D position and time We will be particularly interested in that part of the probability associated with nding the electron at various distances from the nucleus the proton not worrying about the probabilities associated with different directions from the nucleus this part is associated with the angular information and is more complicated both mathematically and visually We are thus interested in the radial part of the hydrogen probability function related to the radial part of the wavefunction R n lr we will denote this radial part of the probability function as P I I r We know from the previous lecture that the radial part of the probability function PH 1 is related to the absolutesquare of the radial wavefunction Rn 1 However due to some intricacies associated with spherical 2 note the unexpected factor of r2 coordinates it actually turns out that Pquot 17 r2 R n 7 multiplying the absolute square This detail is not important for our discussion here but is just mentioned in case you are interested The next section will discuss some of the properties of the radial probability function P r nl httpmtsueduphy52020LecturesPart2L6 Ll 1Lecturem9HWavchtnbodyhwav 262008 nauiai rroos Page 1 0 The Hydrogen Radial Probability From the previous section we know that the radial part of the probability function depends on the principal quantum number n and the orbital angular momentum quantum number 1 P n 139 This function tells us the probability of detecting an electron in a hydrogen atom at some distance r from the nucleus the proton Let s take a bit of time to discuss some of the results found for the electron in a hydrogen atom associated with P1110 The following graphs show plots of P1110 vs r for various quantum states that is for various values of the quantum numbers n and I In particular the graphs show all of the quantum states for n 1 n 2 and n 3 Remember that for each value of n the orbital angular momentum quantum number I can take on values only from 0 up to n 1 Also recall that a quantum state with l 0 is called an s state one with l 1 is called a p state and one with l 2 is called a d state We can thus say that graphs of P7110 vs r are shown below for the quantum states Is 2s 2p 35 3p and 3d for atomic hydrogen A brief discussion of the properties of Pquot 1 suggested by these graphs then concludes this section Radial Probability Function for it a I 1 0 1210 1 10 810 F 1 Sr 610 4109 2mquot 0 1 2 3 4 5 6 r a 9 to Radial Distance r x 10quot m httpmtsueduphys2020LecturesPart2L6L1lLecture9RadialProbsbody radial 262008 1au1al r1005 Page 2 Of7 Radial Probability Function for n x 2 I 0 1 M C Sc M W A UT 3 D MM 51138 D 1 2 3 4 5 8 7 8 9 1 Radial Distance r x 10quotquot m httpmtsueduphy32020LecturesPart2L6L11Lecture9Radia1 Probsbodv radial 262008 KaCllal 1quot I ODS 410 35m9 310 3 25109 a5 2 19 15109 115 51c8 httpmtsueduphy52020LecturesPart2L6 L1 1Lecture9Radia1Probsbodyradial Radial Probability Function for n 2 I I M 3 Radiai Distance 139 x 10quot m 4 5 6 7 262008 mutual rroos Page 4 of 7 Radial Probability Function for n 3 I 0 392 10 in m l I 0 1 2 3 4 s a 7 B 9 10 Radial Distance r x 10quot m httpmtsueduphy32OZOLecturesPart2L6L11Lecture9RadialProbsbody radial 262008 namal rroos Page 5 of 7 Radial Probability Function for n 3 I 1 15109 l f 2m9 Page 110 51ca D 1 2 3 4 5 6 7 B 9 10 Radiai Distance r x 10 m httpmtsueduphy52020LecturesPart2L6L11Lecture 9Radia1 Probsbodv radial 262008 naumi 1391 005 Page 6 0 Radial Probability Function for n 3 I 2 25 m 2109 8 15109 2 9 1 110 I 5m8 M N If D 1 2 3 4 5 6 7 8 9 10 Radial Distance r x 10quot m We close this section with a few comments about the radial probability function graphs shown above 0 Remember that the probability function P 1 shows the probability of locating the electron in a hydrogen atom as a function of the distance form the nucleus the proton Thus we will most likely nd the electron at the distance from the nucleus where the probability graph of Pnl 39 vs r has a maximum Note that for a given quantum state the electron has a relatively sizeable probability of being at many different distances from the nucleus it does not simply orbit the nucleus in a set trajectory with a speci c radius Most hydrogen atoms are in their ground state which is the Is 11 1 l 0 quantum state see the next two sections concerning the energies of the quantum states Note that the graph of P 1S0 vs r has a maximum at about r 05 X 1039 m This gives us the approximate size radius of the typical hydrogen atom This radius of the ground state is given a special namemit is called the Bohr radius named after Niels Bohr who rst proposed a model of the atom which suggested this radius in 1913 denoted r1the 1 subscript is for n I A more precise