chapter9 experiment7: theelectron’schargetomassratio · chapter9:experiment7 of the solenoid is...

Chapter 9

Experiment 7:The Electron’s Charge to Mass Ratio

9.1 Introduction

Historical AsideBenjamin Franklin suggested that all matter was made of positively and negativelycharged fluids that flowed under the influence of electric forces. In the 1800s Englishchemist John Dalton established that matter was instead made of particles that he andthe ancient Greeks called “atoms”. We learned to make vacuum tubes, and in 1869German physicist Johann Hittorf observed that “cathode rays” could jump across thevacuum between two electrodes.

J. J. Thomson, using a device similar to a cathode ray tube discovered the electronand measured the ratio of the size of its electric charge (e) to its mass (me). Thomson’sexperiment was concerned with observing the deflection of a beam of particles in a combinedelectric and magnetic field. Its result established:

1) the existence of the electron;2) the fact that the electron has a mass (me);3) the fact that the electron has an electric charge (−e);4) that both charge and mass are quantized;5) that the ratio of e/me is constant;6) that all electrons might be identical;

7) that e/me = 1.759× 1011 C/kg.

In this lab we will repeat Thomson’s measurement by observing the deflection of an electronbeam by a magnetic field.

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CHAPTER 9: EXPERIMENT 7

General InformationThe electron’s charge to mass ratio is the first datum ever measured about thefundamental particles and is important in and of itself; however, the method of usinga beam of particles in experiments is at least as important. Without this method wewould know nothing about the nuclei or the unstable fundamental particles.

The electrons orbiting around nuclei are what matter is made from, what allows chemiststo develop new compounds, and what allows biological phenomena to be explained. Theelectrons oscillating in a radio transmitter’s antenna enable them to transmit information.The controlled flow of electrons through a semiconductor allows a computer to process data;it is a beam of electrons that allows specimens to be seen with an electron microscope. It iswith a beam of very high energy electrons that scientists have established that nucleons aremade up of components which are called quarks and gluons. Possibly, these new componentsof matter will shape the way mankind will live 100 years from now in a way similar to thatin which Thomson’s discovery is responsible for the way we are living today.

There are two basic physical phenomena which play a significant role in the experimentcarried out in this lab: the existence of a magnetic field associated with an electric current

Figure 9.1: A vector schematic diagram illustrat-ing the vector directions for the Biot-Savart law.

and the deflection of a moving chargedparticle in a magnetic field. Beforewe discuss the experiment itself we willbriefly review these two phenomena andwe will discuss the fact that the earthhas its own magnetic field.

9.1.1 Magnetic Forces

The forces between electric currentsare called magnetic forces; because,the same phenomenon accounts for theforces acting between magnetic materi-als such as pieces of magnetized iron.William Gilbert, Queen Elizabeth I’sphysician, noted that a magnet has twopoles at which magnetic effects seemto be concentrated. He also showedthat like poles repel each other, whereasunlike poles attract each other. Todaywe explain the forces between magneticpoles in exact analogy with the elec-trostatic forces between charges (2nd

lab in Physics 136-2) by introducing amagnetic field represented by field lines.

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As we shall see later, the relation between magnetic field lines and magnetic forces is morecomplicated than in the case of electrostatic forces.

Historical AsideIn 1820 Hans Christian Oersted observed that a magnetic field was created by anelectric current; during a lecture he observed that a magnetic compass was deflectedfrom pointing north when he completed a nearby circuit and caused current to flow.This stimulated French mathematician Andre-Marie Ampere to study the phenomenonmore closely. In addition to Ampere’s observation that two currents interact via amagnetic force (circa 1820), these studies also led to Ampere’s law of magnetostatics.Parallel currents are attracted and anti-parallel currents repel. Ampere also publisheda magnetic force law similar to Coulomb’s law of electrostatics using electric currents assources; however, the Biot-Savart law below seems to be more fundamental. MichaelFaraday’s law of induction (published in 1831) was later proved to be due to themagnetic force on electric charges in conductors due to the conductor’s motion. Thecombined electric and magnetic forces on a moving charge in both fields is calledthe “Lorentz force” though it was published earlier by Oliver Heaviside in 1889 andperhaps by James Clerk Maxwell himself in 1865.

