Transcript Motion in a Straight Line

to Atomic and Nuclear
Physics
Purpose of the course:
To provide basic knowledge of the atom, its
constituents and nuclear processes
All these slide presentations are at:
http://www.hep.shef.ac.uk/Phil/PHY008.htm
Phil Lightfoot, E47, (24533) [email protected]
Most Important Thing !!!!!!
I’m always available to help with
any aspect of the course
Stop me if you’re confused
My contact details are on the top of
your lecture notes
Review of last lecture
Does an electron have positive or negative charge ? Negative
Because of its charge, in which direction does the electron move in the
diagram bottom left ? Opposite charges attract so moves towards positive 400V
What is the electric field in the diagram bottom left ?
E
What is the force on an electron in the diagram bottom left ?
V 400

 20 kV/m
d 0.02
Felectric  qE  (1.61019 )  20000 3.21015 newtons
In which direction does the force act ? Towards the positive voltage
How do we define the direction in which current flows? Other way to electron flow
What is Fleming’s left hand rule ? See diagram middle bottom
In which direction does the force on the electron act due to the magnetic field ?
Always at 90 degrees to both the field and the flow of current and here down page
What is equation for force on electron moving with velocity v in magnetic field ?
Fmagnetic  Bqv
Cathode ray tubes (CRT)
Before we go on we should mention a major use of ‘cathode rays’.
The screen of old fashioned TVs is coated on the inside surface with dots of
chemicals called phosphors. When a beam of electrons hits a dot, it glows.
These phosphor dots are in groups of three: Red, Green, and Blue which then
create all the other colours by combining which dots are illuminated.
There are 3 signals that control the 3 electron
beams, one for each RGB colour, each touching
the dots that the signal tells it to light. This is
then repeated for all the other pixels on the
screen by scanning across the screen.
Here we see a photo of a water droplet
acting as a magnifying glass on a TV screen.
See how the three colours of dots create the
overall pink.
Reminder of stuff covered in last lecture
In 1897 Thomson had for the first time measured two fundamental properties
of the electron using a combination of electric and magnetic fields. Of course
the best he could do was to measure the charge and mass as a ratio of one
another but it was better than nothing!!!!
q
yE

m DLB 2
Two of the most important features of any particle are its mass and charge.
Although this was a great measurement it’s clearly not very useful without
knowing each value individually!!!!!
Millikan’s oil droplet experiment
In 1909 Millikan had an idea how to find the charge on an electron
In order to understand what he did we need to remember a
little physics first….
Electric field : If we have two parallel plates a distance d apart
and the voltage difference between them is V, we can say that the
electric field is E where V is in volts, d in metres and E in Vm-1.
V
E
d
Force on a charged particle in an E field : If the particle has a F
 qE
charge of q, then the force on it is qE in newtons when q is given electric
in coulombs and E is given in Vm-1.
Force due to gravity : Force in newtons is equal to the mass of
an object in kg multiplied by ‘g’ which is 9.81 ms-2.
F  mg
Newton’s 1st law : An object at rest stays at rest and an object in motion stays in
motion with the same speed and direction unless acted on by an unbalanced force.
Millikan’s oil droplet experiment
All that stuff should be pretty clear apart from the last bit maybe…
Newton’s 1st law : An object at rest stays at rest and an object in motion stays in
motion with the same speed and direction unless acted on by an unbalanced force.
If you jump out of a plane what happens to your velocity
as you fall to earth and why?
What are the forces acting on your body as you fall?
If you are at terminal velocity what can you say about these forces?
The net force is zero so the force of air resistance and weight are the same.
Millikan’s oil droplet experiment
Let’s have a look at what Millikan did ….
Instead of jumping out of a plane he squirted some oil vapour, watched it fall,
and measured its terminal velocity by timing it between two points.
The drag force on the drop is Fdrag  6 r  v1 where v1 is the terminal velocity of
the falling drop, η is the viscosity of the air, and r is the radius of the drop.
The volume of a sphere is : V 
Density is :  
M
V
4 3
πr
3
4
and its weight is : W  Mg   r 3  g
3
When the droplet is at terminal
velocity we can write: F
W
drag
4
3
So : Fdrag  6 r  v1  3  r  g
and by rearranging we can say:
r2 
9  v1
2 g
Millikan’s oil droplet experiment
Since r 
2
9  v1
4 3
we can find the weight of the oil droplet since W  Mg   r  g
3
2 g
Now as the oil droplets were slowly falling down, he decided to switch on an electric
field with the positive voltage at the top.
As he increased the strength of the field he noticed that some of the oil droplets
slowed down and some actually started rising!!!!
The only way this could be explained
was by saying those droplets had a
negative charge on them.
What is the force due to the electric
field on an oil droplet of charge q ?
Felectric  qE
What can we say about the forces on
the droplet when the electric field is
increased so that the droplet remains
stationary?
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=357.0
Millikan’s oil droplet experiment
What can we say about the forces on the droplet when the electric field is increased so
that the droplet remains stationary?
Since the droplet is stationary there is no drag force and so weight must be exactly
equal to the force due to the electric field.
4
W  mg   r 3  g
3
Felectric  qE
Since we have already worked out that r 2 
9  v1
2 g
3
2
we can say that W  m g  4   9 v1   g
3  2  g 
Since
Felectric  W when the droplet is stationary …
3
2
3
2
4  9 v1 
4  9 v1 
  g and therefore … q 
  g
qE   
 