value of the Bohr radius is BohrRadius r1 0529 gtlt10 m m 92 0 Look at the graphs associated with the quantum states for which I n I that httpmtsuedulphy52020LecturesPart2L6L11Lecture9Radia1Probsbodyradia1 262008 Ka lal 1quot I ODS is those quantum states for which the orbital angular momentum quantum number is just one less than the value of the principal quantum number this is the largest possible value for 1 given that value of 12 These are the quantum states Is 21 and 3d for the graphs shown All of these have only one maximtun see the next note showing us at what distance from the nucleus we are most likely to nd the electron If we let r denote the value of r at which the graph of Pquot 1r vs 7 has a maximum value for the quantum state n l n Z then it turns out that Approximate atomic radii rn n2 r1 93 where r I is the Bohr radius given in Eq 92 Thus as It gets bigger so does the approximate size of the atom actually the approximate radius of the atom increases as 172 Note that the number of maxima or bumps in a given Puff vs r curve is equal to 11 For example the 3p graph has 3 2 bumps while the 1s graph has only 1 0 I bump This section has discussed the radial aspect of the distribution of the possible electron positions around the nucleus in a hydrogen atom that is it has discussed in part what the atom looks like The next two sections will discuss the energies associated with these various electron distributions along with important consequences of those allowed energies for the quantum states of the electron in atomic hydrogen http mtsueduphy52OZOLecturesPart2L6 Ll 1Lecture 9RadialProbsbodyradial 262008 nieCIron energies Hydrogen Electron Energies We have seen that a solution of the Schrodinger equation for the hydrogen atom results in wavefunctions that describe a probability of nding the electron at various positions and times around the nucleus We saw in the last section that the size of the atom tends to increase as the principal quantum number 12 increases rn n2 139 Eq 93 The various quantum states not only have different sets of quantum numbers and different wavefunctions or probability functions but they also have different energies It turns out from the solution of the Schrodinger equation that the energy of an electron in an isolated hydrogen atom one that is not interacting with other atoms or electric or magnetic elds depends only on the principal quantum number n of the quantum state occupied by the electron In particular the energy of the electron in a hydrogen atom is given by 11164 FOR HYDROGEN ONLY E 94 quot 8599119113 where m is the mass of the electron m 911 x 10 31 kg e is the electroncharge magnitude e 16 X 10 quot19 C 80 is a constant called the electric permittivity of free Space so 885 x 10 2 CZN m2 h is Planck s constant h 6626 x 10 34 J s and n is the principal quantum number leew The energy of the electron in an isolated hydrogen atom therefore depends only on the value of the quantum number n all the other quantities in the equation are known constants We can thus simplify this equation signi cantly by combining a number of the constants in Eq 94 above into a single constant In particular it is convenient to de ne a constant called the Rydberg constant R as follows 4 R Jn fT 10973730 3 Rydberg Constant 95 880 h c m With this definition of the Rydberg constant it follows that Eq 94 above for the energy of the electron in the nth quantum state can be written in the simpler form th For Hydrogen ONLY E 96 n 2 n Substituting in for the values of the constants 11 c and R in the equation above nally allows us to write the equation for the hydrogen electron energies in the simplest form 21sxm 181 136w 2 2 n n En 97 where we have shown the energy value in bothjoules J and electron volts eV at small http mtsueduphy52020LecturesPart2L6 L1 l Lecturea9ElectronEnergiesbodyiel 262008 El CII39OI unergles rage L OI 5 unit of energy often used in atomic and nuclear physics 1 eV 16 X 10 19 J Note that the energies given by the equation above are all negative this is a result of the fact that the electron is bound to the proton in the atom Lower energies correspond to states with larger negative numbers for energy values be careful To help us understand more clearly what these energies are telling us let s compute a few of these energies and the construct an energy level diagram see Lecture 8 The lowest energy given be Eq 97 is the energy for the quantum state having it I Using this value of n in Eq 97 gives us that E I 436 eV Since this energy is the lowest energy possible for the electron bound in the hydrogen atom it must be the ground state energy for the electron so that the n 1 quantum state must be the ground state Recall from Lecture 8 