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Title: all summary notes
Description: these are all summary notes for physics all type of classes
Description: these are all summary notes for physics all type of classes
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Higher Physics: Mechanics and the properties of matter
Summary Notes
Vectors
Page 2
Equations of motion
Page 7
Newton’s Second Law, energy and Power
Page 14
Momentum and Impulse
Page 18
Density and pressure
Page 26
Gas Laws
Page 30
Vectors
A physical quantity is a measurable quantity
...
There are two kinds of physical quantity
...
They have to be recognised and treated
differently
...
Scalar quantities are combined using simple arithmetic
...
Vector quantities are added together in triangles
...
velocity
mass
10 N
displacement
speed
5 km
VECTORS
momentum
distance : A distance is simply the length of a journey
...
Speed / velocity
speed : The speed of an object is the rate at which distance
is increasing
...
Velocity should always be quoted with a direction
...
The combined effect of a number of vectors can be found by adding the vectors together
...
The resultant vector is a single vector which represents the combined effect of a number
of vectors and can be used to calculate the effect of those vectors
...
Find the resultant displacement of a journey consisting of the following
displacements: 100 m North; 200 m East; 500 m South
...
Length of resultant = 17
...
The air it
O
is flying through has a velocity of 90 knots in a direction 240
...
N
000
270
300 kt
090
90 kt
180
resultant
300 kt
Sketch
Scale 5 cm = 100 kt
length of resultant = 11 cm
magnitude = 220 kt
nt
re
lta
su
90
300
61O
Resultant velocity is 220 knots 061
O
Points to Note :
1
...
Draw a sketch to work out how the
scale drawing will look
...
Choose a scale to make best use of
the available paper
...
4
...
5
...
Higher Physics : Mechanics and the Properties of Matter
Page 3
Rectangular Components of a vector
...
This simplifies the
analysis of the effect of vectors on objects
...
The sum of the component vectors will have a resultant equal to the original vector
...
y
y
Ay
A
B
B
x
C
Bx
By
A
Ax
x
Cx
C
Cy
Vectors replaced by
rectangular components
A sin q
y
A
q
A cos q
x
The process of replacing a vector by its
rectangular components is known as
resolution
...
Once the vector has been resolved, the
original vector is replaced
...
Example : An object is projected with a velocity of 60 m
...
Calculate the vertical and horizontal components of the
velocity
...
s
-1
vertical component = 60 sin 60 O
= 52 m
...
s
O
60 cos 60 O
O
horiz
Higher Physics : Mechanics and the Properties of Matter
Page 4
Inclined Plane
...
The forces acting on an object, moving
down an inclined plane, are shown in
Fig
...
N is the force of the plane on the object
...
N is always found acting at right angles to
a surface
on
ti
ric
F
q
q
To solve this problem we resolve
the weight W into two rectangular
components
...
The problem then reduces to
W
Unbalanced force F, acting
down the plane
F = W sin q - Friction
Also N = W cos q
W
N
ion
ict
r
W cos q
F
W sin q
nq
W
si
Higher Physics : Mechanics and the Properties of Matter
sq
w
co
Page 5
Equations of motion
Measurement of acceleration
Acceleration (a) : Acceleration is the change in velocity per unit time
...
Measurement of acceleration : To measure the acceleration of an object we require to measure
the velocity of the object at two points in its journey and the time taken to travel between the
two chosen points
...
er
tim
Light gate 2 provides time
for final velocity
...
change in velocity
acceleration =
time
er
tim
1
ate
ht g
lig
e2
gat
ght
li
Double obstructor and single
light gate
The motion computer
measures the time taken for the
first segment of card to pass the
light gate and the time taken
for the second segment
...
From this information and the
length of the segment, the
computer works out the
acceleration
...
As we are only
dealing with linear motion, direction will be indicated by ‘+’ and ‘-’ signs
...
The motion of an object can be described using a displacement - time graph, a velocity - time
graph or an acceleration - time graph
...
For Higher work you will be expected to derive an acceleration - time graph from a velocity - time
graph
...
