5: Thermionic Emission

5: Therionic Eission Purpose While we think of quantu echanics being best deonstrated in processes that show discontinuous change, historically quantu echanics was first revealed in systes where a large nuber of particles washed out the jups: blackbody radiation and therionic eission. In this lab you will investigate these two phenoena in addition to classical space-charge liited electron eission: Child s Law. Introduction Metals, as deonstrated by their ability to conduct an electric current, contain obile electrons. (Most electrons in etals, particularly the core electrons closest to the nucleus, are tightly bound to individual atos; it is only the outerost valence electrons that are soewhat free.) These free electrons are generally confined to the bulk of the etal. As you learned in E&M, an electron attepting to leave a conductor experiences a strong force attracting it back towards the conductor due to an iage charge: e 2 F x = 4πǫ 0 (2x) 2 (5.1) where x is the distance the electron is fro the interface and e is the absolute value of the charge on an electron. Of course, inside the etal the electric field is zero so an electron there experiences zero (average) force. You can think of these valence electrons as bouncing around inside a box whose walls are provided by the iage-charge force. (Odd to think: the walls are non-aterial force fields; the inside of the box is filled with solid etal.) Since teperature is a easure of rando kinetic energy, if we increase the teperature of the etal, the electrons will be oving faster and soe will have enough energy to overcoe the iage-charge force (which after all becoes arbitrarily sall at large distances fro the interface) and escape. This is electron evaporation. The higher the teperature the larger the current of escaping electrons. This teperature induced electron flow is called therionic eission. Starting in 1901, Owen Richardson studied this phenoenon and in 1929 he received the Nobel prize in Physics for his work. A hot wire will be surrounded by evaporated electrons. An electric force can pull these electrons away fro the wire the larger the electric force, the larger the resulting current of electrons. The precise relationship between the voltage and the resulting current flow 101

102 Therionic Eission V = 0 electric current density J A V A cathode etal x = 0 vacuu anode x = b accelerating electrons x Figure 5.1: A planar cathode and a planar anode are separated by a distance b. A positive potential difference V A attracts electrons fro the cathode to the anode, so the speed of the electrons v(x) increases as they approach the anode. The oving electrons constitute an electric current fro anode to cathode. The resulting steady current density is called J A. is called Child s law 1 (or the Child-Languir law, including Languir who independently discovered it while working at G.E.). In this experient you will easure both Child s Law and the Richardson Effect. Child s Law Consider a planar interface between a etal (x < 0) and vacuu (x > 0). Vacuu is in quotes because this region will contain escaped electrons a space charge rather than being totally epty 2. The nuber of electrons per volue (i.e., the nuber density) is denoted by n. In this experient, the etal will be heated (i.e., its a hot cathode or filaent) which will result in a supply of electrons evaporated fro the etal into the vacuu. An additional conducting sheet (the anode) is located at x = b. A positive potential difference, V A, between the cathode and the anode plane provides a force pulling these electrons fro the vicinity of the cathode towards the anode. The result is a strea of oving electrons (a current); the nuber density n(x) and speed v(x) of these electrons will depend on location, x, between the plates. The negatively charged electrons oving to the right constitute a steady electric current density to the left, i.e., a steady conventional electric current fro the anode to the cathode: J = en(x)v(x) = J A (5.2) Since the electrons leave the etal with (nearly) zero speed at zero potential, we can calculate their speed along the path to the anode using conservation of energy: 1 2 v2 ev (x) = 0 (5.3) 2e v = V (x) (5.4) 1 Cleent Dexter Child (1868 1933) Born: Madison, Ohio, A.B. Rochester, Ph.D. Cornell 2 In fact a perfect vacuu is not possible, so the word vacuu actually refers siply to a region with relatively few particles per volue

Therionic Eission 103 where V (x) is the potential difference ( voltage ) at x and is the ass of an electron. Because the accelerating electrons constitute a steady current (i.e., J A doesn t depend on position), n(x) ust decrease as the electrons speed toward the anode. The varying space charge density affects the electric potential in the vacuu according to Poisson s equation 3 : 2 V x 2 = ρ(x) = en(x) (5.5) ǫ 0 ǫ 0 Putting these pieces together with have the differential equation: d 2 V dx 2 = J A ǫ 0 v(x) = J A ǫ 0 V (x) (5.6) Since the electric field will be zero at the interface, we have a pair of initial conditions: V x x=0 = 0 (5.7) V x=0 = 0 (5.8) This differential equation looks a bit like Newton s second law: as you can see if in Newton s second law you substitute: d 2 x dt 2 = 1 F(x(t)) (5.9) t x(t) 1 J F(x(t)) x V (x) A ǫ 0 V (x) Recall that force probles are often ost siply solved using conservation of energy and that conservation of energy was proved using an integrating factor of dx/dt. If we try the analogous trick on our voltage proble, we ll ultiply Poisson s equation by dv/dx: dv dx d2 V dx 2 = ( [ ] ) 1 dv 2 = 2 dx [ ] 1 dv 2 = 2 dx J A ǫ 0 J A ǫ 0 J A ǫ 0 V 1 2 dv dx ( V 1 2 V 1 2 1 2 1 2 (5.10) ) (5.11) + constant (5.12) The initial conditions require the constant to be zero, so 1 2 [ dv dx ] 2 = J A ǫ 0 3 Poisson s equation is derived in the Appendix to this lab. V 1 2 1 2 (5.13)

104 Therionic Eission or This differential equation is separable: dv dx = 4J A ǫ 0 dv V 1 4 V 3 4 3 4 = = 4J A ǫ 0 4J A ǫ 0 V 1 4 (5.14) dx (5.15) x (5.16) where again the initial conditions require the constant of integration to be zero. Finally: 2 V (x) = 9J A 4ǫ 0 3 x 4 3 (5.17) Of course, V (b) is the anode voltage V A, so we can rearrange this equation to show Child s law: [ ] 4ǫ 0 2e J A = 9b 2 V 3 2 A (5.18) Much of Child s law is just the result of diensional analysis, i.e., seeking any possible diensionally correct forula for J A. Our differential equation just involves the following constants with diensions (units) as shown: b : L (5.19) E V A : Q = ML2 /T 2 (5.20) Q 2e ǫ 0 k : Q 2 Q 1 2 Q 5 2 = (5.21) EL M 1 2 M 3 2L 3 /T 2 Q/T J A : L 2 (5.22) where the diensions are: L=length, T =tie, M =ass, E=energy, and Q=charge. To ake a diensionally correct forula for J A, we just eliinate the M diension which we can only do with the cobination: V A k 2 3 : We can then get the right units for J A with: ( V A k 2 )3 2 3 Q 2 3 T 2 3 b 2 = k b 2 V 3 2 A : Thus the only possible diensionally correct forula is (5.23) Q/T L 2 (5.24) J A k b 2 V 3 2 A (5.25)