that the next lowest energy that is the rst allowed energy above the ground state is called the rst excitedstate energy Inspection of Eq 97 shows that this must correspond to the state with principal quantum number n 2 2 for which E 2 36 eV22 3 40 eV Likewise the second excited state 72 3 has an energy given by E 3 3 6e V32 51 eV Again note that these negative numbers are getting closer and closer to zero so they must be getting larger and larger they are increasing in energy We can therefore construct an energylevel diagram for the electron in the hydrogen atom Such a diagram is shown in Fig 91 below 0 eV n 1 await Ionized Electron m 0544 eV n m 5 FourthExcited State 0359 3V a m 4 ThirdExcited Siam Lil eV n e 3 Seconda xcited State v 340 eV m n 2Fimt1 xeitcd State efeemm in gimme state at reusfug Enesgv n 3 Ground State Fig 91 36 eV We will use this energylevel diagram in the next section to help us understand the emission and absorption spectrum for atomic hydrogen httpmtsueduphys2020LecturesPart2L6 Ll 1Lecture9ElectronEnergiesbody el 262008 1 1 DPCULI Hill 1 asp L UL r The Emission Spectrum for Hydrogen The electron in a hydrogen atom is ordinarily in its lowest energy state the ground state correspondong to the quantum state having principal quantum number n I l 0 and energy E I remember that an s stare is one for which I 0 the value of I does not affect the energy within this formalism for hydrogen however The electron can be promoted to a higher energy state either by means of a collision or by absorbing a photon of energy Ephomn hf where the energy of the absorbed photon must match exactly the energy difference between the state the electron is currently in usually but not always the ground state and some higher quantum state to which it will make a transition once the photon is absorbed We will only consider transitions involving the absorption or emission of photons Once in the higher energy state the electron will drop down to some lowerenergy state not necessarily its original state in an attempt to return to the ground state In making such a transition two things must be true The rst is that the transition must obey the selection rule which says that a transition from some initial state to some nal state whether that nal state is of higher or lower energy than the initial state doesn t matter can only occur if the orbital angular momentum quantum number changes by 1 or 1 That is only those transitions for which A1 I are allowed transitions The probability for a transition in other cases is zero There are other selection rules in atomic transitions that we will not mention here The second thing that must be true for a transition to take place is that the energy difference 1AEifl lEf E is emitted in the form of a singlephoton for the case in which the electron makes a transition to a lower energy state Eflt E On the other hand a single photon of energy AEif must be absorbed in the case that Efgt Ei From the Einstein relation for the energy of a photon E hf hck it then follows that the energies of possible photons photon emitted by the hydrogen atom must satisfy the equation TRUE FOR ANYATOM 53 Ef Eihf 98 Since this equation does not rely on the details of the form of the equation for the energies as given in Eq 96 this equation is in general true for any atom or molecule As soon as we use Eq 96 however the result becomes true for hydrogen only Such is the case for the following discussion Combining Eqs 96 and 98 gives us that for a photon absorbed or emitted by a hydrogen atom he EmTi5Er iziEr Eiiz hrR th n 1 quot39239 n1 nf th quot17 99 2 nf where the Rydberg constant R is as de ned in Eq 95 Solving Eq 99 for one over the wavelength 1 this makes writing easier than solving for the wavelength itself then gives us that httpmtsueduphy52OZOLecturesPartn2mL6 Ll lLecture 9Emissionbodyemissionh 262008 LL UFVULL 1111 I 1IAE he 1 397quot l 910 n R Wavelengths for Hydrogen gt21 n Equation 910 is a very important equation It can be used to compute the wavelength of a photon that must be absorbed by an electron in a hydrogen atom in quantum state 111 if it is to make a transition to the state nf Likewise it can be used to compute the wavelength of a photon that will be emitted by an electron in a hydrogen atom as it makes a transition from the quantum state 11 to the state 17 To possibly help clarify things let s consider a speci c example Let s say that we have a hydrogen atom in its ground state so that 111 1 The electron in the atom absorbs a photon and makes a transition to the state 11 3 We have already computed in the previous section that E E 136 eV and that Ef E3 I51 eV The change in energy is then lAEifl AEME3E1l 51eV136 e101 11209 6V 1209 eV 193x1a 8 