5
velocity / ms
-1
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
acceleration / ms -2
+5
0
-5
constant
acceleration
Higher Physics : Mechanics and the Properties of Matter
Page 8
Equations of motion
The equations of motion are used to describe motion in a straight line involving uniform
acceleration
...
Equation 1 :
a =
v-u
t
v = u + at
velocity
v
Area B
u
Area A
time
t
Equation 2 :
displacement, s = area under velocity - time graph
= area A + area B
= ut + ½ t (v - u )
= ut + ½ t
...
(v - u) = at
= ut + ½ at
from Eq
...
2
a
2
s = u (v - u) + ½ a (v -2u)
a
a
2
s = u (v - u) + (v - u)
2a
a
substitute t =
multiply both sides by 2a
2
2as = 2u(v - u) + (v - u)
2
2
= 2uv - 2u + v -2uv + u
= v
2
- u
2
2
rearranging
v
2
= u
2
+ 2as
Higher Physics : Mechanics and the Properties of Matter
Page 9
-1
Example: A trolley is projected up a slope with a velocity of 12 ms
...
Find (a) velocity of the trolley after 6 s
...
(a)
-1
v = u + at
v = 12 - 3 x 6
= - 6 ms -1
u = 12 ms up the slope
a = - 3 ms -2 (down the slope)
t = 6s
velocity after 6 s is 6 ms -1 down the slope
...
5 x 3 x 6 x 6
= 18 m
displacement after 6 s is 18 m up the slope
...
Projectile motion is motion in two dimensions
...
v
vertical component vV = v sin q
v
vV
q
vH
q
horizontal component vH = v cos q
The horizontal component is not affected by gravity and remains constant throughout
the flight
...
-1
O
Example: An object is projected with a velocity of 50 ms at an angle of 20 to
the horizontal
...
Find horizontal and vertical components of initial velocity
...
8t
v = 0
u = 17 ms
-1
t = 1
...
8 ms
-2
time to maximum height = 1
...
Higher Physics : Mechanics and the Properties of Matter
Page 11
s = ut + ½ at
(a) continued
2
= 17 x 1
...
5 x 9
...
73 2
= 29
...
7
= 14
...
7 m
(b) maximum height is reached at mid point of journey so that the time taken for the
complete flight is twice the time taken to reach the maximum height
...
73
= 162
...
6 m
If there was no requirement to find the time of flight we could have used equation of
motion no
...
v 2 = u 2 + 2as
= 17 2 - 2 x 9
...
8
0
= 14
...
Higher Physics : Mechanics and the Properties of Matter
Page 12
Newton’s Second Law, energy and power
...
Newtons’ Second Law deals with the effect of forces on moving objects, and provides
a means to define the unit of force
...
a
resultant
force
mass
acceleration
-2
If the mass is measured in kg, and the acceleration is measured in ms , then force
is measured in Newtons (N)
1 Newton is the resultant force which will accelerate an object with mass 1 kg at a
rate of 1 ms -2
...
A free body diagram is a diagram which shows all the forces acting on an object
...
Example
...
Calculate its acceleration at lift off
...
8
= 52000 N
10000 kg
F = m
...
2 ms -2
acceleration at take off is 5
...
A block, mass 2 kg, slides down a slope which is inclined at an angle of 20 O to
the horizontal
...
5 ms -2 down the slope
...
tion
ma
nor
ac
l re
fric
Free body
diagram
...
We have now to deal with 4 forces but
the arrangement is less complicated
...
This means the resultant force acts down the
slope
...
So the Resultant force, F = w sin 20 - friction
m
...
a
2 x 9
...
5
6
...
0
friction = 4
...
7 N
Higher Physics : Mechanics and the Properties of Matter
Page 14
Work Done
...
The Work
Done is given by:
work done = force applied x distance moved
...
F
q
F cos q
s
E W = F cos q x s
Work Done is a scalar quantity measured in Newton metres ( Nm)
When work is done on an object, energy is transferred to the object (and the surroundings
if friction is involved)
...
One Joule of energy is transferred for each Newton metre of work done
...