Therionic Eission 105 cathode: hot filaent, radius a anode 2 anode l cathode 1 4 b Figure 5.2: Coaxial cylinders: an inner wire (radius a) and outer cylindrical anode (radius b), for a vacuu tube diode. The cathode is heated so electron evaporation is possible, and a potential difference V A attracts electrons fro the cathode to the anode. The speed of the electrons v(r) increases as they approach the anode. The oving electrons constitute a steady electric current fro anode to cathode. Since the sae current is spread out over larger areas, the current density, J, between the cylinders ust be proportional to 1/r. The exact proportionality constant, found fro the differential equation, is (as usual) is not hugely different fro 1. We have derived Child s law for the case of infinite parallel plates, but you will be testing it in (finite length) coaxial cylinders. The inner wire (radius a) is the cathode; the outer cylinder (radius b) is the anode. Your cylinder with have soe length l, but we will below consider infinite length coaxial cylinders. Note that diensional considerations require that the anode current per length should be given by a forula like: I/l j k b V 3 2 A (5.26) although we could have an arbitrary function of the radius ratio: b/a on the right-hand-side. Fro Poisson s equation 4 we have: 2 V = J ǫ 0 v(r) = I 2πrlǫ 0 v(r) = Using the Laplacian in cylindrical coordinates we find: 2 V r 2 + 1 r V r = j 2πrǫ 0 j 2πrǫ 0 V 1 2 (5.27) V 1 2 (5.28) There is no known forula for the solution to this differential equation, but we can ake considerable progress by writing the differential equation in ters of diensionless quanti- 4 Poisson s equation is derived in the Appendix to this lab.

106 Therionic Eission ties: r/a = ρ (5.29) V = ja 2 3 f(ρ) (5.30) 2πǫ 0 yielding: with initial conditions: 2 f ρ 2 + 1 ρ f ρ = f (ρ) + 1 ρ f (ρ) = 1 ρ f 1 2 (5.31) f(1) = 0 (5.32) f (1) = 0 (5.33) We can nuerically solve this differential equation using Matheatica: NDSolve[{f [p]+f [p]/p==1/(p Sqrt[f[p]])}, f[1]==0, f [1]==0, {f},{p,1,200}] It s actually not quite that siple. The cathode, at ρ = 1, is actually a singular point of the differential equation (i.e., f (1) = ). However the situation very near the cathode is well approxiated by the planar case, where we ve shown: V (x) = = 9J A 4ǫ 0 [ ]2 9 3 4 2 3 x 4 3 = 9I 2πal4ǫ 0 2 3 ( )4 ja r a 3 a 2πǫ 0 So, near the cathode (i.e., ρ slightly larger than 1): f(ρ) 2 3 (r a) 4 3 = 9ja 2π4ǫ 0 2 3 ( r a a )4 3 (5.34) [ ]2 9 3 4 (ρ 1) 3 (5.35) 4 We can use this approxiation to start our nuerical differential equation solution at a non-singular point (like ρ = 1.00001). Real devices are designed with b/a = ρ anode 1. The behavior of f for large ρ can be deterined by finding A and α for which f = Aρ α is a solution to the differential equation. One finds: ( )2 9 f = 4 ρ 3 (5.36) A useful approxiation for the range: 100 < b/a < 1000 is: f = ( )2 9 4 ρ 3 + 2 (5.37) (For exaple, the device used in lab has b/a = 121.5. For this value, the differential equation gives f = 44.136; the above approxiation gives: f = 44.130.)

Therionic Eission 107 10 f(ρ) for ρ 1 f 60 50 8 f(ρ) for ρ 40 6 30 20 10 4 2 f(ρ) exact 50 100 150 200 ρ 2 4 6 8 10 12 14 ρ Figure 5.3: The plot on the left displays the diensionless voltage f obtained by nuerical solution to the differential equation. The plot on the right copares various approxiations for f to this nuerical solution. We recover Child s law by rearranging (5.30): 2πǫ 0 a [ ]3 VA 2 = j = I/l (5.38) f(b/a) Note: Languir s original work (Phys. Rev. 22, 347 (1923)) on this subject is expressed in ters of β where: ρ 1 (ρ 1) 2 So: β 2 (ρ) 4 9 β 2 = 1.072 for the device used in lab. f 3 2 ρ = ρ 1 (5.39) 2e 8πǫ 0 l 9bβ 2 V 3 2 A = I (5.40) Richardson s Law Most any theral process is governed by the Boltzann factor: ( exp E ) = e E/kT (5.41) kt where k is the Boltzann constant. Approxiately speaking the Boltzann factor expresses the relative probability for an event requiring energy E in a syste at (absolute) teperature T. Clearly if E kt, the probability of the event happening is low. If an electron requires an energy W (called the work function) to escape fro the etal, The Boltzann factor suggests that this would happen with relative probability e W/kT. Thus you should expect that the current eitted by a heated etal would follow: I e W/kT (5.42)

108 Therionic Eission Clearly you should expect different eleents to have different work functions, just as different atos have different ionization potentials. What is surprising is that the proportionality factor in the above equation includes a universal constant that is, a constant that just depends on the properties of electrons (and, ost iportantly, Planck s constant, h) and does not depend on the type of aterial. (This situation is siilar to that of blackbody radiation, in which photons rather than electrons are leaving a heated body, which was Planck s topic in discovering his constant. We will take up this topic on page 113.) Therionic eission probes the quantu state of the electrons statistically, whereas the photoelectric effect probes uch the sae physics electron by electron. (The next level proble is to explain why this universal constant (the Richardson constant, A) in fact does depend a bit on the aterial.) To show: J = AT 2 e W/kT (5.43) where A = 4πek2 h 3 = 1.2 10 6 A/ 2 K 2 (5.44) Quantu Theory: Free Electron Gas Instead of thinking about electron particles bouncing around inside a box, de Broglie invites us to consider standing waves of electron probability aplitude: ψ = N exp(ik x x)exp(ik y y)exp(ik z z) = Ne ik r (5.45) Recall 5 that vector k is the oentu, p = v, of the electron and = h/2π. Periodic boundary conditions on the box (which we take to be a cube with one corner at the origin and the diagonally opposite corner at the point r = (L,L,L)) require each coponent k i to satisfy: k i = 2π n i L (5.46) where each n i is an integer. Thus each wave function is specified by a triplet of integers: n = (n x,n y,n z ), the n-vector. Applying Schrödinger s equation, we find that this wavefunction has energy: E(n) = 2 k 2 2 = (2π )2 n 2 2L 2 = (2π )2 (n 2 x + n 2 y + n 2 z) 2L 2 (5.47) Notice that there is a definite relationship between the velocity vector of the electron and the n-vector. v = 2π L n (5.48) Another way of saying the sae thing is that allowed quantu-state velocities for a cubic lattice with cube-side 2π /L. The nuber of states with electron velocities in soe specified region (for exaple a velocity-space parallelepiped with sides: v x v y v z ) can be found fro the nuber of 2π /L sided cubes that fit into the region, which is the 5 For a review see: http://britneyspears.ac/physics/dos/dos.ht