Figure 9.2: A pictureillustrating Right HandRule #1.

Moving electric charges create a magnetic field. In the case ofa continuous electric current this field is described by the Biot-Savart Law in Equation (9.1) reported by J. B. Biot and F. Savartto the French Academy.

dB = µ0I

4πdl× r̂r2 (9.1)

The vector dB is tangent to the magnetic field lines, whichin the case of a straight wire ( dl ) conductor are represented byconcentric circles in a plane perpendicular to the conductor. r̂ = r

r

is a unit vector directed from the current element toward theposition where dB is computed. Figure 9.2 shows such a fieldand also illustrates (together with Figure 9.5) the “right handrule,” which gives the relation between the directions of the 3vectors dB, dl, and r̂. If the thumb of the right hand is pointedalong the direction of the current I, then the fingers curl in thedirection of B. To increase the strength of B in a given volume,one usually uses a solenoid. In this case the B generated byeach winding of the coil will add up inside the solenoid to the Bof the other windings, as shown in Figure 9.3. The same figurealso shows the similarity between the fields of a solenoid and ofa permanent magnet. Notice that the magnetic field inside thesolenoid is nearly uniform. (This is true only when the length

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of the solenoid is large compared to its diameter.) The solenoid is the classic device forgenerating a magnetic field. It is the analog of the parallel plate capacitor, which is theclassic device for generating a uniform electric field throughout a given volume. Magneticfields of a specific shape can be generated by a system of coils.

Two parallel coils separated by a distance equal to the radius of the coils (Figure 9.4) areknown as Helmholtz coils. They are frequently used because they generate a magnetic fieldthat is uniform over an appreciable region about its midpoint. We will be using a system ofHelmholtz coils to carry out this lab’s experiment. If each one of the coils, with radius R,has N turns and carries a current I, then the field at the center of the system is,

Table 9.1: Symbol definitions for the Helmholtz coil’s magnetic field equation.

Figure 9.3: Illustrations of the magnetic field lines generated by current flowing through asolenoid coil and by a bar magnet.

BC = 0.714µ0NI

R(9.2)

Let us now look at the force exerted by a magnetic field on a moving charged particle. Itwas the American physicist H. A. Rowland who first observed that a particle with charge qand velocity v moving in a magnetic field B will be subjected to a force F . The direction ofthis force is perpendicular to v and to B. If a particle enters a volume with an electric fieldE and a magnetic field B then the total force on the particle is given by the Lorentz law

F = qE + qv×B. (9.3)

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CheckpointWhat is the Biot-Savart law? What is the Lorentz force law?

CheckpointWhy are Helmholtz coils useful in physics experiments? What relation between radius,position, and separation distinguishes Helmholtz coils from other coil pairs?

B

R

RR

II

xy

z

Figure 9.4: A schematic diagram of one pairof Helmholtz coils. The radius, R, of bothcoils equals the distance between the centers.Helmholtz coils generate uniform magneticfields near the geometric center at the origin.

This is an extremely important relationpartly because it connects mechanics (forceF) to electromagnetism (fields E and B).The magnetic force is the cross productof two vectors. Figure 9.5 reminds youonce more of the “right-hand rule” whichis defined for positive charge. You must befamiliar with this rule in order to carry outthis experiment.

As mentioned in the introduction, in thisexperiment we will observe the deflection ofelectrons in a magnetic field. To simplify theexperiment we will choose the direction ofthe electron beam (qv) to be perpendicularto B. In this case the force acting on theelectrons (each with charge q) is simply,

Figure 9.5: A picture of theright hand rule needed to deducethe direction of the magneticforce on positive charge movingin a magnetic field.