3  2 g 
3E  2  g 
We can measure all values on right hand side and so can get a value for q, the charge
on the droplet.
Millikan’s oil droplet experiment
We can measure all values on right hand side and so can
get a value for q, the charge on the droplet.
q
3
2
4  9 v1 
  g
 
3E  2  g 
That’s really interesting and everything ….. But how does this help find the charge of
an electron ? After all, what you’ve just found is the total net charge on an oil droplet !!!
What Millikan found when he looked at the values of q for many many oil droplets was
that they were all integer multiples of 1.6 × 10-19 C.
The fact is that when you spray oil droplets
(he used his mother’s perfume atomiser) you
don’t charge them much. You just add one or
two electrons to the droplet (not more than 10
at most). So the charge represented integer
multiples of the charge of a single electron
which Millikan stated was 1.6 × 10-19 C
Remember that Thomson had calculated the
charge to mass ratio? This was used to find the
value for the mass of an electron as 9.1 × 10-31 kg
http://physics.wku.edu/~womble/phys260/millikan.html
Electron-Volts
This is such a tiny topic but it is also one of the most important ones. You will use
electron-volts in two main ways; (i) as a convenient unit of energy, and (ii) to
calculate the velocity of a charged particle passing through an electric field.
Definition: The change in energy of a charged particle q as it moves through a
potential difference V is given by :
Energy qV
1 electron-volt =1eV =1.610-191 = 1.610-19 J.
(i) In particle physics we often deal with very small energies and it is often
therefore more convenient to refer to 3.210-19 J as 2 eV for example.
(ii) Imagine a particle of mass m and charge q is accelerated from rest through a
voltage V. The energy given to the electron is therefore
Energy qV
If all this energy is converted into motion then this will be equal to the final kinetic
energy of the particle. So
1
mv 2  qV and the final velocity of electron is:
2
v
v is in metres per second, V is in volts, q is in coulombs, m in kilograms.
2qV
m
Let’s try a couple of potential exam questions
Electrons are emitted from the cathode in an evacuated tube. The electrons start from
rest and are accelerated through a potential difference of 1150V.
What is the speed of the electrons when then arrive at the anode?
(Charge on an electron is 1.6 × 10-19 C and mass of electron is 9.1 × 10-31 kg).
Answer: The velocity of the electrons when they reach the anode is
2qV
2 1.61019 1150
7
1
v


2

10
ms
m
9.11031
v
2qV
m
Let’s try a couple of potential exam questions
a) E 
V
310

 62000 Volts per metre
d 0.005
b) Sphere is stationary so we know that
Felectric  W so qE  m g
m g 3 1015  9.81
q

 4.81019 C This is the charge of 3 electrons.
E
62000
Let’s try a couple of potential exam questions
a)
Fmagnetic  Bqv Felectric  qE
When the electron travels in a straight line then forces must balance.
E
Bqv  qE so B 
v
so
E 1.1104
3
B 

1.1

10
teslas
7
v
110
Let’s try a couple of potential exam questions
b) If the electric field is switched off then the only force is
Fmagnetic  Bqv
This force always acts at 90 degrees to the magnetic field and direction of travel.
The electron feels a centrifugal force as it starts to curve. Once the force due to the
magnetic field has been balanced by the centrifugal force, the electron rotates in a
perfect circle.
m v2
Forces balance and so :
Bqv 
R
31
7
mv
9.1

10

1

10
So R 
so R 
 0.051m
3
19
Bq
1.110 1.610