Therionic Eission 109 unoccupied levels vacuu etal U W vacuu iage charge potential occupied levels 0 zero force (flat) potential inside etal E F Figure 5.4: Electrons in the etal experience a constant confining potential of depth U. Possible quantu echanical states for these electrons are displayed as horizontal lines. Electrons fill all the available states up to the Feri energy, E F. The work function, W, is defined at the iniu energy needed to reove an electron fro the etal. As shown above: W = U E F. volue of that velocity-space region divided by (2π /L) 3. Hence: nuber of states with velocity between v and v + v = v x v y v z (2π /L) 3 (5.49) nuber of states per volue with velocity between v and v + v = v x v y v z (2π /) 3 ( ) 3 = vx v y v z = N v x v y v z (5.50) 2π where N is the (constant) density of states in velocity space. Quantu Theory: Feri Energy Ferions (half-integer spin particles), in contrast to bosons (integer spin particles), cannot group together. Since the electron is spin 1 2, each of the above states can hold at ost 2 electrons: one spin up and one spin down. The probability that a particular ferion state with energy E will be occupied is given by a generalization of the Boltzann factor called Feri-Dirac statistics: 1 f(e) = ( ) (5.51) 1 + exp E EF kt where E F is called the Feri energy. The Feri energy is basically a disguise for the nuber of electrons, as, approxiately speaking, it is the dividing line between occupied states and unoccupied states. (If the Feri energy is high, there ust be lots of occupied states and hence lots of electrons.) Note that if E E F, the exponential factor is huge and we can neglect the +1 in the denoinator so ( f(e) exp E E ) F (5.52) kt

110 Therionic Eission v x t A Figure 5.5: Consider just those electrons with soe particular x-velocity, v x. In order to hit the wall during the coing interval t, an electron ust be sufficiently close to the wall: within v x t. The nuber of such electrons that will hit an area A will be equal to the nuber of such electrons in the shaded volue (which is the perpendicular extension of A a distance v x t into the volue). Note that any electrons in that volue will not hit A because of large perpendicular velocities, but there will be atching electrons in neighboring volues which will hit A. To find the total nuber of hits, integrate over all possible v x. that is, if E E F Feri-Dirac statistics approxiate the Boltzann factor. Classical Theory: Electron Escape The density of states cobined with the Boltzann factor gives us the nuber of free electrons per unit volue with a particular velocity. In order for an electron to escape during soe tie t, it ust have v x sufficient to overcoe the iage-charge barrier and it ust be sufficiently close to the wall. All the electrons with v x > 2U/ within a distance v x t, will escape, where U is the depth of the potential well for the electrons. Thus the nuber of electrons escaping through area A during t is: dv x dv y dv z 2Nf(E) Av x t 2U/ = 2Ne E F /kt A t e v2 x/2kt v x dv x e v2 y/2kt dv y e v2 z/2kt dv z 2U/ = 4π(kT)2 (2π ) 3 ( A t exp (U E ) F) kt where we have used the Gaussian integral: e αz2 dz = π α The electric current density is the electric charge escaping per tie per area: ( e nuber escaping J = = 4πe(kT)2 A t h 3 exp W ) kt which is Richardson s equation, with work function W given by U E F. (5.53) (5.54) (5.55)

Therionic Eission 111 Table 5.1: G.E. FP-400 Specifications Filaent (W) length 3.17 c (1.25 ) Filaent diaeter 0.013 c (0.005 ) Anode (Zr coated Ni) I.D. 1.58 c (0.620 ) Maxiu filaent voltage 4.75 V Maxiu filaent current 2.5 A Maxiu anode voltage 125 V Maxiu anode current 55 A Maxiu anode dissipation 15 W Experient: Richardson s Constant is Material Dependent! Experientally it is found that Richardson s constant depends on the aterial 6 Why? 1. The work function depends on teperature (due to, for exaple, theral expansion of the lattice of atos). If the data analysis assues it s constant, the resulting A will be grossly in error. 2. Classically reflection requires a turning point (where v x = 0), whereas quantu echanical reflections are possible just due to sharp changes in potential. Quantu echanical reflection at the etal boundary was not included in our calculations; we assued every energetic electron headed toward the wall went through the wall. 3. Surface containation can affect eission probability. In fact, it was originally thought that therionic eission was 100% due to surface containation. (You can think of surface containation as a locally varying work function.) 4. Even in the absence of surface containation, in typical experients, the etal is polycrystalline and different crystal surfaces have different work functions. Experient This experient involves therionic eission fro the hot tungsten filaent of a G.E. FP-400 vacuu tube. Teperature Deterination Often teperature easureent in physics experients is easy. In the noral range of teperatures there are any types of transducers which convert teperature to an electrical quantity (e.g., Pt resistance theroeters, therocouples, theristors, ICs). However at the extrees of high and low teperature, easureent becoes tricky. Questions like 6 This should reind you a bit of the aterial dependent eissivity, ǫ T, for blackbody radiation to be discussed on page 113.

112 Therionic Eission 2 anode Keithley 2400 source eter FP 400 GND 4 1 cathode Keithley 2420 source eter Keithley 192 volteter Figure 5.6: Circuit diagra (note tube pin labels) for therionic eission experient. What exactly defines teperature? ust then be answered. This experient requires high teperatures in a vacuu, so we do not have direct contact with the aterial whose teperature we seek. In addition the FP-400 tube was not built by us, so we have liited direct inforation about the device. One coon way to easure teperature is by using the teperature dependence of resistance. The resistance of etals is approxiately proportional to teperature. Jones and Languir 7 have published a table of resistance vs. teperature for tungsten, fro which Kirkan has found an approxiating forula: where x is the ratio of the hot resistance to that at 293 K. T r = 112 + 202x 1.81x 2 (5.56) A proble with this approach is that the easured resistance, R easured, will include both the resistance of the tungsten 8 filaent, R W and the wires supporting it in the vacuu tube, R support. Thus the quantity we seek (tungsten filaent resistance, R W ) ust be calculated as the sall difference between two nubers: R W = R easured R support (5.57) a situation that produces big relative errors. Additionally, we have no independent way of easuring R support (we can t take the tube apart); In the end you will easure R support at roo teperature and then assue it is constant 9. There is a further proble with any easureent of voltage when parts of the syste are at different teperatures: therally induced efs (therocouples). If the ends of the tungsten filaent are at different teperatures, there will be a voltage induced approxiately proportional to the teperature difference between the two ends. This additional voltage source confuses the resistance deterination. The phenoena can be detected and corrected by reversing the direction of current flow (which reverses the Oh s law voltage, but does not affect the sign of the theral voltage.) Theral voltages are generally less than a V, and so are negligible once our easured voltages approach one volt. 7 GE Rev 30, 310 (1927) 8 The cheical sybol for tungsten is W fro the Geran Wolfra 9 See Assuing Away Variation page 18 in Chapter 0

