Thermionic Emission. Introduction. Child s Law

Transcription

1 Therionic Eission Introduction Metals, as deonstrated by their ability to conduct an electric current, contain obile electrons which are not strongly bound to individual atos. (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 = 4πɛ 0 (2x) 2 (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 just 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 is called Child s law (or the Child-Languir law, including Languir who independently discovered it while working at G.E.). In this experient you will easure both of these phenoena. Child s Law Consider a planar interface between a etal (x < 0) and vacuu (x > 0). Electrons for a cloud of escaped electrons surrounding the hot etal filaent (called the cathode, which we take to be at zero potential). Thus the vacuu adjacent to the etal is not epty, rather it is filled with electrons (with nuber density n(x) i.e., a space charge). A positive potential,v A, on the anode plane located at x = b provides a force pulling these electrons fro the vicinity of the cathode towards the anode. The electrons oving towards the right constitute a steady electric current density towards the left, i.e., a steady electric current fro the anode to the cathode: J = en(x)v(x) = J A (2) where v(x) is the electron velocity. 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 1

2 V = 0 electric current density J A V A cathode vacuu etal x = 0 anode x = b accelerating electrons x Figure 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. conservation of energy: 1 2 v2 ev (x) = 0 () v = V (x) (4) 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: 2 V x 2 = ρ(x) ɛ 0 Putting these pieces together with have the differential equation: d 2 V dx 2 = J A ɛ 0 v(x) = = en(x) ɛ 0 (5) J A ɛ 0 V (x) (6) Since the electric field will be zero at the interface, we have a pair of initial conditions: V x x=0 = 0 (7) V x=0 = 0 (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)) (9) t x x(t) V (x) 1 F (x(t)) J A ɛ 2e 0 V (x) 2

3 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 ɛ 0 J A ɛ 0 J A ɛ 0 V 1 2 dv dx ( V 1 2 V (10) ) (11) + constant (12) The initial conditions require the constant to be zero, so or This differential equation is separable: [ ] 1 dv 2 = J A 2 dx ɛ 2e 0 dv dx = 4J A ɛ 2e 0 dv V 1 4 V 4 4 = = V (1) V 1 4 (14) 4J A dx (15) ɛ 2e 0 4J A x (16) ɛ 2e 0 where again the initial conditions require the constant of integration to be zero. Finally: V (x) = 9J A 4ɛ 2e 0 2 x 4 (17) Of course, V (b) is the anode voltage V A, so we can rearrange this equation to show Child s law: [ ] 4ɛ0 2e J A = 9b 2 V 2 A (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 as shown: b : L (19) V A : E Q = ML2 /T 2 Q (20)

4 cathode: hot filaent, radius a anode 2 anode l cathode 1 4 b Figure 2: Coaxial cylinders: an inner cathode (radius a) and outer anode (radius b), for a vacuu tube diode. A potential difference V A attracts electrons fro the cathode to the anode, so 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. ɛ 0 k : Q 2 Q 1 2 Q 5 2 = (21) EL M 1 2 M 2 L /T 2 Q/T J A : L 2 (22) To ake a diensionally correct forula for J A, we just eliinate the M diension which we can only do with the cobination: We can then get the right units for J A with: ( ) V A k 2 2 V A k 2 : Q 2 Thus the only possible diensionally correct forula is T 2 (2) b 2 = k b 2 V 2 A : Q/T L 2 (24) J A k b 2 V 2 A (25) 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 cylinder (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 2 A (26) 4

5 although we could have an arbitrary function of the radius ratio: b/a on the right-hand-side. Fro Poisson s equation we have: 2 V = J ɛ 0 v(r) = I 2πrlɛ 0 v(r) = Using the Laplacian in cylindrical coordinates we find: 2 V r r V r = j 2πrɛ 0 j 2πrɛ 0 V 1 2 (27) V 1 2 (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 quantities: yielding: with initial conditions: 2 f ρ ρ r/a = ρ (29) 2 ja V = f(ρ) (0) 2πɛ 2e 0 f ρ = f (ρ) + 1 ρ f (ρ) = 1 ρ f 1 2 (1) f(1) = 0 (2) f (1) = 0 () 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 at 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) = = 2 9J A x 4 9I = 4ɛ 2e 0 2 [ ] 2 ( 9 ja r a 4 2πɛ 2e a 0 2πal4ɛ 0 ) 4 So, near the cathode (i.e., ρ slightly larger than 1): f(ρ) 2 (r a) 4 = 9ja 2π4ɛ 0 2 ( r a a ) 4 (4) [ 9 4]2 (ρ 1) 4 (5) We can use this approxiation to start our nuerical differential equation solution at a non-singular point (like ρ = ). 5

