Discovery of the Electron

Question 1

What did Crookes discover?

Answer 1

That gases at a sufficient low temperature will conduct electricity and glow with a specific colour.

Question 2

What does the magnet do in a Crookes discharge tube?

Answer 2

Deflects the positive column

Question 3

How much P.D is applied in cathode rays?

Answer 3

Enough to produce a field that ionises atoms.

Question 4

What do electrons do in cathode rays?

Answer 4

Accelerated towards the anode, exciting gas molecules and causing the glow.

Question 5

What do positive ions do in a cathode ray?

Answer 5

Move towards the cathode, recombining to give off a negative glow.

Question 6

How is an electron beam produced in thermionic emission?

Answer 6

Heating a cathode (via a filament);
Accelerating the beam using a high p.d anode
Evacuating the glass tube so there are no collisions between electrons and gas molecules.

Question 7

What is the k.e of an electron accelerated through a p.d V?

Answer 7

eV = 1/2 m v^2

Question 8

What did Millikan determine electronic charge e to be?

Answer 8

1.6 x 10^-19

Question 9

What is special about electric charge?

Answer 9

Is quantised, so only exists in integer values of e. Suggests existence of electron.

Question 10

What are the equations showing an oil drop in stasis?

Answer 10

F(e) = QE = QV/d = mg

Question 11

What is stokes’ law, and what are the symbols?

Answer 11

Fd = 6 (pi) (eta) rv, r is radius, v is velocity, eta is viscosity of air.

Question 12

What is m for a spherical droplet?

Answer 12

m = 4/3 rho pi r^3, rho is density

Question 13

What is the weight of a spherical droplet?

Answer 13

mg, so 4/3 rho pi g r^3

Question 14

When does an oil drop fall at a steady speed?

Answer 14

With no electric field, as mg is cancelled by an upward viscous drag Fd.

Question 15

What is the radius of an oil drop, using Stokes’ law?

Answer 15

6 (pi) (eta) r v = 4/3 (pi) (rho) g r^3
so
r^2 = 9(eta v) / 2(rho g)

Question 16

What did Thompson find?

Answer 16

That e/m for the electron was much larger than for that of a proton (hydrogen ion).

Question 17

What did Thompson deduce from his e/m values?

Answer 17

That electrons were much much less massive than protons.

Question 18

How does deflection by a magnetic field for a cathode ray work?

Answer 18

Am electron beam moves in a circle, by a magnetic field out of the diagram - no work is done as force is perp to the velocity, forms circular motion.

Question 19

Do electric fields have effects on cathode rays?

Answer 19

No, discovered by Hertz

Question 20

What equations are linked to mag. field deflection for cathode rays?

Answer 20

F = Bqv = Bev.
F = mv^2 / r

Question 21

What do you get from combining the equations in mag. field deflection for cathode rays?

Answer 21

Bev = mv^2 / r
v^2 = (ber/m)^2, so
eV = 1/2 m(ber/m)^2
e/m = 2v/b^2r^2

Question 22

What should you measure to find e/m for mag. field deflection for cathode rays?

Answer 22

Anode V
Mag. B
Radius of circle r

Question 23

How does deflection by a magnetic AND electric field for a cathode ray work?

Answer 23

Electron beam is undeflected, as mag. and electric fields will cancel. Mag field into page, positive plate of e. field at top.

Question 24

What are the equations for deflection by an electric AND magnetic field for a cathode ray beam?

Answer 24

Fmag = Bev
Felec = ev
E = Vplates/d (plate separation)
v = E/B = Vplates/Bd

Question 25

What do you get from combining the equations for deflection by an electric AND magnetic field?

Answer 25

F = mv^2 / r = Bev
mv/r = Be so
e/m = v/Br

Question 26

What do you have to measure to find the e/m for an electric AND magnetic field?

Answer 26

Vplates
d, plate separation
B, mag field
r

Question 27

How does deflection by an electric field in a cathode ray work?

Answer 27

Electrons deflected, due to a vertical force. V should be separately measured when beam is not bent. Positive plate at top when electrons coming from left.

Question 28

What are the equations for deflecting a beam of electrons by an electric field?

Answer 28

Felec = eV
E = Vplates / d (plate separation)
a = F/m = eVplates/md

Question 29

How do you find a using equations for the deflection of a beam of electrons by an electric field?

Answer 29

deflection y = 1/2at^2, a = 2y/t^2
t = L/v

Question 30

How do you find v using equations for the deflection of a beam of electrons by an electric field?

Answer 30

v = E/B = vplates/Bd
also e/m = da/Vplates

Question 31

What do you have to measure to find the e/m for an electric field deflection?

Answer 31

d
Vplates
y
L
B