J Again we have used the conversion factor from electron volts to joules 1 eV 16 X1049 J Thus the electron would have to absorb a photon of energy Ephomn l ml 1209 eV in order to undergo a transition from the ground state n I to the second excited state n 3 This process is shown schematically in Fig 92 below Remember that this diagram shows only energies of the allowed quantum states The actual electron distribution around the nucleus in the various states is as described by the wavefunctions and the corresponding probability functions as described earlier 014 httpmtsueduphy52020LecturesPart2 L6L1 1Lecture9Emissionbodyemissionh 262008 r1 opecrru m rage 5 or 4 n w infinity Ionized Electron 836 n e 5 Fourthdecited State n m 4 ThirdExcited State OeV 0544 8V 0850 eV 151 eV 139 m 3 Second Excited State 340 eV 11 a 2 FirstExcited State 1 4 t a S a W n 1quot 1 Ground State 136eV Increasing Eire Fig 92 Likewise if the electron in a hydrogen atom were already in the n 3 state and were to undergo a transition down to the ground state 12 I note that it could also make a transition down to the n 2 state and then at possibly a later time another transition from n 2 down to n 1 it would emit a photon of energy Ephmon AE131 1209 eVin the process Fig 93 below shows a schematic diagram of this process http rntsueduphy52020LecturesPart2L6Ll lLecture9Emissionbodyemissionh 262008 Page 4 of 4 L L UPUULI ulll 0 eV n ia ntzy leniaed Electron m etc 0544 eV W n at 3 Fourth Excited State 0850 eV m a w 4 ThirdExcited State 151 eV 11 3 Second Excited State 340 eV m M a a 2 Firstxeited State 3 as 1 it if Wag 3 W Jigs Increasing Erie n e 1 Ground State 136eV Fig 93 As electrons in excited states of various hydrogen atoms make transitions down to lower energy states they give off photons of various wavelengths as dictated by Eq 910 The group of various wavelengths thus emitted form what is called the emission spectrum for atomic hydrogen Likewise the same wavelengths would be absorbed from a beam of radiation containing all possible wavelengths in the electromagnetic spectrum such as a blackbody emission from a hot star if that radiation were to pass through a gas of hydrogen atoms that were in various excited states The corresponding spectrum would then be called the absorption spectrum The next example computes the wavelengths in the emission or absorption spectrum of atomic hydrogen for wavelengths in the visible region of the electromagnetic spectrum httpmtsueduphy52020LecturesPart2 L6 Ll lLecture9Emissionbodyemissionh 262008 Page 1 of 3 Solution to Example 91 The Balmer series refers to the emission of photons corresponding to transitions in atomic hydrogen which give wavelengths in the visible region of the electromagnetic spectnim The relevant transitions are given special names the HOL transition corresponds to a 3p to 2s transition HB corresponds to 4p to 25 H y corresponds to Sp to 25 and H5 corresponds to 6p to 25 a Show that the selection rule for l is satis ed for each of these transitions Note that all of the transitions under consideration are from a pstate to an sstate Recall that a pstate means that the angular momentum quantum number has a value I 1 while an s state means that the angular momentum quantum number has the value I 0 Therefore in each of the transitions the electron is jumping from an 1 1 state to an 1 0 state The change in the angular momentum quantum number nal value minus initial value is therefore which satis es the angular momentum quantum number selection rule which states that in any allowed transition Al must equal either I or 1 b Find the wavelengths for the given transitions This is actually a very easy problem From Eq 910 we have that 1AEif he 910 1 l quotaquot 11 13911 Wavelengths f or Hydrogen R gt4 i The symbols 2 and nf stand for the values of the principal quantum number associated with the initial and nal states of the electron undergoing the transition in the hydrogen atom and the absolutevalue sign tells us to ignore any minus signs that may appear from the subtraction of the principal quantum number teims Using the value of the Rydberg constant given in Eq 95 R 10973730 1 Rydberg Constant m we can use Eq 910 to construct a table of wavelengths corresponding to various values of Hz and nf for the transitions of interest ones that yield wavelengths in the visible region of the electromagnetic spectrum Such a table is shown below Note that since the equations above came from Eq 910 which in turn came from Eq 96 all of the formalism developed here is true only for the hydrogen atom httpmtsueduphy52020LecturesPart2L6 Ll1Lecture9Example 9 1Sol9 1body 262008 Page 2 of 3 lName Hi Hf Hk nn Color 139 L HB L4 2 4860 Blue 1 l ll y 5 2 4339 Violet l