Therionic Eission 113 Teperature Dependence of Resistance: W 3000 2500 Teperature (K) 2000 1500 1000 500 0 5 10 15 Resistance Ratio: R/R293 Figure 5.7: The teperature dependence of tungsten-filaent resistance fro the data of Jones & Languir with an approxiating curve. The x-axis is the ratio of the hot resistance to that at 293 K. Another approach is suggested by Jones & Languir. In a vacuu the teperature of a current-carrying wire is the result of an equilibriu between electrical power dissipated in the wire and energy lost in the for of radiation. (We assue that energy lost through conduction either through the wire-supports or residual air in the vacuu is negligible.) According to the Stefan-Boltzann law, the power radiated fro the surface of a hot black-body is given by: P = σt 4 A (5.58) where σ is the Stefan-Boltzann constant, T is the teperature of the body, and A is the surface area of the body. (In fact tungsten is not a black-body, so when applied to tungsten the above should be ultiplied by a fudge factor, the total eissivity ǫ T, about 0.3.) Using just crude approxiations, we equate the electrical power dissipated to the power radiated: I 2 T l d 2 I2 T l π 4 d2 I2 R = ǫ T σt 4 πdl T 4 dl (5.59) where d is the diaeter of the wire and l is the length of the wire. On the right hand side we ve assued that the resistivity of tungsten is proportional to teperature, and on the left hand side we ve assued ǫ T doesn t depend on teperature. We conclude: I 2 T 3 d 3 (5.60) [ I ]2 3 T (5.61) d 3 2 Absent the above approxiations we can hope that teperature is a function of a I/d 3 2. Once again Jones & Languir provide us with calibrating data for this expected relationship. For teperatures 400 K < T < 3000 K, Kirkan finds: T i = 117 + 56a 0.5 + 0.00036a 1.8 (5.62)

114 Therionic Eission 3000 Teperature vs. Current for W Filaents 2500 Teperature (K) 2000 1500 1000 500 0 500 1000 1500 2000 a=i/d^1.5 (A/c^1.5) Figure 5.8: The teperature of an in-vacuu tungsten filaent as a function of current fro the data of Jones & Languir with an approxiating curve. The x-axis is the current divided by the diaeter of the wire (in c) raised to the 3 2 power. Finally, attaining theral equilibriu 10 is a proble that affects ost any teperature easureent. The balance between electrical power in and heat lost is not iediately achieved. The parts of the syste with large heat capacities (particularly the filaent supports and other large structures in the tube), will only gradually approach equilibriu. Thus soak tie is required following each jup in heating power. The effect of this theral inertia is seen in hysteresis : teperatures obtained while increasing the teperature disagree with those found while decreasing the teperature. This will be an iportant source of uncertainty. Hands-on Electrical Measureents Support Resistance: R support As shown in Figure 5.6, the tungsten filaent (cathode) is powered by a Keithley 2420 source-eter. The filaent+support voltage is easured directly at the socket with a Keithley 192 volteter. By cobining the current through the filaent (easured fro the 2420) with the voltage across the socket (fro the 192), a resistance, R easured, can be deterined. At roo teperature, the filaent resistance, R W, can be calculated fro the filaent geoetry (see Table 5.1) and the resistivity of W-filaent aterial at roo teperature: ρ 293 = 5.49 µω c. (You should calculate: R W.1 Ω.) Then: R support = R easured R W (5.63) Because roo teperature R easured is sall (and hence error prone), you will ake 10 See Special Proble: Teperature page 17 in Chapter 0

Therionic Eission 115 three distinct easureents of it. Begin by sourcing 1 A and then 10 A into the roo teperature filaent (using the 2420), reading the resulting voltages on the 192, and calculating the corresponding R easured. Follow up those two easureents with a fourterinal resistance easureent just using the 192. (If R easured is substantially larger than.2 Ω, confir that you have fir, low-resistance contacts between the socket and the tube. Working the socket-tube contact ay be required to find the lowest resistance spot.) Maxiu Filaent Current, 2420 Voltage Copliance Liit Following your deterination of R support in a roo teperature tube, check out the conditions required to stay just below the tube s axiu ratings (4.75 V, 2.5 A). Using the 2420, successively source filaent currents of 2.0 A, 2.1 A, 2.2 A,... to directly deterine the axiu current you can use without exceeding the 4.75 V liit across the tube s filaent. Note that at just below axiu conditions, the 2420 will probably need to produce voltages above 4.75 V (because of the resistance of the external wires: the voltage drop across the connecting wires is not zero). Record the axiu 2420 voltage and tube current allowed by the tube s ratings; you will need these nubers in step #2 of your coputer progra. Data Collection Plan You will be collecting two types of data at the sae tie: theral characteristics of the filaent and the therionic properties of the tube (anode current/voltage relationship). Starting at a filaent current of 0.9 A, increase the current flowing through the filaent in steps of 0.1 A to a axiu current (found as described above, about 2.4 A) and then reverse those steps down to a filaent current 1.0 A. The up-sweep in filaent current followed by the down-sweep will allow you to test for hysteresis. At each step in current, allow the filaent to approach theral equilibriu (wait, say, 15 seconds) and then easure the voltage across and current through the cathode/anode. Calculate filaent teperature two ways (Equations (5.56) and (5.62)). Average the two to estiate the teperature, and use half the absolute value of the difference to estiate the uncertainty. You see above a classic exaple of systeatic error. The teperature is easured two different ways. Direct application of error propagation forulas to these teperatures calculated fro 6-digit eter values would suggest sall uncertainties. However the two teperatures in fact disagree. If only one ethod had been used to easure teperature, we would have badly underestiated the error. T 4 vs. Power: Testing Stefan-Boltzann By conservation of energy we know that the power duped into the filaent (ostly fro electrical heating, but also fro other sources like radiation fro the roo teperature environent to the filaent) should equal the power out of the filaent (fro black-body

116 Therionic Eission radiation and other factors like conduction down the supports). Thus: ǫ T σat 4 = I 2 R W + constant (5.64) T 4 1 = ǫ T σa I2 R W + constant (5.65) y = bx + a (5.66) A graph of T 4 vs. power should be a straight line fro which you will deterine ǫ T. (Note that the error in power is quite sall, so we have properly used it as an x variable.) In order to test this relationship you will want to ake a file containing the filaent power, T 4 (use the average of the two teperatures: (Ti 4 + Tr 4)/2), and the error in T 4 (use half the difference fro the two teperatures: Ti 4 Tr 4 /2). I A vs. V A : Testing Child and Richardson You will collect anode current vs. voltage curves for each different filaent (cathode) teperature. Use the Keithley 2400 to sweep the anode voltage logarithically fro 2 V to 120 V. (Note the axiu liits for the anode: 0.055 A or 125 V. Do not exceed either!) According to Child s law, the anode current, I A, should increase in proportion to V 3 2 A. Of course, at sufficiently high voltage the current will be liited by the axiu electron evaporation rate, and a current plateau fors at a level given by Richardson s law. At the axiu filaent current (corresponding to the axiu filaent teperature and evaporation rate), plateau foration occurs at very high voltage and you have the longest run of data following Child s law. Make a file containing V A, I A, and δi A which you can use to fit to the Child s law functional for: I A = k 1 (V A k 2 ) 3 2 (5.67) In addition, you will want to ake a big continuous file containing: V A, I A at every teperature tested. The current plateaus can be extracted fro this file and fit to the Richardson relationship: I A = k 1 AT 2 e k 2/T (5.68) Coputer Data Collection As part of this experient you will write a coputer progra to control the experient. Plagiaris Warning: like all lab work, this progra is to be your own work! Progras strikingly siilar to previous progras will alar the grader. I understand that this will often be a difficult and new experience. Please consult with e as you write the progra, and test the progra (with tube disconnected!) before attepting a final data-collecting run. Your progra will control all aspects of data collection. In particular it will: 0. Declare and define all variables. 1. Open (i.e., create integer nicknaes i.e., iunit for) the enets gpib2 and gpb1.