6 10 f(ρ) for ρ 1 f f(ρ) for ρ f(ρ) exact ρ ρ Figure : 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 the nuerical solution. Real devices are designed with b/a 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 (6) A useful approxiation for the range: 100 < b/a < 1000 is: ( ) 2 9 f = 4 ρ + 2 (7) (For exaple, the device used in lab has b/a = For this value, the differential equation gives f = 44.16; the above approxiation gives: f = ) We recover Child s law by rearranging (0): 2πɛ 0 a β 2 (ρ) 4 9 [ ] VA 2 = j = I/l (8) f(b/a) Note: Languir s original work (Phys. Rev. 22, 47 (192)) on this subject is expressed in ters of β where: ρ 1 (ρ 1) 2 So: β 2 = for the device used in lab. f 2 ρ = 8πɛ 2e 0 9bβ 2 V 2 A ρ 1 (9) = j = I/l (40) Richardson s Law Most any theral process is governed by the Boltzann factor: e E kt (41) 6

7 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 (42) 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. That is to say that therionic eission probes the quantu state of the aterial 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 (4) where A = 4πek2 h = A/ 2 K 2 (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 (45) Recall that vector hk is the oentu, p = v, of the electron and h = 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 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) = h2 k 2 2 = (2π h)2 n 2 2L 2 = (2π h)2 (n 2 x + n 2 y + n 2 z) 2L 2 (47) Notice that there is a definite relationship between the velocity vector of the electron and the n-vector. v = 2π h L n (48) Another way of saying the sae thing is that allowed quantu-state velocities for a cubic lattice with cube-side 2π h/l. The nuber of states with electron velocities in soe specified region (for exaple a parallelepiped with sides: v x v y v z ) can be found fro 7 (46)

8 unoccupied levels vacuu etal U W vacuu iage charge potential occupied levels 0 E F zero force (flat) potential inside etal Figure 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. the nuber of 2π h/l sided cubes that fit into the region, which is the volue of that velocity-space region divided by (2π h/l). Hence: nuber of states with velocity between v and v + v = v x v y v z (2π h/l) (49) nuber of states per volue with velocity between v and v + v = v x v y v z (2π h/) ( ) = v x v y v z = N v x v y v z (50) 2π h 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 state with energy E will be occupied is given by Feri-Dirac statistics: f(e) = exp ( E EF kt ) (51) 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 so ( f(e) exp E E ) F (52) kt 8

9 v x t A Figure 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 electrons in the shaded 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. i.e., 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 to calculate the nuber of electrons escaping through area A during t we ust dv x dv y dv z 2N f(e) Av x t 2U/ = 2N e 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 = 4π(kT )2 (2π h) A t exp 2U/ ( (U E F ) kt The electric current density is the electric charge escaping per tie per area: J = e nuber escaping A t = ) 4πe(kT )2 h ( exp W ) kt which is Richardson s equation, with work function W given by U E F. (5) (54) Experient: Richardson s Constant is Material Dependent Experientally it is found that Richardson s constant depends on the aterial. Why? 9

10 etal E iage charge potential vacuu 0 electric field potential reduction in energy required to escape zero force (flat) potential inside etal total potential Figure 6: When an electric field E is present, the total potential outside of the etal will be the su of the iage-charge potential and the electric field potential ( eex). The result is a reduction in the energy needed to escape the etal. If the classically disallowed barrier is sufficiently thin, quantu echanical tunneling through the barrier can be easured. 1. The work function depends on teperature (due to, for exaple, theral expansion). 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.. 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. 5. Real experients ay only approxiate the idealized theory. For exaple, since the hot cathode needs to be supported and heated, it is generally not at a unifor teperature. Schottky Eission For sall anode voltages, the space charge partially shields the cathode, and the current is specified by Child s law. At higher voltages, the anode collects every electron eitted, and the current is specified by Richardson s law. At higher fields yet, the strong electric field reduces the barrier to eission effectively reducing the work function. This is Schottky eission. At very strong fields the electrons can quantu echanically tunnel through the barrier even though they lack the energy required classically to overcoe the iage-charge 10