H 5 6 2 4101 Violet 7 l 7 2 3969 Deepvioletl i ll 8 H 2 H 3888 I Note that only the rst four transitions in the Balmer series result in wavelengths in the Visible region of the spectrum some people may also be able to see the other transitions listed The approximate colors for those wavelengths in the Visible are also given Note T he following values are the result of experimental measurements Ha 6563 nm HIB 4861 nm H 4340 nm H5 4102 nm The agreement of the classical quantum mechanical results with the experimental values is seen to be very good In the theory whose results are presented in this lecture an approximation was made in which the proton was assumed to be motionless as the electron orbits it In other words we assumed that the proton is in nitely more massive than the electron While the proton is almost 2000 times as massive as the electron it is not in nitely more massive Taking the nite mass of the proton into account so that the electron and proton then orbit their mutual centerofmass improves the agreement between theory and experimental results c What is the ionization energy for the electron in the n 4 state of atomic hydrogen The Loniz tion energy of an electron simply means that amount of energy required to make the electron undergo a transition from an initial quantum state 111 to the nal quantum state with n f in nity An electron in the n in nity quantum state is said to be an ionized electron Such an electron is no longer bound to the nucleus and is free to move away from the nucleus As we ll see below such an electron has zero energy An electron with a positive energy is one that is not only freed from the nucleus but also has extra energy in the form of kinetic energy or energy of motion In this case we want to nd the ionization energy for the electron in the n 4 quantum state so that nl 4 Noting that lin nity 0 we get from Eq 97 that 136eV 435w E 4 43 16 httpmtsueduphy52020LecturesPart2L6 Ll lLecture9Example9lSol9lbody 262008 3of3 and that EDD13I58V 136eV ZOBV 002 00 The energy to undergo the transition from 111 4 to Hf in nity is then simply A E4 in nity 2 IEinfinity atomic hydrogen E Ionization E4 0850 eV This is the ionization energy for the n 4 quantum state in 0850 eV for n 4 http mtsueduphy52020LecturesPart2L6 L1 1Lecture9Example91Sol91body 262008 VLJLUL lbUlllO Iquot g A Multielectron Atoms We have so far only discussed the simplest commonly occurring atom hydrogen What about all of the other atoms in the periodic table of the elements It turns out that the hydrogen atom is the only atom that can be solved exactly using either the Schrodinger or the Dirac formalisms Everything we know about the structure and energy levels of all of the other neutral atoms in the periodic table came from either experimental wavelength measurements from emission spectra of those atoms or from approximate models of those atoms usually elaborate computer models There are also models which combine theory and experiment called semi empirical models There are two important things to remember about all of these models and results they all use terminology and thinking that relates back to the hydrogen atom since that s the only one that we can really solve and they are all approximate models only the results for hydrogen are exact at least as far as our current knowledge ot physics is concerned It should be noted that if an atom other than hydrogen is ionized to an extent that it only has one electron left then it behaves like a hydrogen atom in that all of the hydrogen atom equations will still hold true for the ionized atom if you replace the quantity 82 in these equations by ZeZ where Z stands for the number of protons in the nucleus Note that if Z I we will just get the hydrogenic equations back again This means for example that if you see the term e4 in a hydrogen equation then you would replace it with Z2e4 since e4 e22 which should then be replaced by Ze22 2294 Atoms which have been ionized until they only have one electron left are referred to as hydrogenic atoms Examination of Eq 94 for the energy of the electron in hydrogen 4 FOR HYDROGEN ONLY En 842 2 94 an n shows that if we have a neutral atom with Z protons in the nucleus and some number of neutrons that we don t care aboutmwe are only concerned with the total charge here and Z electrons orbiting the nucleus and we remove all of the electrons but one we will be left with an atom having a nucleus with Z protons with a single electron orbiting it This is a hydrogen atom Since Eq 94 above contains a term e4 we see that we should replace that term with Z264 so that the equation would now look like mz e 135 av Hydrogemc atoms En 8502112112 n 911 Equation 911 can be used to compute electron energies