Therionic Eission 117 2. Initialize eters requires knowing the gpib priary address i.e., iadd of the eter and the iunit it is attached to. Get the status of each eter after you have initialized it. (a) Each source-eter ust be told the axiu voltage and current it ust produce during the experient. Initialize the 2400 (anode voltage/current) for the tube axiu ratings. (b) Initialize the 2420 (filaent voltage/current) for the near tube-axiu ratings found above 11. In the following I assue the current axiu is 2.4 A, but it ay be different for your tube. (c) Initialize the 192 for autorange DC voltage easureents. 3. Open the files: (a) filaent.dat (intended for: I f,v f,t r,t i of filaent). (b) stefanb.dat (intended for: power, T 4, δt 4 of filaent). (c) VI.dat (intended for: all V A, I A of anode, with coents (! ) for filaent I f,t r,t i ). (d) child.dat (intended for: V A, I A, δi A of anode at axiu filaent current). (e) child-.dat (like above but intended for a downsweep of anode voltage). (f) rich.dat (intended for: T i, T r, I A i.e., the estiated teperatures and the corresponding axiu anode current for Richardson s Law). 4. Tell the 2420 source-eter to source a filaent current of 0.9 A. 5. Let the syste sleep for 60 seconds to approach theral equilibriu. 6. Do a sequence of easureents where the filaent teperature is sequentially increased (i.e., a teperature up-sweep) with filaent currents ranging fro 0.9 A to soe axiu (e.g., 2.4 A) current in steps of 0.1 A. (a) Tell the 2420 source-eter to source a filaent current (I f ). (b) Let the syste sleep for 15 seconds to approach theral equilibriu. (c) Request a logarithic sweep of the anode voltage (controlled by the 2400 sourceeter) fro 2 V to 120 V. Receive the resulting arrays: V A and I A. (d) Turn off the anode voltage. (e) Repeat (a) thus receiving an updated version of the filaent current (it will be very close to the requested current). (f) Read the 192 to get the filaent voltage (V f ). (g) Using Eqs. (5.56) and (5.57), calculate T r based on the calculated tube resistance R easured, the calculated roo teperature filaent resistance and R support. (h) Calculate T i fro Eq. (5.62) (i) Write a line to the file filaent.dat reporting: I f,v f,t r,t i. (j) Write a line to the file stefanb.dat reporting: filaent power (R W I 2 f ), T 4, and δt 4 (see p. 116). 11 See Hands-on Electrical Measureents, p. 114. Recall: the 2420 axiu voltage will need to be a bit above 4.75 V. If you have not yet copleted those easureents, teporarily initialize with 4.75 V.

118 Therionic Eission (k) Write a coent line (i.e., starts with! ) to the file VI.dat reporting filaent data: I f,t r,t i. (l) Write the anode sweep data (one data-pair per line) in the file VI.dat. () Write a line to the filerich.dat reporting T i, T r, I A. Use I A at V A =120 V as the estiated plateau current. (When the experient is over, you will need estiate δi A based on hysteresis, and ay need to delete I A values if, for exaple, the current did not plateau or if cold eission substantially added to the plateau current.) (n) Increent the filaent current by 0.1 A and return to (a). 7. Collect data for Child s Law. Begin by repeating a noral anode sweep at the axiu filaent current; follow all the steps (a) () outlined in 6 above. In addition, write the anode sweep data (V A, I A, δi A ) in the file child.dat (one data-triplet per line). In this case, δi A will be calculated fro the anufacturer s specs: percent+digits; the fortran function eak2400 can do this error calculation autoatically. Now check for hysteresis by doing a reverse anode sweep: fro 120 V down to 2 V. Write this reverse anode sweep data (V A, I A, δi A ) in the file child-.dat. 8. Do a sequence of easureents where the filaent teperature is decreased (i.e., a teperature down-sweep) by sequentially sourcing filaent currents fro one step down fro axiu (e.g., 2.3 A) back to 1.0 A. Follow steps (a) () used in the teperature up-sweep (part 6 above) and then: (n) decreent the filaent current by 0.1 A and return to (a). 9. Turn off the output of the 2420. 10. Close all files. Note that the 0.9 A filaent current data is just trash collected so the 1.0 A filaent current data is taken on a pre-wared filaent. Observations While the coputer collects the data observe the light fro the filaent. (There is a 1 16 diaeter hole in the anode allowing light fro the id-point of the filaent to escape.) Initially the filaent will not appear to be incandescent (i.e., not a source of light at all: dark) so it ay help to turn off the lab lights to better observe the beginning of visible incandescence. According to the Stefan-Boltzann law the light energy radiated depends on T 4, so sall changes in T produce big changes in light intensity. In addition to changes in light intensity, you should look for the ore subtle change in light color: fro a dull red to a brilliant yellow. Record your observations! At what filaent teperature did you first see the filaent producing light?

Therionic Eission 119 FP 400 Vacuu Tube.01.001 Ia (A) 1.E 04 1.E 05 1.E 06 1.E 07 1 10 100 Va (V) Figure 5.9: The teperature dependence of therionic eission in a FP-400 vacuu tube. Each curve shows the anode current-voltage relationship at a particular filaent teperature. At high filaent teperatures and low anode voltages the anode current follows Child s law: an upward sloping straight line on this log-log plot. At sufficiently high anode voltage, the filaent current plateaus at a value given by Richardson s law. At low filaent teperatures and high anode voltages, cold eission ay augent (distort) the plateau.

120 Therionic Eission Data Analysis Beginnings The ain result of this experient is a plot siilar to Figure 5.9 showing the anode currentvoltage relationship at various filaent teperatures. Production of such a ulti-plot is a bit coplex, and you will alost certainly need further detailed instructions 12 fro your instructor on using the progra Nplot. Each curve in the ulti-plot represents a anode voltage sweep, which for sall anode voltage V A follows the nearly power-law relationship discovered by Child (power laws look linear in the log-log plot) but then levels out ( plateaus ) at a current given by Richardson s Law. The curves are paired: each filaent current was done once when the teperature was being increase and once when the teperature was being decreased. The first check you should ake is: is a proper plateau achieved for every filaent current? If the high-teperature V I sweep reaches a plateau, then Child s Law will not apply at high V A so child.dat will require editing; if it does not reach a plateau, then Richardson s Law does not apply to that sweep so rich.dat will require editing. The anode current, I A at the axiu anode voltage V A = 120 V has been stored in the file rich.dat as a candidate plateau along with teperature estiates T r and T i (all currently lacking error estiates). While the actual Richardson s Law relationship (Eq. 5.69) is not on the list of WAPP + functions, we can arrange a quick but useful approxiation. According to Richardson s law, these plateau currents should satisfy: I A = AAT 2 e W/kT = k 1 AT 2 e k 2/T (5.69) where A is the tungsten filaent surface area. Clearly the largest (if unknown) error is in T, so the standard approach would be to put T on the y-axis and I on the x-axis: the opposite of what is iplied by the above equation. However, we can t siply solve the above equation for T without soe approxiations. It turns out that e k 2/T is the significant factor in the above equation, so we start by ignoring the T 2 and assue: I = K e k 2/T (5.70) for soe constant K. Now if we take log e of both sides: or log(i) = log(k) k 2 1 T (5.71) 1 T = log(k)/k 2 1 k 2 log(i) (5.72) This equation is now in a for 13 known to WAPP +. Thus you can quickly WAPP + (I,T) data fro rich.dat (of course still lacking x-errors and y-errors) and find an approxiating 12 Docuentation on the plot progra can be found in Appendix A. It is also worth noting that since a log-log plot is requested, negative values coonly negative anode currents ust be reoved fro the file VI.dat. As a general rule: do not destroy your original data, in this case siply renae the edited file soething like VI2.dat. 13 Inverse-Natural Log: 1/y = A + B log(x)