11 Table 1: G.E. FP-400 Specifications Filaent (W) length Filaent diaeter Anode (Zr coated Ni) I.D. Maxiu filaent voltage Maxiu filaent current Maxiu anode voltage Maxiu anode current Maxiu anode dissipation.17 c 0.01 c 1.58 c 4.75 V 2.5 A 125 V 55 A 15 W force. We seek to estiate the extent to which Schottky eission affects our deterination of the work function. In the presence of an electric field, the potential near the cathode is given by: U(x) = e2 eex (55) 4πɛ 0 4x where E is the electric field. We seek the aount the electric field reduces the barrier-top potential. The axiu potential is found by finding where du/dx is zero: du(x) dx = The potential at this axiu is: e 2 4πɛ 0 4x 2 ee = 0 x = e/16πɛ 0 E (56) e U(x ) = E (57) 4πɛ 0 For ost vacuu tubes this shift in work function is sall. However, icroscopic peaks on the surface of the etal will enhance the electric field near the peak beyond that calculated for a sooth surface resulting in locally enhanced Schottky eission. 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 teperature 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 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 11

12 2 anode Keithley 2400 source eter FP 400 GND 1 4 cathode Keithley 2420 source eter Keithley 192 volteter Figure 7: Circuit diagra for therionic eission experient. 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 1 have published a table of resistance vs. teperature for tungsten filaents, fro which Kirkan has found an approxiating forula: where x is the ratio of the hot resistance to that at 29 K. T r = x 1.81x 2 (58) The proble with this approach is that the easured resistance, R easured, will include both the resistance of the filaent and the wires supporting it in the vacuu tube. (Take care that your easureent of it does not include further coplicating factors, like the resistance in the external wires.) Unfortunately the wire-support resistance, R support, is larger than the filaent resistance at low teperatures. Thus the quantity we seek (filaent resistance, R W ) ust be calculated as the sall difference between two nubers: R W = R easured R support (59) a situation that produces big relative errors. Additionally, we have no independent way of easuring R support (we can t take the tube apart), so we can only calculate it based on a calculated value for R W at roo teperature. (This calculated value for R W will depend on the spec sheet values for the diensions of the filaent and the resistivity of W-filaent aterial at roo teperature: ρ 29 = 5.49 µω c.) You should be quite uncofortable that so uch of this easureent depends on book values rather than things you can directly check. In addition we have no choice but to assue that the found R support is a constant. This is surely false as the filaent will heat the ends of the support, and the resistance of essentially every aterial depends on teperature. 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 1 GE Rev 0, 10 (1927) 12

13 Teperature Dependence of Resistance: W Teperature (K) Resistance Ratio: R/R29 Figure 8: 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 29 K. 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 a Volt. 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 (60) 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 ɛ T, about 0..) 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 = ɛ T σt 4 πdl T 4 dl (61) 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 ɛ T doesn t depend on teperature. We conclude: I 2 T d (62) 1

14 Teperature vs. Current for W Filaents Teperature (K) a=i/d^1.5 (A/c^1.5) Figure 9: 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 2 power. [ I ] 2 T (6) d 2 Absent the above approxiations we can hope that teperature is a function of a I/d 2. Once again Jones & Languir provide us with calibrating data for this expected relationship. For teperatures 400 K < T < 000 K, Kirkan finds: T i = a a 1.8 (64) Finally, attaining theral equilibriu 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. Electrical Measureents As shown in Figure 7, the filaent is powered by a Keithley 2420 source-eter. The filaent voltage is easured directly at the socket with a Keithley 192 volteter. By cobining a current easured fro the 2420 with the voltage fro the 192, the resistance 14

