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Demonstration Experiments: The charge to mass ratio of the electron 27 September 2012 The charge to mass ratio of the electron In the standard model of particle physics, the fundamental properties of the electron are its charge and mass, which are fundamental constants, and its spin and lepton number, which are quantum numbers. It is not easy to measure the electron’s mass directly. It is usually easiest to apply an electric or magnetic force to the electron to measure its charge to mass ratio and deduce its mass from an independent measurement of its charge. In this experiment you will determine this fundamental ratio. The apparatus that we have to measure the charge to mass ratio of the electron consists of an electron source in an evacuated tube and a set of Helmholtz coils* that produce a uniform magnetic field. The tube contains a trace amount of hydrogen gas that is ionized through electronmolecule collisions and the resulting emission allows you to see the trajectory of the electron beam. By varying electrical potentials applied to the source you can vary the energy and focus of the electron beam. Background A particle of mass m with a charge q moving with velocity v in a magnetic field B experiences a force given by the Lorentz for law, (1) F = q v×B . Because of the cross product, the force is always at right angles to the motion, thus the magnitude of the velocity never changes.† For some radius r, the magnitude of the Lorentz force will just equal the centripetal force needed to keep the particle in circular motion, v2 (2) FC = m . r Equating the magnitudes of these forces, we can solve for the charge to mass ratio, q v . (3) = m rB Our apparatus produces a beam of electrons, so we make the substitutions q → −e , m → me . The electron beam is accelerated by an electrical potential U, so an individual electron has kinetic energy eU. This allows us to determine the electron’s velocity through the equality (4) eU = 12 me v 2 . Substituting this into eq. (3), it is easy to show (you should verify this!) e 2U . (5) = me (rB )2 The experimental strategy should be clear: you can control U through the voltage applied to the electron source and B through the current in the Helmholtz coils. By measuring the radius r for different values of these parameters you can determine the charge to mass ratio of the electron. ( ) Experimental procedure Safety warning: The electron tube operates at high voltage, up to 500V. All connections should be made using the special shielded banana cables, not ordinary cables. The tube itself is made of very thin glass. Take care not to bump it or drop things on it, as it will implode if broken. * † The magnetic field produced by the Helmholtz coils is discussed in the appendix. See Young and Freedman, §27.4. Blackett Laboratory, Imperial College London 89 Demonstration Experiments: The charge to mass ratio of the electron 27 September 2012 The electrical connections to the electron tube are shown in figure 1. It should already be connected to its power supply, but it is worth understanding how the tube works. The cathode is heated by the current from a 6.3VAC supply. The hot cathode emits electrons via thermionic emission. The cathode is held at ground potential, 0V. It is surrounded by a metal cylinder with a small hole in its end, called a Wehnelt cylinder. The Wehnelt cylinder is held at a negative voltage with respect to the cathode and helps to focus the electron beam. You’ll probably find a bias voltage of about -20V to be optimal. The anode is the conical metal structure you can see in the tube. It is held at a large positive voltage, typically about 250V. The electrons born at the cathode are accelerated towards the anode and emerge from the hole in its tip. There are two small deflector plates near the anode; they should be electrically tied to the anode using the terminals on the tube base. One meter should be set up as a voltmeter to monitor the anode voltage. The Helmholtz coils are powered by the separate DC supply, you can use its built in ammeter to monitor the coil current. The B field at the center of the coils is proportional to the coil current I. It has the theoretical value 3/ 2 ⎛4⎞ n (6) I ≡ β I, B = µ0 ⎜ ⎟ ⎝5⎠ R where µ 0 = 4π × 10 −7 N A −2 , n = 130 is the number of turns in each coil and R is the coil radius, nominally 150mm. For convenience we have defined β as the proportionality between the B field and the current. We can rearrange equations (5) and (6) to get e r 2β 2 2 (7) U= I , me 2 in other words, a plot of the anode voltage U vs. I 2 for an electron beam with a given radius r should give a straight line whose slope is proportional to the charge to mass ratio. Starting with U = 250V, adjust I to given an electron orbit with diameter about 8cm. Use the slides and the mirror reflector to measure the edges of the electron beam. Make 5 or so additional measurements with U between 200V and 300V, always adjusting the coil current to keep the beam diameter constant. Figure 1: Electrical connections to the electron beam tube. The electron beam loses energy due to collisions with the background hydrogen gas. According to eq. (5), lower energy electrons follow orbits with a smaller radius. This implies that you should measure the outside diameter of the beam to measure the highest energy electrons. On the other hand, the electrons in the beam repel each other, causing the beam to expand. At the same time the ionized background gas tends to cancel this space charge and therefore focus the beam. You might think this would mean you should measure from the center of the beam. The manufacturer of the equipment does suggest using the outside edge, but you should at least consider this as a possible source of systematic error. Blackett Laboratory, Imperial College London 90 Demonstration Experiments: The charge to mass ratio of the electron 27 September 2012 Analyse your (U,I) data to extract a value of e/me. Remember the diameter of the orbit is 2r! You should also calculate the uncertainty in your measurement. There are a few different approaches to this, so you might want to discuss your analysis strategy with your demonstrator. Details The expression for the B field given by eq. (6) is only exact for the single point in the center of the coils. You should take additional sets of measurements for smaller beam diameters to see if your e/me ratio varies systematically with the orbital diameter. The earth’s magnetic field adds an offset to the field generated by your coils. If you rotate the apparatus by 180º and repeat your measurement for a particular diameter, you can subtract off this background. In fact, you can also use your results to get the value of the component of the earth’s field parallel to the coil axis. Assuming the spreading in the beam is due entirely to scattering energy loss, can you determine how much energy the beam loses? Hint: how much voltage is required to move the beam by its width? It probably makes sense to use the center of the windings on the coils to define R in eq. (6). How much difference would it make to your results if you took the inner or outer edge instead? Can you think of any other factors which could systematically alter your results? The accepted value of e/me is 1.7588×1011 C kg-1. Blackett Laboratory, Imperial College London 91 Demonstration Experiments: The charge to mass ratio of the electron 27 September 2012 Appendix: Helmholtz coils It is well known‡ that the magnetic field at a distance z along the axis of a current loop of radius R lying in the x-y plane is R2 . (a1) Bz = µ 0 I 3/ 2 2 R2 + z2 We can simplify the notation by measuring all distances in units of R. The field from two coils separated by their radius, i.e. located at ±½, is then ⎞ µ I⎛ 1 1 ⎟. (a2) + Bz = 0 ⎜ 3/ 2 3/ 2 ⎟ ~ 1 2 1 2 2 ⎜ 1 + (~ ) ( ) + + − z 1 z 2 2 ⎝ ⎠ ~ At z = 0, this reduces to ( ( ) ) ( ) 3/ 2 ⎛4⎞ (a3) Bz = µ 0 I ⎜ ⎟ , ⎝5⎠ which is exactly eq. (6) when we take account of the number of turns and put back in the radius R. The magic of Helmholtz coils is that both dBz / dz and d 2 Bz / dz 2 vanish at the center.§ This means the magnetic field the coils produce is very uniform. In fact, it is possible** to derive a general expression for all of the field components. It turns out the field is also very uniform as you move in the x and y directions around the origin, though the mathematics becomes a bit messier. B. E. Sauer, Sept. 2008 ‡ See Young and Freedman, §28.5. See also problem 28.67. The first non-constant term in a series expansion of eq. (a2) around z = 0 goes as z4. ** See §5.5 in J. D. Jackson, Classical Electrodynamics, 2nd Ed., (Wiley, New York, 1975). § Blackett Laboratory, Imperial College London 92