Therionic Eission 121 1000 Work Function Systeatic Error Teperature (K) 1500 2000 2500 1.E 08 1.E 06 1.E 04.01 Anode Current (A) Figure 5.10: Siplified analysis suggests Richardson s Law data can be fit to the Inverse- Natural Log relationship of Eq. 5.72. The filled-square points use y = T i, the unfilled-square points use y = T r. Systeatically different teperature easureents yield systeatically different B (slopes in the above plot) and hence systeatically different work functions W = k/b. Note the inexact pairing of the data due to teperature hysteresis. Note that the rightost datapoint is aberrant.: a non-plateau. curve. Retain hardcopy 14 of the fit results. Request (and hardcopy) a linearized plot of your data with x-scale: log and y-scale: inverse as in Fig. 5.10. This should allow you to check for aberrant data points. (Usually the high teperature curve has not plateaued, and so the high teperature data point ust be reoved. Occasionally the low teperature data points are also aberrant.) But what should be used for T?...You have two values (T i and T r ) and they are not the sae! The answer is both: fit/plot once with T i and then again with T r. The systeatically different teperature scales will result in different fit paraeters in particular you will get an upper-liit and lower-liit for the work function fro the two B values of your two fits, as W/k = k 2 and k 2 1/B. δw can then be taken as half the difference between these two extree values. (Note an oddity: you ve deterined a value for δw, but not yet found the best value for W.) Take this opportunity to delete the aberrant data fro rich.dat (and, say, save as rich2.dat). We can also a quick but useful check of the Child s law data using WAPP +. Once again the proper functional for (Eq 5.76) is not available in WAPP +, but a power law is soewhat siilar. In this case the files child.dat and child-.dat include a y-error estiate based on the k2400 specifications. (In fact these book-based errors very uch underestiate the deviations you will experience: expect a horrendous reduced χ 2 ). Check that WAPP + s B is nearly 1.5, record the A value (you will need it as an initial estiate for k 1 when you do the fit to the proper functional for), ake a log-log plot (no hardcopy yet!) to check for 14 You should retain a hard and/or soft copy of every successful fit. Hardcopy ends up taped into your lab notebook. I retain softcopy by copy&paste into a generic text file, e.g., using kwrite.

122 Therionic Eission Stefan Boltzann Law 4.E+13 3.E+13 T^4 (K^4) 2.E+13 1.E+13 0 2 4 6 8 10 Power (W) Figure 5.11: A test of the Stefan-Boltzann law: that power radiated is proportional to T 4. Note that the fit line hits well within each error bar. The χ 2 for this fit will be sall. Evidently the average teperature is a better easure of teperature than you ight expect fro the deviations between T i and T r. aberrations (like a plateau at large V A ). Finally, the Stefan-Boltzann Law can be properly evaluated using WAPP +, as a line is the expected function. The file stefanb.dat has the required data with y-errors (x-errors are sall). Produce a plot (and hardcopy fit report) siilar to Fig. 5.11. I expect you ll find a sall reduced χ 2 due to the large systeatic error in teperature, so a bit of additional work will be required to estiate δǫ T. Stefan-Boltzann Law II You should have already checked for an approxiately linear relationship between electrical power in and T 4 : Power = ǫ T σat 4 (5.73) and found a reduced χ 2 indicative of large systeatic uncertainty in teperature. We now seek an estiate for ǫ T (with error) at the highest teperature. The best value for ǫ T can be found by plugging in our best estiates for T 4 and the easured power (found in the file stefanb.dat), and A (calculated based on the diensions recorded in Table 5.1). Alternatively ǫ T could be calculated based on slope as suggested by Eq. 5.66. But how should we incorporate the large systeatic errors in T 4 and the unknown systeatic error in A? For the surface area A, all we know is the book values for the diensions of the filaent. Based on the sigfigs in the reported values, I estiate: δa A 10% (5.74)

Therionic Eission 123 Child s Law.01 Ia (A).001.0001 1 10 100 Va (V) Figure 5.12: A plot of the space-charge liited therionic eission in a FP-400 vacuu tube (Child s law). The data was taken at a filaent current of 2.4 A. Every other data point has been eliinated so the fit line is not obscured. Note that the fit line systeatically isses the data, soeties a bit high others a bit low. The easureent errors are tiny, so these sall isses do result in a too-large χ 2. Nevertheless, the law provides an excellent suary of the data over a huge range of variation. (ostly due to the filaent diaeter, where a sall uncertainty leads to a large fractional uncertainty). We can then use the high-low ethod or use the proper forula in Appendix E to estiate the range of possible values for ǫ T, given the range of possible values for T 4 and A (assue the error in power is negligible). Child s Law II At sufficiently high filaent teperature (and low anode voltage), Child s law governs the anode current-voltage relationship: I A = 8πǫ 2e 0l 9bβ 2 V 3 2 A (5.75) (see vicinity of Eq. 5.40, page 107, for a definition of sybols) At the highest filaent teperature (e.g., filaent current of 2.4 A) you have saved the (V A,I A,δI A ) data in the file child.dat. Now fit 15 this data to the functional for: I A = k 1 (V A k 2 ) 3 2 (5.76) (where k 2 represents a sall offset between ground and the actual average voltage of the filaent). The progra fit will required an initial guess for k 1 ; use the WAPP + value for A 15 The progra fit is docuented in Appendix B. Note that plotting and fitting now are handled by two different progras, not a cobined progra like WAPP +