To measure the
energy we need to convert all the energy to work done and measure the work done
...
The quantity of energy transferred to
heat is equal to the work done overcoming the force of friction
...
F x s = ½ mv 2 + mgh + Ffrict x s
typically
...
The Conservation of Energy applies to situations involving work done
...
Higher Physics : Mechanics and the Properties of Matter
Page 15
Example
...
He moved
-1
the wheelbarrow a distance of 20 m reaching a speed of 4 ms
...
Work done = kinetic energy gained + work done overcoming friction
...
5 x 50 x 16 + F x 20
2000 = 400 + F x 20
F = 80 N
500
kg
Force of friction = 80 N
A 500 kg mass is dropped a distance of 1 m to
strike the top of a concrete pile
...
Find
the average driving force on the pile
...
1m
mgh = F
...
8 x 1
...
2
note: potential energy
lost by falling mass is
given by:
F = 29400 N
E = 500 x 9
...
2
Average force = 29400 N
Higher Physics : Mechanics and the Properties of Matter
The mass dropped a
further 20 cm driving
the pile
Page 16
Momentum and impulse
Momentum
The momentum of a moving object is the product of its mass and velocity
...
momentum = mass x velocity
p = m
...
A resultant force will cause a change in velocity (F = ma)
...
Collisions
When two objects collide, the collision generates forces which act on both objects
...
If there are no external forces acting on the objects and the only forces are those created by
the collision, the vector sum of the momentums before the collision is equal the vector sum
of the momentums after the collision
...
Conservation of Momentum
...
A
VA
A
UA
UB
before the collision
B
B
VB
mA uA + mB uB = mA vA +
after the collision
mB vB
vector addition
External forces which might be involved include gravity, friction and electric forces
...
After the collision both vehicles are
locked together
...
Sum of the momentums before the collision
...
(800 + 1500) x v = 39000
-1
v = 17 ms ( rounding up )
Example:
1
...
2 g, is fired horizontally to strike a piece of modelling clay
mounted on a stationary air track vehicle
...
67 ms-1
...
Calculate the speed of the air gun pellet
...
2 x 10-3 x v + 350 x 10-3 x 0 = 1
...
2 x 10 + 350 x 10 ) x 0
...
2 x 10-3 + 350 x 10-3 ) x 0
...
2 x 10-3 x v
v = 196 ms-1
Higher Physics : Mechanics and the Properties of Matter
Page 18
Explosions
...
With no external forces, the Conservation of Momentum applies
...
Mv
mV
Recoil of a gun when fired
Mv
mV - Mv = 0 ( original momentum of gun )
Rocket engines create a continuous explosion
...
mV
Elastic and inelastic collisions
An elastic collision is one where there is no loss of Kinetic Energy during the collision
...
Kinetic energy is lost during an inelastic collision
...
Total Energy is always conserved during collisions
...
0 ms-1
0
Example
A
B
400 g
Before
600 g
1
...
0 ms
...
6 ms-1
...
(b) Show that this was an elastic collision
...
4 x 2
...
6 x 0 = 0
...
4 x VA + 0
...
6 kg ms
-1
-1
Apply the conservation of momentum
0
...
6 x 1
...
8
0
...
8 - 0
...
16
VA = - 0
...
4 ms-1 to the left
...
5 x 0
...
0) = 0
...
5 x 0
...
4)2 + 0
...
6 x (1
...
032
+
0
...
8 J
No kinetic energy has been lost during the collision so collision is elastic
...
This is a direct result of the conservation of momentum
...
mA uA +
rewritten
mB uB =
mA ( vA - uA )
m A vA +
m B vB
= - mB ( vB - uB )
change in momentum A = - change in momentum B
If objects are in contact for time t seconds
...
They are equal and
opposite at any time during the collision
...
F = m(v-u)
t
If we rewrite this:
Ft = mv - mu = change in the momentum of the object
...
Impulse
is measured in Newton seconds (Ns)
Impulse ( J ) = Force x time for a constant or average force
...
Where the force is not constant
...
4 m
Time between lightgates = 0
...
4
= 0
...