15 of the tube alone an be deterined. Of course, only part of the tube resistance is filaent resistance. The roo teperature filaent resistance (R W.1 Ω) can be calculated fro the specifications of the FP-400 tube. Using a four-terinal easureent, easure the roo teperature resistance of the tube (R easured. Ω). Confir this easureent by sourcing 1 A and 10 A into the tube and reading the resulting voltage on the 192. Use these easureents to deterine a value for R support. At the sae tie the resistance of the external wires can be found. The voltage difference between the 192 and the 2420 ust be dropped across the wires. Fro the voltage across the external wires and the current flowing through the, calculate the wire resistance. Use this resistance to calculate the 2420 voltage liit. (That is the voltage the 2420 would need to produce to have have the axius (4.75 V, 2.5 A) on the filaent.) You will be collecting two types of data at the sae tie: theral characteristics of the filaent and the therionic properties of the tube. Starting at a filaent current of 0.9 A, increase the current flowing through the filaent in steps of 0.1 A to a near-axiu current (e.g., 2.4 A) and then reverse those steps down to 1.0 A. (Note the axiu liits of the filaent: 2.5 A or 4.75 V. Do not exceed either! You will probably need to do soe experienting to deterine the axiu current you can use without exceeding the 4.75 V liit.) 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 tube. Calculate filaent teperature two ways (Equations (58) and (64)). Average the two to estiate the teperature, and use half the absolute value of the difference to estiate the error. By conservation of energy we know that the power duped in the wire (ostly fro electrical heating, but also fro other sources like radiation fro the roo teperature environent to the wire) should equal the power out of the wire (fro black-body radiation and other factors like conduction down the wire supports). Thus: ɛ T σat 4 = I 2 R W + constant (65) T 4 = 1 ɛ T σa I2 R W + constant (66) y = ax + b (67) A graph of T 4 vs. power should be a straight line fro which you will deterine ɛ 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). In the second part of the experient, you will collect anode (voltage, current) curves for each different cathode teperature. Use the Keithley 2400 to sweep the anode voltage logarithically fro 1 V to 120 V. (Note the axiu liits for the anode: A or 125 V. Do not exceed either!) According to Child s law, the anode current, I A, should increase in proportion to V 2 A. At sufficiently high voltage the current will be liited by the axiu electron evaporation rate given by Richardson s law. At the axiu teperature (corresponding to the axiu filaent current) you will want to ake 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 ) 2 (68) 15

16 In addition, you will want to ake a big continuous file containing: V A, I A at every teperature tested. 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: 1. Declare and define all variables. 2. Initialize the source-eters which ust be told axiu voltage and current to occur during the experient. In particular you will need to know the external wire resistance to estiate the axiu voltage that the 2420 can safely output at axiu current and keep the tube voltage less than the axiu rating.. Display the status of all devices before starting data collection. 4. Open files: (a) filaent.dat for I, V, T r, T i of filaent. (b) stefanb.dat for power, T 4, δt 4 of filaent. (c) VI.dat for V A, I A of anode, with notes for filaent I, T r, T i (d) child.dat for V A, I A, δi A of anode at axiu filaent current. 5. Start an up-sweep in the filaent current: 0.9 A to axiu (e.g., 2.4 A). (a) Request the 2420 source-eter to source a filaent current. (b) Let the syste sleep for 15 seconds to approach theral equilibriu. (c) Request a logarithic anode voltage sweep fro 1 V to 120 V. (d) Turn off the anode voltage. (e) Read the source-eter to get the exact filaent current (it will be very close to the requested current). (f) Read the 192 to get the exact voltage across the tube. (g) Calculate the two filaent teperatures. (Requires knowledge of R W and R support and use of Kirkan forulas.) (h) Add a line to the file: stefanb.dat containing filaent power, T 4, and δt 4. (i) Add a line to the file: filaent.dat containing I, V, T r, T i of the tube. (j) Add a coent (!) line to the file: VI.dat containing I, T r, T i of filaent. (k) Record the anode sweep data in the file: VI.dat. (l) Increent the filaent current and return to (a). 16

17 FP-400 Vacuu Tube Ia (A) 1.E 04 1.E 05 1.E 06 1.E Va (V) Figure 10: 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 is the doinant effect. 6. Take an additional special sweep at the axiu filaent current, following all the steps outlined above. In addition record the anode sweep data in the file child.dat. 7. Start a down-sweep in the filaent current: 2. A to 1.0 A, following all the steps used in the up-sweep. 8. Turn off the output of the 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. Data Analysis The ain result of this experient is a plot siilar to Figure 10 showing the anode currentvoltage relationship at various filaent teperatures. Note that each curve is a pair: one fro the up-sweep in filaent teperature, one fro the down-sweep. See that hysteresis is largest at low filaent teperature. Note that for the lowest filaent teperatures, no current plateau is in evidence. Starting at about I f = 1.2 A, plateau currents can be 17