124 Therionic Eission found above. (In fit the default initial guess for paraeters is zero; that is reasonable for k 2.) Do not be surprised if you get a huge reduced χ 2. Find an estiate for k 1 error either by a fudged fit or a bootstrap. As always retain a copy of your full fit report, including the covariance atrix. Plot your data and best-fit function using the progra plot, producing a result siilar to Fig. 5.12. Follow exactly the sae process for the downsweep data in the file child-.dat. Often you will find that the k 1 values for the two sweeps differ by ore than coputer-based value of δk 1. Systeatic error (here a type of hysteresis) is responsible. Note that the usual reduced χ 2 alerts us to a proble, but easuring twice (in different ways) provides an estiate (perhaps still an under-estiate) for δk 1 : half the difference between the two values of k 1. We expect that k 1 = 8πǫ 2e 0l 9bβ 2 (5.77) so using the tube diensions in Table 5.1, the electron charge-ass ratio e/ can be calculated. Since the FP-400 is a finite length cylinder (rather than the infinite cylinder discussed in our theory) use the effective length 16 = 0.7 l as the length of the cylinder. But what can we use as errors for the book values for b and l? Theral expansion of the filaent should, by itself, change l by about 1% (notice the spring-tensioned support for the filaent in the FP-400). Based on the sigfigs in the reported values, I estiate: δ(b/l) (b/l) 3% (5.78) Calculate e/ and its error. Copare to the known value (citation required!). Richardson-Dushan Law II You should have already checked for an approxiately exponential-inverse relationship between T and I A, edited out aberrant data (e.g., not yet plateaued), and have a range of possible values (due to systeatic error in teperature) for k 2 in the expected relationship: I A = k 1 AT 2 e k 2/T (5.79) We now seek a treatent incorporating the hysteresis error in I A and the proper functional for, to obtain the best possible value for k 2. We will need to anipulate 17 the data in rich2.dat to bring together equivalent data fro the teperature upsweep and downsweep. An easy way to get the data into the gnueric spreadsheet, is to type it to the screen using the linux coand 18 cat: cat rich2.dat 16 This effective length corrects for current loss through the ends of the cylinder under space-charge situations. A saller correction should be applied when large anode voltages are designed to sweep up all evaporated electrons, i.e., for Richardson s Law, where 90% electron collection sees appropriate. The details of handling such corrections to theory ay be found in reference 1. 17 This processing could very easily have been done within the progra itself. I ve instead opted to ake the progra as siple as possible at a cost of additional by-hand processing in a spreadsheet. 18 fro concatenate coonly this coand is used to cobine several files

Therionic Eission 125 Richardson s Law.01 Anode Current (A).001 1.E 04 1.E 05 2600 2400 2200 2000 1800 Teperature (K) 1600 Figure 5.13: A Richardson Plot of therionic eission in a FP-400 vacuu tube. Each data point displays the plateau anode current at a particular filaent teperature. The curve through the data fits for the work function; the slightly steeper curve uses the book value for the work function. You can then copy and paste 19 this data into gnueric. Our ai to put together data with the sae filaent current, i.e., to cobine the first line of data with the last; to cobine the second line of data with the next-to-last; etc. This is easily accoplished by typing the data botto-to-top to the screen using the linux coand 20 tac: tac rich2.dat The results can be copy and pasted next to the previous data so that the data we ai to cobine (e.g., both I f = 1 A) is on the sae line. The best estiate for T is the average of the four Ts (the two T i should be nearly identical); the best estiate for I A is the average of the two I A s; for δi A use half the difference between the two values of I A ( I A1 I A2 /2); for δt use half the difference between T r and T i (average the two differences). The result of this data reduction is half as any data points, but with a value for δi A based on hysteresis. (The alternative for δi A would be the eter anufacturer s specifications, and we know they are way to sall fro analysis of Child s Law.) Report the relative iportance of hysteresis and calibration in teperature uncertainty deterination. For I f = 1.2 A record the difference between the two T r due to hysteresis and the difference between T r and T i which is a teperature calibration uncertainty. Copy and paste this reduced data into a new file, and fit it to Eq. 5.79. (As always retain a copy of the full fit report.) You should have an estiate for k 2 fro WAPP + ; use the book 21 value for A, 0.72 10 6 A/ 2 K 2, as an initial guess for k 1. Produce a plot siilar to Fig. 5.13; include lines both for your best fit and the book values of k 1 and k 2. 19 Note use of See two two separators as one : Alt-e 20 clever or what? 21 Blakeore, Solid State Physics

126 Therionic Eission The work function is an atoic quantity, and it is usually expressed in the atoic scale unit ev 22. Calculate the work function fro your value of k 2 in joules and ev and copare to the book 20 value of 4.5 ev. Since our teperatures are so uncertain, particularly at the low end, the best estiate for the Richardson constant A coes fro substituting the book value of the work function and the plateau (T,I A ) easureent for the highest valid filaent teperature into Eq. 5.69. We can then estiate the systeatic uncertainty in A by using the high-low ethod of 191 (A + I +,T, A ). Report Checklist 1. Write an introductory paragraph describing the basic physics behind this experient. For exaple, why did higher cathode currents produce ever higher plateaus of anode current? (This anual has any pages on these topics; your job is condense this into a few sentences and no equations.) 2. Calculations (no errors) of roo teperature R W. Measureents (4-wire oheter and direct voltage/current) of R easured at roo teperature. Calculation of R support. 3. Observations of the light intensity and color as a function of filaent teperature. 4. Data files and coputer progra: Leave the in your linux account. Print out a copy of your progra, the file filaent.dat, and the data you used to fit Richardson s Law; Tape the into your lab notebook. Docuent how your Richardson s Law data was calculated in your spreadsheet. (I would self-docuent the spreadsheet, then File Print Area Set Print Area to select just the relevant 4 coluns, including the self-docuentation of those coluns, and then print.) 5. Plots siilar to Figures 5.9, 5.13 (with fit curve Eq. 5.68 and also the Richardson function using book values for k 1 and k 2 ), 5.12 (with fit curve Eq. 5.67), 5.11 (with line Eq. 5.66) and 5.10 (two separate WAPP + plots one with T r data and the other with T i data). Note that Figure 5.9 is coplex to produce. Use a file of Nplot coands and feel free to talk to e about how to go about doing this. Carefully note the use of log and inverse scales (which requires positive data edit to achieve this!). Include a fit report for each fit curve. 6. Experiental values (with error range) for: W (in ev), A, e/ and ǫ T. 7. Show the issing steps leading to Equations (5.31) and (5.36). Substitute the ρ 1 approxiation (Eq. 5.35) into the differential equation Equation (5.31). Show that while we do not have an exact solution to the differential equation, the singular parts (i.e., those that approach infinity as ρ 1) cancel. 8. Make a final results table, recording your nuerical results fro #6 (with proper units and sigfigs) adjacent to the corresponding book values. Don t forget to record your tube s identifying letter! 22 1 ev = 1.6022 10 19 J is the energy an electron gains in going through a potential difference of 1 V.