in atoms other than hydrogen as long as the atom contains only one electron For example a neutral beryllium atom has 4 protons in its nucleus in addition to usually 5 neutrons and 4 electrons orbiting the nucleus We do not have any equations that will allow us to compute the energy of any of the electrons in this neutral beryllium atom However httpmtsueduphy32020LecturesPart2L6 L1 lLecture9OtherAtomsbodyother 262008 utner Atoms Page 2 of 2 if we were to ionize three of the electrons that is remove them from the neutral atom then we could use Eq 911 to compute the allowed energies of the single remaining electron For this hydrogenic beryllium atom in the n 3 state for example we have for Z 4 4 protons in the nucleus of beryllium 2 136ev4 1365V162412ev39 E 3 3 2 9 This is of course just 16E339H where E 3 H is the value of E 3 for the hydrogen atom If we wished we could thus write Eq 911 in the alternate form Hydrogenic atoms En 22 EWham 912 For those of you who will go on to study chemistry or atomic physics you will learn much more about the structure of multi electron atoms It will always be important to remember that all of this structure is approximate and represents the culmination of much experimental and theoretical work that is continually ongoing httpmtsueduphys2020LecturesPart2L6Ll1Lecture 9Other Atomsbody other 262008 uuxllylv ulb 7 1 Sample Quiz 9 l The principal quantum number in the hydrogen atom tells us about the of the electron a charge b speed c velocity 1 energy e color 2 If the principal quantum number has the value 71 Z 4 then the largest possible value of the angular momentum quantum number is 210 b1 02 d3 64 3 The spin quantum number is a purely effect a relativistic b chemical c orbital d energetic e magnetic 4 When an electron in an atom makes a transition from a higher energy state to a lower one it a cools down b decreases its gravitational potential energy c emits a photon d absorbs a photon e causes the nucleus to decay 5 The electron states in multi electron atoms are all based on a the proton b the electron c the atom d hydrogen e the positron An swam3 httprntsueduphy32020LecturesPartg2L6Ll 1Lecture9SQ9body sq9html 262008 ogynns Page 1 of 1 Answers to Sample Quiz 9 httpmtsueduphy52OZOLecturesPart2L6L11Lecture9SQ9SQ9Ansbody sq9ans 262008 numeworx y Page 1 of3 Homework 9 l a Use Eq 94 to show that the groundstate energy of the electron in an isolated hydrogen atom is equal to the well known result 136 eV suggested by Eq 97 This is the socalled ionization energy for the electron that is the minimum amount of energy that would have to be given to the electron in order to remove it from the ground state of the hydrogen atom This is the energy required to induce a transition in the electron from the n 1 state to the n in nity state b What wavelength photon would have to be absorbed by the electron in the ground state of hydrogen in order to just ionize the atom 2 The graph shown below in Fig 94 is the same as the graph shown earlier of P 1 vs r for the 1s state of hydrogen but the horizontal axis is expanded near the origin to more clearly see the maximum in the probability function Use this graph to more precisely estimate the Bohr radius r1 the position of the maximum in the probability function for the electron in the 1s state of atomic hydrogen Radial Probability Function fora 1 1 0 1213 1 10 81C9 6189 P180 4199 i 2 109 M U 02 04 06 03 1 12 14 16 18 2 22 24 26 28 3 Radial Distance 3 x 113 m D Fig 94 3 It was noted after the probability graphs of P r vs r that for states for which I n httpmtsueduphy52020LecturesPart2L6L1 1Lecture9HW9bodywhw9html 262008 LLLLLLLL v U11 1 1 the position of the maximum of the probability function r n satis es the relation Eq 93 r 112139 where r is the Bohr radius the position of the maximum in the 1 S graph Use the graphs shown previously to nd the positions of the maxima r and then show that these positions do indeed satisfy the equation above Eq 93 to Within uncertainties associated with your position determinations from the graphs 4 Electron capture is a process by which the nucleus of an atom actually captures an electron from its orbit around the nucleus A classical picture of an electron orbiting the nucleus like a planet orbits the sun makes it very dif cult to explain the process of electron capture However a quantum mechanical description makes such a process much more plausible The graph shown below is another plot of Pnlr vs r for the Is state of hydrogen greatly expanded near the origin Given that the approximate radius of the proton in the hydrogen nucleus is about 12X1015m use the graph shown below to explain how the nucleus might possibly capture an electron in the 13 quantum state