Fmag = qv×B. (9.4)

According to the right-hand rule #2, Fmag is perpen-dicular to v and B; consequently the electrons would movein a circle as shown in Figure 9.6 if they had a positivecharge. Since they have a negative charge, the electronswill actually rotate in the opposite direction from thatshown in the figure. (Remember from mechanics: if F ⊥ v,then it must be circular motion.) The centripetal forceresponsible for motion in a circle instead of a line is givenby

mv2

r= mar = Fmag = qvB. (9.5)

We can calculate the radius of curvature of the circularmotion of a charged particle moving perpendicularly to a

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magnetic field using

r = m

q

v

B. (9.6)

This simple relation is the basic equation that we will use to carry out the measurement ofe/me. By measuring the radius of curvature r of an electron beam’s trajectory, by knowingeach electron’s velocity, and by having the beam deflected by a magnetic field, we cancalculate the ratio q/m.

Before we describe how to do this experiment, there is one last topic to cover: the earth’smagnetic field! This field will have an effect on our electron beam, hence we must properlytake account of it - otherwise, our measurement of e/me will be wrong.

CheckpointDescribe the angles between the electron’s velocity, the magnetic force, and themagnetic field.

9.1.2 The Earth’s Magnetic Field

The earth’s magnetic field is the field of a magnetic dipole, which means that it is equivalentto the external field of a huge bar magnet. The lines of force of such a field are directednot towards the geographic poles but rather towards the magnetic poles. (The magneticnorth pole is located near the geographic south pole.) They are also directed (except at theequator) towards or away from the center of the earth - as shown in Figure 9.8.

The intensity of the field at the surface is on the order of one Gauss. Sediments ofmagnetic materials (iron, cobalt, nickel) can drastically change the local pattern of this fieldwhich has been carefully mapped, most recently with the use of satellites. After centuries ofresearch, the earth’s magnetic field remains one of the best described and least understoodof all planetary phenomena.

The history of the earth’s magnetic field has been traced back 3.6 million years, andit has been established that during this time the earth’s field has reversed nine times. Toestablish such a fact two elements were necessary: the magnetic “memory” of volcanic rocksand the presence in the same rocks of atomic clocks that begin to run just when theirmagnetism is acquired. The memory elements themselves are magnetic “domains”, tinyvolumes of magnetic material in which magnetism is uniform. These bodies consist of ironand titanium oxide.

At temperatures above a few hundred degrees (depending upon the chemical composition)these domains are not ferromagnetic. When a domain cools it becomes magnetized in thedirection of the surrounding magnetic field. The atomic clocks that record the time of the

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lava solidification are based on the radioactive decay of potassium 40 into argon 40. Thisradioactive decay (transformation of potassium into argon) takes place similar to the decayof an RC circuit (see 5th lab last quarter) but with a much longer time constant or half-life.

Figure 9.6: An illustration of the cyclotron motionresulting from a positive charge moving in a magneticfield. All other forces are assumed to be negligible.

The argon is trapped withinthe crystal structure of the min-erals and if the minerals are notheated or changed in some way, itaccumulates there. The amountof trapped argon is a function ofthe amount of potassium presentand the length of time since thedecay and entrapment process be-gan. The potassium-argon dat-ing method has now been success-fully applied to rocks from nearly100 magnetized volcanic forma-tions, with ages ranging from thepresent back to 3.6 million yearsago; nine earth magnetic field re-versals were observed during thistime. You should not worry aboutthe earth’s magnetic field chang-ing during your experiment; thedata of volcanic rocks shows thatit takes about 5000 years for afield reversal to complete once ithas begun. You must be aware,however, that there is an earthmagnetic field which affects thisexperiment.

CheckpointHow do we know that earth’s magnetic field has reversed itself? How do we know howlong ago these geologic changes occurred?

9.1.3 Radius vs. Magnetic Field

Measure the ratio of the electron’s charge to the electron’s mass in C/kg. As describedabove, the basic relation for this measurement is given by Equation (9.6)

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Figure 9.7: A cutaway schematic illustration of our charge to mass ratio apparatus.

r = m

q

v

B.

This relation tells us that in carrying out the measurement of e/me we need three basicelements:

1) a beam of electrons with known velocity.2) a magnetic field (uniform over the region where the electrons will describe a circular

trajectory).

3) a way to observe the electron’s path, so that we can measure the radius of curva-ture. Figure 9.7 and Figure 9.10 show schematic drawings of the equipment used tomeasure e/me.

CheckpointIf v ⊥ F, what kind of motion results? Why doesn’t the speed change?

