Topic 25: Charged Topic 25: Charged ParticlesParticles

25.1 Electrons25.1 Electrons

25.2 Beams of charged particles25.2 Beams of charged particles

Charge of an ElectronCharge of an Electron

ee--

1.6 × 10-19 C

How do we manage to determine such a small value?

Thanks to Robert Millikan

Millikan’s Oil Drop ExperimentMillikan’s Oil Drop Experiment

1909

1868-1953Robert Millikan

A Glimpse into Millikan’s EndeavourA Glimpse into Millikan’s Endeavour

http://video.google.com/videoplay?docid=2799052432147926032&ei=Q3unS4q4MIiEwgON9vn2CA&q=Millikan+Experiment+CalTech+12&hl=en#

Millikan’s EndeavourMillikan’s Endeavour

Millikan’s ApparatusMillikan’s Apparatus

Millikan’s ResultsMillikan’s Results

Millikan found that charge is quantised into multiples of 1.6 × 10-19 C

ExampleExample

Between the PlatesBetween the Plates

Weight of the droplet

Upward force by the electric field

Tiny oil drop

Weight of a droplet = Electric force on it

mg = qE

charge on oil droplet:

m = mass of dropletq = charge on droplet

E

mgq

Weight of the Oil DropletWeight of the Oil Droplet

Weight of oil droplet Weight of oil droplet

= = mgmg – upthrust – upthrust

= = VVoil oil gg – – VVair air gg

= = VgVg ( (oiloil – – airair) )

= 4/3 = 4/3 rr33 g ( g (oiloil – – airair) )

How to find r r ?

Determining radiusDetermining radius

To determine the radius of the oil droplet, To determine the radius of the oil droplet, rr, the electric , the electric field is switched off and the oil droplet allowed to fall with field is switched off and the oil droplet allowed to fall with its terminal velocity. its terminal velocity.

The terminal velocity, The terminal velocity, vv is measured by recording the is measured by recording the time, time, tt, for the droplet to fall through a measured distance, , for the droplet to fall through a measured distance, yy. .

When the oil droplet is falling with its terminal velocity, its When the oil droplet is falling with its terminal velocity, its Weight = viscous drag on itWeight = viscous drag on it

4/3 4/3 rr33 g ( g (oiloil – – airair)) = = 66rrvv

2 9

2 oil air

vr

g

ExampleExample

An oil drop of mass 2.0 An oil drop of mass 2.0 10 10-15-15 kg fall with its terminal kg fall with its terminal velocity between a pair of parallel vertical plates. When velocity between a pair of parallel vertical plates. When a potential gradients of 5.0 a potential gradients of 5.0 10 1044 Vm Vm-1-1 is applied is applied between the plates, the direction of fall becomes between the plates, the direction of fall becomes inclined at an angle of 21.8inclined at an angle of 21.8oo to the vertical. to the vertical.

Sketch vector diagram to show the forces on the oil dropSketch vector diagram to show the forces on the oil drop(a) before and(a) before and(b) after the electric field is applied(b) after the electric field is applied

Write down expressions for the magnitudes of the vectors Write down expressions for the magnitudes of the vectors involved. Calculate the charge on the oil drop.involved. Calculate the charge on the oil drop.

Production of Electron BeamProduction of Electron Beam

Cathode Ray Tube

Electrons in an Electric FieldElectrons in an Electric Field

0 V

V V

d

Electric Force, Fe = e E

ma = e (V / d)

aa == ee V V // m d m d

Deflection of Electrons in an Deflection of Electrons in an Electric FieldElectric Field

2 2

22

2 2

1 1

2 2

1

2 2

eVy at t

md

eV x eVx

md v mdv

22 2

2 2

1 1

2 2 2

eV x eVy at x

md v mdv

Apply dynamic equations to motion of charged particles in an electric field.

Its verticle displacement:

Its verticle speed:

At any point:

The direction of the beam: The speed of the beam:

tan = vy / v vr = (v2 + vy2)

y

eV x eVxv at

md v mdv

ExampleExample

SolutionSolution

Electrons in a Magnetic FieldElectrons in a Magnetic Field

Magnetic Force, F = Bev

As it is making a circular motion:

Centripetal force = Magnetic force

mv2 / r = Bev

v = Ber / m

OR

r = mv / Be

Combined Electric & Magnetic Combined Electric & Magnetic FieldsFields

For zero deflection:

Magnetic force = Electric force

Bev = eE

v = E / Bv = E / B OR v = V / Bdv = V / Bd

Particles with the right speed will move undeflected in the combined electric and magnetic fields.