56
= 0
...
114
F x 25 x 10
F
-3
= 0
...
6 N
Average force on ball = 4
...
Objects which may
be involved in collisions need to be designed to cope with the situation
...
This has the effect of decreasing the magnitude of the forces involved
...
Calculate the average force acting on the car during the collision
...
The maximum force will
be at least double this value
...
This reduces the average and maximum forces acting
...
If the collision time was increased to 4 seconds
F t = change in momentum
F x 4 = 1200 x 13
F = 3900 N
The force acting on a passenger, mass 70 kg, would be reduced from 970 Newtons
to a safer 243 Newtons
...
A seat-belt is worn!!
Higher Physics : Mechanics and the Properties of Matter
Page 24
Pressure and density
Density
The density of a substance is the mass per unit volume of that substance
...
density =
r
mass
volume
= m
V
r
Greek letter ‘ro’
Example: Find the density of mercury, if 200 cm3 of mercury has a mass of 2
...
72
-6
200 x 10
-6
m
3
m = 2
...
Air is pumped into a RIGID container then valve is closed
...
Container is detached from pump and its mass measured using
a SENSITIVE balance
...
The extra volume of air inside the container is measured by bubbling
the air into a measuring cylinder as shown
...
The mass of the container is measured again
...
2 kg m
-3
Higher Physics : Mechanics and the Properties of Matter
Page 25
Solid, Liquid and Gas
When comparing the densities of the different phases of a substance, we find that the
densities of the solid and liquid phases are roughly equal, but the density of the gas
is 1000 times less than the liquid or gas
...
The difference in density is due to the difference in the spacing between
the particles
...
10
10
solid / liquid
1
1
1
gas
volume = 1 x 1 x 1
10
volume = 10 x 10 x 10
Solid carbon dioxide can be collected
from a fire extinguisher and formed
into a 1 cm cube
...
3
Approximately 1000 cm of gas is
collected
...
Pressure is the force
per unit area, when the force acts normal to the surface
...
Pressure is measured in pascals, where 1 pascal = 1 newton per square metre
...
Calculate the pressure on the
surface due to the block
...
4
0
...
3 Pa
F = weight of block
= mg
= rVg
= 11343 x 20x15x30 x 10 -6 x 9
...
3 Pa
= 1000
...
06 m 2
O
Calculate the pressure when the surface is inclined 30 to the horizontal
...
4 x cos 30
0
...
5 Pa
Higher Physics : Mechanics and the Properties of Matter
w cos 30
W
30
O
Page 27
Pressure in Liquids
The pressure in a liquid or gas increases with depth
...
Pressure at a point in a liquid or gas is created by particle collisions on
a surface
...
This means that the pressure
on the surface will be the same for any orientation of the surface
...
Pressure is a scalar quantity
...
An object, immersed in a liquid or
a gas experiences a force which
acts vertically upwards
...
Pressure increases
with depth
...
p = rgh
= 1030 x 9
...
8 N kg -1
h = 400 m
= 4
...
Higher Physics : Mechanics and the Properties of Matter
Page 28
Gas Laws
The Kinetic Model Gases are composed of freely moving particles, moving around
randomly a high speeds
...
The only energy carried by the particles is kinetic energy which can
only be changed by changing the temperature of the gas
...
The Gas Laws define the relationship between the pressure, volume and temperature
of a fixed mass of gas
...
p
1
V
p1 V1 = p2 V2
Higher Physics : Mechanics and the Properties of Matter
1/V
Page 29
Charles’ Law
capillary
tube
ruler
The volume of a fixed mass of gas at constant
pressure varies directly as its temperature
measured on the kelvin scale
thermometer
T ( K ) = T ( OC ) + 273
V
T
mercury
V1
V2
=
T1
T2
trapped air
heat SLOWLY
volume
volume - temperature ( OC )
volume
-300
-200
-100
0
200
300
temperature / K
+100
400
temperature / OC
volume - temperature ( K )
0
100
Higher Physics : Mechanics and the Properties of Matter
Page 30
Pressure Law
thermometer
The pressure of a fixed mass of
gas at constant volume varies
directly as the temperature of
the gas measured on the
Kelvin scale of temperature
...
p1 x V1
p2 x V2
=
T1
T2
Kinetic Model
Pressure The pressure exerted by a gas on the sides of its container is caused by
gas particles colliding with the walls of the container
...