9.2 The Apparatus

The apparatus needed to measure e/me consists of an electron gun to generate a beamof electrons with known velocity, a pair of Helmholtz coils to generate a known uniform

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I

N

S

(a) (b) (c)

Figure 9.8: An illustration of Earth’s magnetic field (a). Note the similarity to the dipolefield of a current loop (b) or a bar magnet (c). Earth’s field protects us from energetic chargedparticles emitted by the Sun.

magnetic field, a thin argon gas to make the electron beam visible, and calibrated positionpegs to allow us to measure the radius of the electron’s circular trajectory.

9.2.1 The Electron Gun

We generate a beam of electrons with an electron gun. It is shown schematically in Figure 9.9.A large current heats a thin filament so that electrons ‘boil’ off. The filament is biased toa large negative potential to accelerate the electrons toward ground. A grid of thin wireswith even more negative potential repels the electrons and allows us to control the electronbeam’s current. An anode is connect to earth ground to accelerate the electrons that passthe grid. The electrons are accelerated to a final velocity that converts the electrons’ electricpotential energy qV into their kinetic energy

12me v

2 = eV and v =√

2 eVme

. (9.7)

Ordinarily, the electrons would strike the anode and transfer their kinetic energy to theanode’s atoms causing the anode to heat up. The anode in the electron gun has a hole in it,however, that allows some of the electrons to continue past into the thin gas of argon atoms.Figure 9.10 shows a view of the electrical connections for the equipment used. The electrongun is controlled by 2 knobs both on the blue Power Supply. The “DC 0. . . 300V” knobsets the accelerating potential, and the “DC 4. . . 10V” knob sets the current in the filament,thereby controlling the electron beam current Ianode.

CheckpointHow can we know the velocity of the electrons?

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9.2.2 The Magnetic Field

The magnetic field is generated by a set of Helmholtz coils. Equation (9.2) gives the valueof the field at the center of the system. The number of turns for each coil is N = 72. Witha ruler you can measure the radius of the coils and/or the distance between them and thencalculate the value of c = 0.714µ0

NR. The magnetic field is aligned along the axis of the

coil system, and its direction is determined by the “right-hand rule” (see Figure 9.2). Asmentioned previously, the earth also has a magnetic field Be, which cannot be neglected inthis experiment. Each apparatus has been individually aligned, with the help of a compassneedle, in such a way that the field of the coil is parallel with the earth’s field but in theopposite direction of the earth’s field. Can you demonstrate this?

Consequently, the magnetic field BT that will deflect the electrons is

BT = cI ±Be. (9.8)

where the ± sign depends upon the directions of the current in the coil and the earth’s field.The knob labeled CURRENT, on the current power supply (see Figure 9.10), will allow youto vary the strength of BT, while the meter above the knob gives a reading of the current Ithrough both coils.

CheckpointHow do we generate a magnetic field? Is this the only field we need to worry about?

9.2.3 Electron Trajectory

Electrons are infinitesimally small objects (radius< 10−16 cm) that cannot be seen by thenaked eye. In order to observe their trajectory without blocking their path, the electron gunis installed in a glass enclosure (25 cm diameter) which contains low pressure argon (Ar) gas.The electrons (with 50 eV kinetic energy) will excite the Ar atoms (requiring 2 eV) whichthen emit an orange light. The electron trajectory can be observed (in a darkened room) asa ray of orange light emanating from the electron gun. The glass bulb also contains, alongone of its diameters, a set of pins. The distances from the anode slit of the electron gun toeach of these pins are:

0.020m, 0.040m, 0.060m, 0.080m and 0.100m.

Some of these pins are still covered with a fluorescent material which emits light when struckby the electron beam; but most of this material has been burned off over the years. By varyingthe “CURRENT” control knob you change the current I in the Helmholtz coils, producingdifferent values of B and forcing the electrons to describe different orbits. Certain values ofB will allow the electron beam to strike the calibrated pins. Knowing B and the radius of

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Table 9.2: A summary of the equations describing the electron’s motion. These will becombined to evaluate a measured e/me.

e-

E

E

Filament

Accelerating

Voltage

Filament

Current

Beam

Focus

e-

e-

e-

e-

e- e

-

e-

e-

e-

Figure 9.9: A schematic diagram showing the electrical connections and electron motionfor our electron gun.

the electron beam’s trajectory will allow you to determine e/me provided the acceleratingpotential is known. The equations summarized in Table 9.10 are used in calculating e/me.

