Turning Points

Question 1

What is meant by thermionic emission?

Answer 1

A filament is heated up using an electric
current, this causes the delocalised
electrons to gain energy. Eventually they
gain enough energy for the electron to leave
the metal surface as a beam of electrons

Question 2

How are cathode rays made in a

discharge tube?

Answer 2

Electrons are released by thermionic
emission. The electrons are repelled by the cathode
and accelerated towards an anode.

Question 3

Why is light emitted from a discharge

tube?

Answer 3

The strong potential difference between the anode and
cathode cause atoms in the discharge tube to ionise. When
this occurs the ionises atoms (which are now positive ions)
are attracted to the cathode. They accelerate towards and
then collide into the cathode. When this happens electrons
leave the cathode and go on to excite other atoms. When
these atoms de excite they release photons of light.

Question 4

What is the speed, v, of each electron leaving

the anode in a cathode ray?

Answer 4

The work done on each electron by the potential difference V between the
anode and the cathode is eV (e - electron charge).
The kinetic energy of each electron, with speed v, passing through the
hole is ½ mv^2
The work done on each electron increases KE at the cathode, then the
speed, v, of each electron leaving the anode is given by ½ mv^2=eV
So v= sqrt(2eV/m)

Question 5

How do you work out the specific charge

of an electron?

Answer 5

The charge of an electron / the mass of

an electron

Question 6

State 3 methods used to work out the specific

charge of an electron

Answer 6

Using a magnetic field
Using m = mv / Be
Using e / m = v / Br

Question 7

Why must electron tubes be evacuated

when working out specific charge?

Answer 7

So the electrons do not collide with air

particles and lose energy in the collisions.

Question 8

If gas is pumped into a electron tube,

why must the pressure be low?

Answer 8

A low pressure means less molecules which is beneficial
because too many gas molecules could disrupt the path
of the electrons. This is because the more air particles,
the more interaction between them and the electrons.
Which could mean the electrons won’t be able to travel
the whole length of the tube.

Question 9

Who was Thomson?

Answer 9

A physicist who conducted experiments to
investigate cathode rays, and the particles
that are produced. This included him carrying
out experiments determining the specific
charge of an electron.

Question 10

Thomson found out that the specific
charge of an electrons was how many
times larger than the specific charge of
an hydrogen ion? And why was the
significant?

Answer 10

1800 times
It was significant because before finding out
the specific charge of an electron, Hydrogen
ions had the largest known specific charge.

Question 11

Why was Thomson’s experiments

important?

Answer 11

Showed that electrons were negatively charged.
Showed that the specific charge of a particle is a
characteristic of that type of particle as all electron
has the same specific charge.
The specific charge was very high which showed
that an electron had to have little mass.

Question 12

What was the aim of Millikan’s

experiments?

Answer 12

To determine the charge of the electrons.

Question 13

In Millikan’s experiment, what forces are
acting on the droplet when it is
stationary?

Answer 13

Gravity and an electric force which is
equal and opposite to the gravitational
force. mg=QV/d

Question 14

In Millikan’s experiments, explain the
journey of a falling droplet when there is
no electric field?

Answer 14

The droplet will begin falling and
accelerating as it does the drag acting on it
will increase. Eventually the weight will
equal drag force and the droplet will fall at
terminal velocity.

Question 15

What is Stokes’ law?

Answer 15

F = 6πηrv
Which is used to work out the force on a
droplet due to drag.

Question 16

How is Stoke’s law used to work out the

radius of the droplet?

Answer 16

As the droplet is moving at terminal velocity the forces acting on
it will balance.
Therefore → 6πηrv = mg
We assume the droplet is a sphere which would suggest its
volume is 4𝛑r^3/3
Mass = density(ⲣ) /volume so 6πηrv = ⲣ x g /(4𝛑r^3/3 )
Which if you rearrange you get r^2
= 9ηv /2ρg

Question 17

How did Millikan use the value of the
radius to determine the charge of an
electron?

Answer 17

He used the radius to determine the mass of the
droplets. That way he could calculate the weight
of a droplet. He then know the pd required for an
object to remain stationary. Therefore the charge
could be worked out using QV/d = mg

Question 18

How did Millikan cause the droplet to
move down from stationary in the
presence of an electric field?

Answer 18

Milikan decreased the pd which would have
reduced the electric field. Therefore
unbalancing the forces and making the
resultant force on the object act downwards.

Question 19

Why was Millikan’s experiments

significant?

Answer 19

It helped him to conclude that charge of an electron, by
assuming that the charge on each droplet would be a multiple
of the charge of an electron. He found the common factor
between the different charges found that there were no
charges less than approx 1.6x10-19 Millikan showed that the
charge on all material is quantised.