The particle bounces off
the wall, the wall experiences a tiny force
...
The magnitude of the pressure depends on the number of collisions per second and
the average force per collision
...
Bring walls closer
together and particle
takes less time to
cover distance between
bounces
gas particle
bouncing
between the
walls of its
container
...
Gas Laws and the Kinetic Model
...
This decreases the number of collisions per
second and with it, the pressure on the walls
...
Charles’ Law Increasing the temperature causes the particles to speed up
...
To maintain a constant pressure, the volume increases to reduce the number of collisions
per second
...
Pressure Law Increasing the temperature causes the particles to speed up
...
If the volume remains constant, the pressure will increase
...
Higher Physics : Mechanics and the Properties of Matter
Page 32
3
Example A syringe contains 50 cm of air trapped air at a
temperature of 20 OC
...
Calculate the new volume of the trapped air
...
trapped
air
3
V1
V2
=
T1
T2
V 1 = 50 cm
O
T 1 = 20 C
= 273 + 20
= 293 K
T 2 = 100 OC
= 273 + 100
= 373 K
50
V2
=
293
373
V 2 = 50 x 373
293
3
= 63
...
2 m 3 Hydrogen gas at a
pressure of 200 x 10 5 Pa
...
How long should the bottle last before needing
to be changed?
H2
Assume the temperature remains constant
...
2 m
5
3
5
200 x 10 x 0
...
2 = 39
...
8 x 10 seconds
100
5
= 3
...
5 hours
...
5 x 105 Pa
...
Find the maximum temperature it can
be safely exposed to
...
5 x 10 Pa
T2
T 1 = 20 OC
5
1
...
0 x 10
=
293
T2
5
= 20 + 273 = 293 K
5
293 x 3
...
5 x 10
p 2 = 3
...
Electric fields are found around charged objects
...
The lines give the
direction of the force on a positive charge
...
-
+
Weak positive charge
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
Strong negative charge
+
-
Dipole field
Uniform field between
two charged metal plates
Higher Physics : Electricity and Electronics
Page 1
Potential Difference
it does not matter
what path is taken
between A and B
+
A
B
A charge in an electric field experiences a force
...
If 1 Joule of work is done moving 1 coulomb of positive charge between A and B,
then the potential difference between A and B is 1 Volt
...
EW = Q V
EW
V =
Q
The potential difference between two points is the work done per coulomb of charge
when charge is moved between those two points
...
The electrons are provided by an
arrangement called an electron
gun
...
The electrons pass through a gap
in the anode to form a beam
...
e
Example: The potential difference between the anode and the cathode of an electron gun
is 20 kV
...
Work done by the field moving the electron is converted to kinetic energy
Q V = ½ m v2
1
...
5 x 9
...
6 x 10
2
C
-31
-19
v = (
-16
1
...
5 x 9
...
11 x 10 kg
V = 20 kV
= 20000 V
= 5
...
By subjecting the electrons to an opposing field it is possible to measure
the maximum kinetic energy
...
In one case the required pd was 6
...
maximum kinetic energy = work done by field in stopping the electron
=QV
-19
= 1
...
3
= 1
...
Conductors contain electric charges which are
easily moved (electrons)
...
+ + + + + + + +
- + - - - + - + - + + + - + +
- - - - - - - Conductor : positive and
negative charges uniformly
distributed
- ---+ + +
-+ + +
-- - - -
+ + + + +
+ + + + +
in an electric field, negative charges move
in response to the force exerted by the field
...
The work done comes from
the electrical energy given to the
charge as it passes through the source
...
m
...
m
...
m
...
Q = I t I is the current flowing in the circuit
Substituting :
W =ItV
This is more familiar as
W
= P = VI
t
P = V I and W = Q V are equivalent expressions
...
d
...
d
...
d
...
m
...
d
...