We can combine these equations and write

cI ±Be = BT = me v

er= 1r

me

e

√2V eme

= 1r

√2V me

e.

In this equation there are two unknowns: (e/me) that we want to measure and thestrength of earth’s magnetic field, Be. It will require a minimum of two measurements (rland r2 for values of V and I to determine both unknowns). The measurement will be carriedout by varying I, measuring r, and plotting BC = cI on the vertical axis and 1

ron the

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Voltmeter

Current

Supply

Electron

Gun

Control

Accelerating

Voltage

Magnetic Field Current

Beam

Focus

Filament

Heater

Figure 9.10: An illustration of the electrical connections between our e/me apparatus andthe power supplies.

horizontal axis. The data must fit a straight line with

intercept = ∓Be and slope =√

2V me

e

so thate

me

= 2V(slope)2 . (9.9)

Make a linear-least-squares fit to the data and calculate the limit of uncertainty.

Checkpoint

How can we extract our measured e/me from measurements of accelerating voltage,Helmoltz coil current, and trajectory radius?

9.3 Procedure

Measure the diameter of the Helmholtz coils with the meter stick. To calculate the magneticfield of the center of the coils you need, according to Equation (9.2), the radius R as seen inFigure 9.4, the number of windings N = 72, and the current in each coil.

BC = 0.714µ0NI

R

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Express this equation as BC = cI and calculate the value of the constant as well as its unitsusing µ0 = 4π × 10−7 Tm/A.

1. The equipment is wired up according to the diagram in Figure 9.10.

2. Turn the power on in all three units. Voltmeter, Black Laboratory Power Source, andBlue Power Source.

3. Turn up the “DC” control to 8V, check to see that there is an orange glowing spot inthe heater of the electron gun and then wait for 10 minutes. .

4. Turn the “DC 0. . . 300V” control to about 100V and note the reading on the voltmeter.You should see the electron beam as a thin orange line emerging from the anode slit.Observe the orange beam in the tube. It should look like it is slightly bent ‘downward’.Why?

5. Turn up the coil current using the Current control on the Laboratory Power Sourceand you should see the electron beam bend into a circle. If the electron beam does notstay flat in a circle, rotate the glass tube slightly on its axis until the electron beamforms a flat circle that hits the pins of the lattice in the tube. The tips of the pins werecoated to fluoresce when the beam hits them, but this coating may no longer exist onyour tube.

6. Dial down the voltage to 70V. Do you still have a beam? If not, raise the voltageuntil you have a beam. You want to have the lowest possible voltage to maximize yourchances of hitting the closest peg.

Your equipment is now ready to take data.

9.3.1 Trajectory Radius vs. Magnetic Field• Vary the current in the Helmholtz coils (Current knob) until the outer edge of theelectron beam matches the outer edge of each pin in the tube. You may not haveample current to bend the beam enough to hit the pins closer to the electron gun.Three pins would be sufficient to get a value of e/me.

• Record in your lab book the accelerating voltage (voltmeter) of the electron gun andthe Helmholtz coil current settings (ammeter) for each measurement correspondingto an electron orbit of radius r. Don’t vary the accelerating voltage, but note anyfluctuations. The distances from the anode to the pins (the diameter of the electronorbit) are 0.020m, 0.040m, 0.060m, 0.080m, and 0.100m.

• Calculate the field BC .

Use Vernier Software’s Graphical Analysis 3.4 (Ga3) to help you graph your data. Asuitable Ga3 configuration can be downloaded from the lab’s website at

http://groups.physics.northwestern.edu/lab/e_over_m.html

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Your notebook and write-up should include a plot of B vs. (1/r) for these measurements.Be sure Ga3 reports the uncertainties in slope M ± δM and intercept b ± δb; if not, right-click the parameters box, “Fit Properties. . . ”, and select ”Show Uncertainties”. Using the fitparameters and other measurement(s), calculate e/me in C/kg; also give the the expectedtolerance limits using

δ(e

me

)=(e

me

)√√√√(δVV

)2

+(

2 δMM

)2

. (9.10)

It would be wise to use the formulas in Section 2.6.3 to derive this result.