Specific Charge, Specific Charge, ee//mm

If a particle carries a charge of e and has a mass m.

Then the ration ee//mm is the specific charge of the particle.

It is measured in C kg-1

To find this value we use a mass spectrometer as shown.

Mass Spectrometer

Determining Determining ee//mm

2eVv

m

2

2

2

mvB ev

re v

m B r

The Electron Gun: Produce the electrons and accelerate them from the cathode to the anode

½ mv2 = eV

The Velocity Selector: Electrons with the right speed will move undeflected in the combined electric and magnetic fields and will pass through a gap S3.

The Deflector: Electrons emerging from slit S3 is deflected through a magnetic field. They are made to strike a photographic plate and the radius of the circular motion measured.

2eVv

m

Ev

B

2

2

2

mvB ev

re v

m B r

Determining Determining e/m e/m (1)(1)1. Ionisation:Thermionic emission:Current is supplied to a cathode (tungsten wire) and heating it up to produce electron beam.

2. AccelerationThe plates are connected to an accelerating +ve voltage V to accelerate the electrons though slit S1 and S2 into a velocity selector of cross E & B fields.

½ mv2 = eV --- (1)

--- (2)

All electrons have the same K.E.

2eVv

m

DeterminingDetermining (2)(2)3. Deflection:Only those with speed

v = E / B will move undeflected in the cross E & B fields, passing through slit S3.

4. Detection: The selected electrons are acted on by Fem and follow a circular path of radius r in the magnetic field B2. The radius can be measured as the electrons darken the photographic plate where they strike.B e v = m v2 / r --- (3)e / m = v / (B r) --- (4)

The electron-volt (eV)The electron-volt (eV)Find the kinetic energy that an electron Find the kinetic energy that an electron acquires if it falls through a potential acquires if it falls through a potential difference of 1 V, and its charge is -1.60 x difference of 1 V, and its charge is -1.60 x 1010-19 -19 C.C.

1.60 x 10-19 J

= 1 electron-volt (eV)

ExerciseExercise

An electron is accelerated by a p.d. of 500 An electron is accelerated by a p.d. of 500 V. Calculate the gain in kinetic energyV. Calculate the gain in kinetic energy

(i) in joules(i) in joules

(ii) in electron-volt (ii) in electron-volt 8.0 x 10-17 J

500 eV

ExerciseExercise

An electron leaves a cathode at zero An electron leaves a cathode at zero potential and travels through a vacuum to an potential and travels through a vacuum to an anode at + 200 V. Calculateanode at + 200 V. Calculate

(i) the kinetic energy that it acquires.(i) the kinetic energy that it acquires.

(ii) the speed when it reaches the anode.(ii) the speed when it reaches the anode.

3.2 x 10-17 J

8.4 x 106 m s-1

ExampleExample

A hydrogen has a charge-to-mass ratio of A hydrogen has a charge-to-mass ratio of qq//mm = 9.65 x 10 = 9.65 x 107 7 C kgC kg-1-1. A Bainbridge mass . A Bainbridge mass spectrometer has spectrometer has BB1 1 = 0.93 T, = 0.93 T, BB2 2 = 0.61 T = 0.61 T

and and EE = 3.7 x 10 = 3.7 x 106 6 V m V m-1-1. Calculate the . Calculate the radius of the paths of each of the following radius of the paths of each of the following ions in the mass spectrometer:ions in the mass spectrometer:

(a) H(a) H++

(b) He(b) He++

(c) He(c) He2+2+

0.0676 m

0.270 m

0.135 m

Example 1Example 1

An electron gun operating at 3000 V is used to An electron gun operating at 3000 V is used to shoot electrons into the space between two shoot electrons into the space between two oppositely charge parallel plates. The plate oppositely charge parallel plates. The plate spacing is 50 mm and its length is 100 mm. spacing is 50 mm and its length is 100 mm.

Calculate the deflection of the electrons at the Calculate the deflection of the electrons at the point where they emerge from the field when the point where they emerge from the field when the plate p.d. is 100 V. Assume the specific charge plate p.d. is 100 V. Assume the specific charge e/m for the electron is 1.76 × 10e/m for the electron is 1.76 × 101212 Ckg Ckg-1-1

Solution 1Solution 1

Example 2Example 2

Solution 2Solution 2

Example 3Example 3

Solution 3Solution 3

Physics is Great!Physics is Great!

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