The behaviour of the source can be predicted if we assume the source consists of a source
of constant e
...
f with a small internal resistor in series with it
...
m
...
0
DV
1
...
d
...
This is termed
the short - circuit current and is
given by
1
2
3
4
5
6
7
current / A
8
9
V = E when no current is being drawn from
source
...
m
...
d
...
E
IS = r
Higher Physics : Electricity and Electronics
Page 5
Example When a voltmeter is connected across the terminals of a battery it reads 9
...
when the battery is connected in series with a 5 ohm resistor, the voltmeter reads 8
...
Find (a) the e
...
f of the battery
...
(a) The e
...
f of the battery is 9
...
m
...
(b)
V = 8
...
2 V
V
8
...
7 A
8
...
2 - 1
...
2 - 8
...
7
= 0
...
d
...
0 = E - Ir
IS =
9
...
41
= 22
...
The
p
...
across the circuit when this happens is ½ E
...
Higher Physics : Electricity and Electronics
Page 6
Resistors in Series and Parallel
Conservation of energy : The energy suppllied to unit charge as it moves round the
circuit is equal to the work done moving the charge round the circuit
...
m
...
s round the circuit
...
The sum of the e
...
f
...
d
...
source
I
A
B
R1
C
D
R2
R3
In the circuit above
VAD = VAB + VBC + VCD
Conservation of energy
Dividing both sides by I, the current flowing round the circuit
...
source
Conservation of Charge
The current flowing into a junction
must equal the current flowing
out of a junction
...
d
...
500 W
500 W
500 W
500 W
Find the combined resistance of the arrangement shown above
...
R T = 500 + 333
500 W
333 W
= 833 W
Total resistance of above arrangement = 833 ohms
...
Higher Physics : Electricity and Electronics
Page 8
Wheatstone Bridge
A Wheatstone Bridge consists of two
potential dividers in parallel
...
RX
V IN
RY
V OUT
Potential Divider
Balanced Wheatstone Bridge
R1
R3
+
V
R2
-
The Wheatstone Bridge is balanced when
the voltmeter reads zero
...
R4
R1
R3
=
R2
R4
The Wheatstone Bridge was originally used as a means of measuring resistance in a
situation where accurate resitors were available but no accurate meters
...
No current measurements are needed
...
With switch S open, the voltmeter is less
sensitive
...
VR is adjusted till voltmeter reads zero
...
Final adjustments are made
to VR
...
Higher Physics : Electricity and Electronics
Page 9
Example
Find the value of R X to
balance the bridge
...
3 kW
V
RX
15
...
3
RX =
830W
RX
15
...
7
10
...
The variable resistor VR is adjusted to balance the bridge
...
Calculate the resistance of VR
...
Condition for Balance
VR
150
=
VR =
1380
1707
150 x 1380
1707
= 121 W
Higher Physics : Electricity and Electronics
Page 10
The Unbalanced Wheatstone Bridge
...
Graph 2
+V
bridge voltage
If a Wheatstone Bridge is balanced then
the value of one of the resistors is changed,
A voltage will be recorded on the voltmeter
...
For small variations: no
more than 5% change in the value of the
resistor, the graph is a straight line through
the origin (Graph 2)
...
These include strain gauges and resistance
thermometers
...
The resistance of the foil (usually 120 ohms)
depends on the length of the foil
...
When stuck on metal, any stretch in the metal
caused by forces changes the resistance of
the strain gauge,
Higher Physics : Electricity and Electronics
Page 11
Alternating current and voltage
Measuring frequency using a CRO
...
This is usually scaled in seconds per screen division
...
Multiplied by the time base setting, this measures the period of the signal
...
200 ms
500 ms
1 ms
100 ms
50 ms
2 ms
ext horiz
...
1
period
frequency =
period = time for 1 cycle
= time to move 3 divisions
= 3 x 200 x 10 -6 s
1
3 x 200 x 10-6
=
= 1667 Hz
AC Supplies
+V
VP
peak voltage
0
time
-V
ac voltage
The voltage of an a
...
supply changes between positive and negative
...