Quantum states with orbital angular momentum quantum numbers other than 0 an 3 state have probability functions very close to zero for distances near the nucleus Radial Probabiiity Function Inside the Nucleus 25 x 20 P130 quot 0 02 04 15 03 1 Radiai Distance r x itquot5 m Fig 95 5 We know that probabilities must be between 0 and I but inspection of the numbers on the vertical axis of the radial probability function curves P7110 vs r show values that are httpmtsueduphy32OZOLecturesPart ZuL L1 1Lecture9HW9bodyhw9html Page 2 of 262008 1 1U111CW U1 1X 7 very large Can this be correct Do these large values make any sense ls it possible that these radial probability functions are valid probability functions Explain 6 What is the ionization energy for the electron in the rst excited state of hydrogen 7 Explain how an emission spectrum is formed for a gas of atoms 8 As was discussed in class the Balmer series in atomic hydrogen is the series of wavelengths emitted by the electron when it makes a transition from an energy level with n gt 2 to the n 2 level Likewise the Lyman series is the series of wavelengths emitted when the electron makes a transition down to the n 1 energy level Find the four largest wavelengths in the Lyman series To what region of the electromagnetic spectrum do these wavelengths belong 9 The Paschen series is the series of wavelengths emitted by the electron in atomic hydrogen as it makes transitions from higherenergy levels down to the n 3 quantum level Find the four largest wavelengths in the Paschen series To what region of the electromagnetic spectrum do these wavelengths belong 10 The four largest wavelengths emitted by a single electron atom not hydrogen as its one electron makes transitions from its excited states down to the n 1 state are as follows 305 nm 257 mm 244 nm and 238 nm The groundstate energy of the electron is 87 x 10 18 J What are the energies of the rst second third and fourth excited states for the electron 11 This one s a bit challenging What is the energy of the fth excited state in the atom discussed in the previous problem rst aware re http mtsueduphys2020LecturesPart2L6L1 1Lecture9HW9bodyhw9html rage 5 or 5 262008 LIvAuU Answers to Homework 9 1 a Show b 914 nm 2 r1 053gtlt103910 m 3 Verz Eq 93 by measuring the positions 0fthe maxima in the curves for the Is 2p and 3d probability functions 4 Look closely at the axis labels for the graph in Fig 95 The entire horizontal axis is inside the radius of the nucleus the proton that is the entire rvaxis is in the region 0 lt r lt 12 X10715 m radius of nucleus Thus the entire graph corresponds to positions inside the nucleus Examination of the probability function for the Is state in atomic hydrogen therefore shows that the electron has a nite probability of being found within the nucleus If it is actually found inside the nucleus then it has been captured by the nucleus electron capture 5 Remember that the probability is the area under the probability function graph While the numbers on the vertical axis of the Pn 1r VS r graphs are surprisingly large we can also see that the numbers on the horizontal axis are surprisingly small If you approximate the given area under the curve by a rectangle of a given width and height you will nd that the area really is something close to I so these values for the probability function really do make sense and the given radial probability functions really are valid probability functions 6 34 eV 545 x10 19 J 7 Energy is put into a gas of atoms by heating them or by sending in a beam of light of many wavelengths such as radiation from a blackbody like a star so that the electrons in the atoms move into various excited states As the excited electrons undergo transitions to various lowerenergy states they emit photons of light of various energies equal to the energy difference of the quantum states involved in the transition and therefore various wavelengths The set of wavelengths emitted by the excited atoms in the gas as their electrons move to lower energy states forms what is called the emission spectrum of that gas Of course if we are looking at the emission spectrum with our naked eyes we will only be able to see those wavelengths in the visible region of the electromagnetic spectrum such as the wavelengths emitted in the Balmer series of atomic hydrogen derived in Ex 91 8 122 run 2 gt 1 102 nm 3 gt 1 973 nm 4 gt 1 950 nm 5 gt 1 These are in the ultraviolet region of the spectrum 9 1880 nm 4 gt 3 1280 nm 5 gt 3 109011m 6 gt 3 1010 nm 7 gt 3 These are in the infrared region of the spectrum 10 E2 218 x 10quot18 J E3 957 x10 19 J 134 544 x103919 J E5 339 x10quot19 J http mtsu eduphysZ020LecturesPart2L6 L1 1Lecture9HW9HWAns9bodyhwa AuDvovAn 262008