Therionic Eission 127 Coent: Classical vs. Quantu I said above that the presence of an in Richardson s constant shows that the process is governed by quantu echanics. It is not quite so siple. The evaporation of water is not fundaentally different fro that of electrons, while the forer (due to the larger ass of a water olecule) is well-approxiated by a classical calculation. Classical evaporation can be calculated uch as quantu evaporation using the Maxwell-Boltzann speed distribution and the nuber density (of water olecules) rather than the disguised version of this: Feri energy (E 3 2 F nuber density). We can write the classical rate in ters of the quantu with no visible: classical flux = 4 [ ]3 3 EF 2 quantu flux (5.80) π kt The different teperature dependence 23 for the classical flux (T 1 2e W/kT vs. T 2 e W/kT ) cannot be detected experientally: the Boltzann factor overwhels all other teperature dependencies. On the other hand, since E F kt, the expected classical rate is uch larger than the quantu rate. This played a role in the istaken idea that therionic eission was due to surface containation: the experiental rate seeed too sall to be theral evaporation. On the other hand a ore fruitful interpretation of the low rate was that only a fraction ( kt/e F ) of the electrons were therally active. This connected with other observations (like the sall specific heat of the electron gas) and provided a link to the idea of a degenerate Feri gas. Coent: Uncertainty Inspection of Figures 5.9-5.13 shows that soething funny is going on. (I can t say abnoral or unusual as it is neither.) Figure 5.11 shows the unistakable signs of sall reduced χ 2 : The fitted line goes nearly dead-center through all 30 error bars, never coing even close to an edge. For error bars sized at one standard deviation (σ), you should expect total isses of the error bar about 1/3 of the tie. In addition recall that each data point is really a double: the sae filaent current was sourced as part of a teperature up-sweep and as part of a teperature down-sweep. These repeated easureent should also frequently iss by a standard deviation, but here they are so close that the two points often look like just one. The answer to this puzzle is that the error bars are not displaying statistical ( rando ) error. Instead the teperature was easured two different ways (T i and T r ), and the error bar represented the deviation between these two easureent ethods. When different ethods of easureent produce different answers for the sae quantity, we have a textbook exaple of systeatic error (in contrast to statistical error). Notice that if we had used the statistical deviation of just one easure of teperature, we would seriously underestiated the error. Furtherore since quite accurately easured electrical quantities were used to calculate the teperature (via Equation 5.62 or Equation 5.56), application of the usual error propagation ethods would also have produced very sall errors in T. The full range of our uncertainty in teperature is ade evident only by easuring T two different ways. (This is typically how systeatic errors are detected.) 23 Saul Dushan ( Phys. Rev. 21, 623 636 (1923)), while working at G.E., provided a general therodynaic arguent for the T 2 dependence and the universal nature of A. The resulting relationship is therefore soeties known as the Richardson-Dushan relationship.

128 Therionic Eission Having detected systeatic errors, we should seek an explanation... In addition to the probles previously cited (particularly use of book values for filaent diensions and the probles associated with R support ), nonunifor filaent teperature due to filaent supports ay be the proble. Koller (p. 89) reports the filaent teperature 0.5 c fro a support is reduced by 15% (and of course the effect is greater closer to the support). Therionic eission is reduced a siilar aount 1.3 c fro a support. Thus the quantity we are seeking (a filaent teperature) does not even exist, so it is hardly surprising that different easureents give different results. (It turns out the T r is soething like the average teperature; whereas T i is soething like the central teperature.) These effects have been investigated, and Koller gives the theory to correct such easureents, but such considerations are beyond the scope of this experient. Figure 5.9 shows the opposite proble, large reduced χ 2 : The fitted line systeatically isses alost every error bar. In this case, the iss ight be called sall, but the error bar is saller still. Once again, errors were not calculated statistically (anufacturer s specifications were used), so reduced χ 2 = 1 should not really be expected. In this case, y guess is that the proble is with our siplified theory (we assued: infinite cylinders, no rando coponent to the electron velocity [zero electron teperature], unifor filaent [teperature, voltage, eissivity,...], perfect vacuu, no incipient current plateau). We could of course test these hypotheses by further experients with different tubes, but such work is beyond the scope of this experient. (Indeed you ost likely have detected V A - sweep hysteresis; correction for this draatically reduces reduced χ 2, but not all the way to 1.) Suary: Very large or very sall reduced χ 2 suggests significant non-statistical errors, a very coon perhaps even the usual situation. Coputer generated errors are soe sort of none sense in this circustance. Presuably your theory and/or easureent ethods are less than perfect. That is, of course, No Surprise. Science next requires you to guess and perhaps further investigate which iperfections are the leading source of the probles, i.e., what changes to the experient would aeliorate the proble. References 1. Jones & Languir GE Review, The Characteristics of Tungsten Filaents as Functions of Teperature 30 (1927) Part I pp. 310 19, Part II pp. 354 61, Part III pp. 408 12 2. Melissinos Experients in Modern Physics, 1966, pp. 65 80 3. Preston & Dietz The Art of Experiental Physics, 1991, pp. 141 47, 152 61 4. Blakeore Solid State Physics, 1974, pp. 188 95 5. Koller The Physics of Electron Tubes, 1937, Ch. I, VI, VIII 6. G.E. FP-400 Description and Rating FP400.pdf found at http://www.tubedata.org

Therionic Eission 129 thickness: x E(x) x Area: A E(x+ x) x Figure 5.14: Gauss Law is used to calculate the charge between two plates of area A separated by a distance x. Since (by assuption) the potential just depends on x, the electric field is in the x direction and is given by E = dv/dx. Appendix Poisson s Equation Equation 5.5 and Equation 5.28 ade reference to Poisson s Equation, which is really a topic for Physics 341, rather than part of Physics 200. In this appendix, Poisson s Equation is derived starting fro two Physics 200 results: Gauss Law: that the electric flux leaving a region depends just on the charge enclosed by that region: E ˆn da = Q enclosed /ǫ 0 (5.81) and the relationship between electric field and electric potential (voltage): E x = dv dx (5.82) Poisson s Equation is a differential equation equivalent to Gauss Law. It is usually written in ters of the Laplacian ( 2 ), which in turn can ost easily be expressed in ters of second derivatives w.r.t. x, y, and z: 2 V = 2 V x 2 + 2 V y 2 + 2 V z 2 = ρ/ǫ 0 (5.83) where ρ is the electric charge density. We need Poisson s Equation only in cases where the electric potential depends on just one variable (x or cylindrical r), which siplifies the required proofs considerably. As shown in Figure 5.14, if V is a function of x alone: V (x), we can find the charge between two plates of area A using Gauss Law: Q enclosed = ǫ 0 A (E(x + x) E(x)) ǫ 0 A de dx Thus the charge density between the plates is given by: x (5.84) dx x ρ = Q enclosed volue = ǫ 0A de A x = ǫ 0 de dx = ǫ 0 d 2 V dx 2 (5.85)

130 Therionic Eission thickness: r E(r+ r) l E(r) r+ r Figure 5.15: Gauss Law is used to calculate the charge between two coaxial cylinders of length l separated by a distance r. Since (by assuption) the potential just depends on r, the electric field is in the r direction and is given by E = dv/dr. which provides what is needed for Equation 5.5. As shown in Figure 5.15, if V is a function of r alone: V (r), we can find the charge between two coaxial cylinders using Gauss Law: Q enclosed = ǫ 0 l {2π(r + r)e(r + r) 2πrE(r)} = ǫ 0 l {2πr(E(r + r) E(r)) + 2π re(r + r)} { ǫ 0 l 2πr de } dr + 2πE(r) r (5.86) Thus the charge density between the cylinders is given by: ρ = Q enclosed volue = ǫ 0l {2πr de/dr + 2πE(r)} r 2πrl r { d 2 V = ǫ 0 dr 2 + 1 } dv r dr which provides what is needed for Equation 5.28. { de = ǫ 0 dr + 1 } r E (5.87) (5.88)