The average
voltage, over time, is zero
...
V2
R
2
average V
P =
R
For dc supplies
P =
For ac supplies
V2
We can find a
average value
for V2
...
c
...
The average value over time is
time
Vp2
2
2
Vp2
V
This means that for ac P =
= rms
R
2R
Where :
Vrms =
VP
2
V P is the peak ac voltage
The r
...
s voltage of an a
...
source is the equivalent d
...
voltage which will produce
the same heating effect when applied across a resistive load
...
m
...
at a frequency of 50 Hz
...
Vrms =
Vpeak
2
Vpeak = 2 x Vrms
= 1
...
c
...
c
...
m
...
voltage and current
...
c
...
m
...
values
...
m
...
voltages and current are unaffected by the frequency of the a
...
supply
...
c
...
The resistance of a resistor is unaffected by the frequency of an a
...
supply
...
Both meters read r
...
s values
...
c
...
c
...
When dealing with a
...
voltage, we need to be aware of the peak value of the a
...
rather than the quoted os
voltmeter measured r
...
s value
...
Diodes are used to convert a
...
current into d
...
current : rectification
...
c
...
Semiconductor diodes can only be reversed biassed up to their peak inverse voltage (PIV)
rating
...
current
+
-
-
Forward biassed
diode conducts
+
Reverse biassed
diode does not conduct
diode
R
cro
cro
rectified a
...
voltage
a
...
voltage
Example Calculate the maximum a
...
supply voltage which can be rectified by a diode with
a PIV rating of 100 V
...
41
= 70
...
7 V r
...
s
...
Most capacitors consist of two sheets
of metal foil separated by a thin layer of dielectric material
...
In this way,
large areas of foil can be incorporated into a
small space
...
The work done
required to move the charge( electrons ) is
provided by an external source
...
This causes the p
...
across the plates
to increase as the stored charge increases (p
...
is the work done moving unit charge)
...
d
...
d
...
v
+
+
+
e
+
+
+
e
Charge cannot pass through the capacitor, it can only be moved externally though a circuit
connecting the plates
...
A charged capacitor acts like a battery, and can be discharged through an external circuit
...
This means that it is possible to discharge enormous currents over
tiny time intervals
...
Higher Physics : Electricity and Electronics
Page 16
Capacitance
Charge and p
...
across a capacitor
R
Switch, S, is set to x, and the
capacitor charged up
...
d
...
The switch is set to y, and the
capacitor discharged through
the coulombmeter
...
This is repeated for
different p
...
s across the
capacitor
y
S
coulombmeter
V
C
Q = It
Q = CV
C is the capacitance of the capacitor
...
One Farad is equal to one coulomb
per Volt
...
charge stored in capacitor
The quantity of charge stored in a
capacitor varies directly as the p
...
across the capacitor
...
d
...
d
...
1 C
Higher Physics : Electricity and Electronics
Page 17
Energy stored in a capacitor
p
...
across capacitor / V
Work must be done to charge a capacitor
...
Work has to be
done by the external source to overcome the repulsion and move charge onto the plates
...
V
Work done moving
small charge DQ
is equal to area of
strip V x DQ
Q
charge stored in capacitor / C
Suppose a capacitor is given a charge Q coulombs and that it has now got a p
...
A tiny amount of charge DQ is now moved from one plate to the other
...
V x DQ Is the area of the small strip on the graph
...
This is simply the total area under the graph,
work done charging a capacitor = ½ Q x V
Energy stored in a capacitor = ½ Q x V
substituting Q = C x V
Q
substituting V = C
= ½ C x V2
=
Q2
2C
The energy is stored in the stretched molecules of the material between the plates
of the capacitor
...
Example How much energy is stored in a 1000 mF capacitor when there is a p
...
5 x 1000 x 10-6 x 6 x 6
= 1
...
The capacitor is charged up so that there is a p
...
of 300 V across it, and then discharged
though the tube to produce a bright flash of light
...
(a)
E = ½ C x V2
-6
= 0
...