Checkpoint

Show that the dimensions of the relation you are using to calculate e/me (Equa-tion (9.9)) are indeed C/kg.

An important relation that allows you to relate units of mechanics (such as kg, m, s) tounits of electricity and magnetism (such as A, V, s) is the energy relation

AV s=CV=J=Nm = kgm2

s2 .

Calculate the earth’s magnetic field (in Gauss) at the location of your e/me apparatus; thisis given by the intercept of the line fitted through your data points and the BC axis. Sketchfor both measurements:

1) the direction of the electron beam,2) the direction of the centripetal force and the Lorentz force acting on the electron

beam,3) the direction of the current in the Helmholtz coils, and

4) the direction of the field; indicate if it was the measurement with (Be + BC) orwith (Be −BC).

The beam of electrons also represents a current and consequently must generate a magneticfield. Does this field point in the same or in the opposite direction to the field generatedby the Helmholtz coils? Remember that an electric current is defined for positive chargecarriers.

9.3.2 Acceleration vs. Magnetic Field

1) Set the voltage control to 100V.2) Adjust the current to have the beam hit a convenient peg.3) Record the readings in your lab book.

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4) Increase the voltage by 10V.5) Find the new current that steers the beam back to the same peg.6) Record the second set of data.7) Repeat for several more voltage readings.8) Calculate the magnetic field from the current as before using Equation (9.2).9) Plot V vs (BC −Be)2.

10) Obtain the slope of the line and note that the slope is equal to 12

(em

)r2.

11) Obtain a value for e/me with error limits and compare it to the value obtained inthe first experiment.

12) Comment on the agreement or lack of agreement.

Note that in this case

δ(e

me

)=(e

me

)√√√√(δMM

)2

+(

2 δrr

)2

. (9.11)

CheckpointWhy is it necessary to perform Section 9.1.3 before it is possible to performSection 9.3.2?

9.4 Analysis

First, check the consistency between your two measurements of e/me. Label the twomeasurements and their uncertainties with subscripts 1, 2.

∆s.c. =∣∣∣∣ e1

me1− e2

me2

∣∣∣∣ (9.12)

andσs.c. =

√[δ(e1

me1

)]2+[δ(e2

me2

)]2. (9.13)

The best estimate ρ± δ is the average of your two measurements ρ1 ± δ1 and ρ2 ± δ2

ρ =ρ1δ1

+ ρ1δ1

1δ1

+ 1δ1

(9.14)

where each measurement is weighted by the reciprocal of its uncertainty. The uncertainty, δ

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in this average is

δ =√

21δ1

+ 1δ2

. (9.15)

Compare the average ρ ± δ of your two measurements to the accepted best estimate ofthe science community e/me = 1.759× 1011 C/kg

∆ = ρ− (e/me). (9.16)

Checkpoint

What should you expect to see if e/me was not a constant? Would the electrons’velocities depend upon accelerating voltage if they had no mass? Would the electronsaccelerate at all in your apparatus if they had no electric charge?

Do your comparisons agree better than 2σ? Worse than 3σ? What other sources oferror have we omitted from consideration before now? Are all of our assumptions valid? Forexample, are the electrons that ‘boil’ off of the filament at rest before we accelerate them?Can you think of other assumptions that are not quite valid?

Have you made ancillary observations that support our hypothesis? For example, do thedata points form the straight line predicted by our theory? What other observations supportour hypothesis?

9.5 Conclusion

Can you conclude that your ratio of e/me is constant? Is e/me sufficiently important toscience that we should report it in our Conclusions? Do our observations support ourhypothesis? Do they contradict it? Answer these questions after allowing for all of yourobservations. Report your answers in a short paragraph or two. Can you think of anyapplications for what we have observed? How might we improve the experiment?

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chapter9 experiment7: theelectron’schargetomassratio · chapter9:experiment7 of the solenoid is...

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