18 Richardson s Law.01 Anode Current (A) E 04 1.E Teperature (K) 1600 Figure 11: 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. deterined. According to Richardson s law, these plateau currents should satisfy: I = AAT 2 e B/T (69) where A is the tungsten filaent area. The errors in I can be estiated fro the hysteresis, but in fact the largest errors are in teperature. A fit to the above relationship allows us to estiate the work function, expressed in units of teperature. Convert this value to joules and ev and copare to the book 2 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 known value of the work function and the data for the highest filaent teperature into the above equation. Do this and copare to the book value: A = A/ 2 K 2. See Figure 11 for typical results. Exaine the relative iportance of hysteresis and calibration in teperature deterination. For I f = 1.2 A record the difference in T r due to hysteresis and the difference between T r and T i which is a teperature calibration uncertainty. At sufficiently low anode voltage (and high filaent teperature), Child s law governs the anode current-voltage relationship: Equation (68). Fit the highest anode-teperature data (e.g., filaent current of 2.4 A) to this functional for. Fro the constant k 1 the electron charge-ass ration e/ can be found. Since the FP-400 is a finite length cylinder (rather than the infinite cylinder discussed in our theory) use the effective length = 0.7 l as the length of the cylinder. Copare to book values. 2 Blakeore, Solid State Physics 18

19 Child s Law.01 Ia (A) Va (V) Figure 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. Nevertheless, the law provides an excellent suary of the data over a huge range of variation. 19

20 Stefan Boltzann Law 4.E+1.E+1 T^4 (K^4) 2.E+1 1.E Power (W) Figure 1: A test of the Stefan-Boltzann law: that energy 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. The Stefan-Boltzann relationship provides a check on our teperature calibration (in fact it is the definition of teperature beyond the elting point of Pt). Fit the data in stefanb.dat to Equation (67). Deterine ɛ T fro the fit slope. Report Checklist 1. Calculations (no errors) of R W. Measureents (4-wire oheter and direct voltage/current) of R easured. Calculation of R support. Calculation of external wire resistance. 2. Data files and coputer progra. Leave the in your UNIX account. Print out a copy of your progra and tape it into your lab notebook.. Plots siilar to Figures 10, 11, 12, 1. Note that Figure 10 is coplex to produce. Use a file of plot coands and feel free to talk to e about how to go about doing this. Carefully note the use of log and inverse scales. Also include a fit report for each fit curve. 4. Experiental values for: W (in ev), A, e/ and ɛ T. Calculate error only in ɛ T. 5. Show the issing steps leading to Equations (1) and (6). Substitute the ρ 1 approxiation Equation (5) into the differential equation Equation (1). 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. 20

21 6. Calculate the electric field at the surface of the FP-400 cathode for V A = 120 V. (Assue zero space charge for this calculation, i.e., that the vacuu has been swept clean of electrons.) Calculate the shift in work function (in ev) due to Schottky effects for this field. 7. Show how the potential energy Equation-(55) is consistent with the iage charge force Equation-(1). Coent: Classical vs. Quantu I said above that the presence of an h 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 a classical process. 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 2 F nuber density). We can write the classical rate in ters of the quantu with no h visible: classical flux = 4 [ ] EF 2 quantu flux (70) π kt The different teperature dependence for the classical flux (AT 1 2 e W/kT vs. AT 2 e W/kT ) cannot be detected experientally: the Boltzann factor overwhels all other teperature dependences. On the other hand, since E F kt, the classical rate coes out 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. References 1. Jones & Languir GE Review, The Characteristics of Tungsten Filaents as Functions of Teperature 0 (1927) Part I pp , Part II pp , Part III pp Melissinos Experients in Modern Physics, 1966, pp Preston & Dietz The Art of Experiental Physics, 1991, pp , Blakeore Solid State Physics, 1974, pp Koller The Physics of Electron Tubes, 197, Ch. I, VI, VIII 6. Saul Dushan (Phys. Rev. 21, (192)), 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. 21