45 Joules
(b)
E
t
0
...
This level of output will generate a powerful but short-lived flash of light, ideal for
photography
...
After 12 seconds
the p
...
across the capacitor is 5 V
...
Q = CV
Q = It
= 2 x 10-3 x 12
0
...
024
5
= 0
...
0048 F
= 4800 mF
Charging / discharging a capacitor
Charging
S
+V
charging p
...
VC
R
Capacitor is fully
discharged at
start
Battery has negligible
internal resistance
V
C
0
charging current
IC
When switch S is closed, the p
...
across the
capacitor is zero as it has no charge
...
d
...
The current flowing
through the resistor is V/R
After a time, the capacitor has accumulated
charge and the p
...
across it has increased
to VC
...
When fully charged, the p
...
across the capacitor
is a steady V
...
V
R
0
+V
R
discharging p
...
S
V
discharging current
0
0
When switch, S, is closed, the capacitor behaves
like a small battery and discharges through the
resistor
...
d
...
The discharge current is in the opposite direction
to the charging current ( hence -ve
...
time
VC
Discharging
The capacitor is fully
charged with a p
...
of V
across it
...
However they do
affect the shape of other
types of signal like the
square wave shown
below
...
Small R + small C
Input Signal
Small R + Large C
Large R + small C
The change in the p
...
across
the resistor R
The signal across the resistor R shows the
changes in the current flowing in the circuit
...
The battery has an e
...
f
...
The capacitor is
initially uncharged
...
+9V
S
5 mF
(a) Calculate the current flowing in the circuit immediately after the switch is closed
...
d
...
Calculate the current flowing
in the circuit
...
(a)
Immediately after the switch is closed, the p
...
across the capacitor is zero
...
d
...
Current flowing in the resistor is given by
V
R
9
=
550
= 0
...
d
...
d
...
V
R
6
=
550
= 10 mA
I =
(c)
When fully charged, the p
...
across the capacitor is 9 V
Q = CxV
-6
= 10 x 10 x 9
= 9 x 10 -5 coulombs
Higher Physics : Electricity and Electronics
Page 22
Capacitor current and frequency
A
A variable frequency a
...
supply is placed across
a capacitor
...
d
...
The current flowing through the capacitor is
measured using an a
...
ammeter
...
d
...
IC
f
frequency
The resistance to a
...
current in a capacitor falls as the frequency of the a
...
increases
...
1mF
C
VC
VC
10 kW
R
0
0
...
c
...
The capacitor filters out the high frequency signal, the resistor the low frequency signal
...
c supply voltage
diode
R
cro
Rectified a
...
voltage
current from
capacitor
diode
R
C
cro
Smoothed rectified voltage
...
Capacitors are also used as supressors
...
These cause noise on radio and TV sets
...
Higher Physics : Electricity and Electronics
Page 24
Coupling capacitors
...
c
...
c
...
c
...
c
...
Capacitors
are used to pass signals from one stage in a circuit to another where the d
...
added by one
stage in the processing is removed
Capacitors in radio
C2
Simple AM
receiver
C1
tuner
modulated carrier wave
from tuner circuit
detector
AF amplifier
rectified carrier wave
smoothed rectified carrier
C1 smooths rectified
carrier wave to restore
af signal
C2 removes d
...
AF signal to amplifier
Higher Physics : Electricity and Electronics
amplified AF for earpiece
Page 25
Analogue Electronics
Operational Amplifier
+ VS
An operational amplifier
is a small integrated
circuit which is frequently
used in measuring
instruments
...
The output current is limited to a
maximum of a few milliamps
...
The basic op-amp, with its massive gain, is very unstable
...
If the op-amp is going to be used
as an amplifier, then the gain has to be reduced
...
R f feedback resistor
input resistor
R1
+VS
+VIN
-VOUT
-VS
0V
With the ‘+’ input connected to 0 V, the equation above becomes
VOUT = A ( 0 - VIN )
= - A VIN
If VIN is positive, then VOUT will be negative
...
This set up is an op-amp used in the INVERTING MODE
...
It
has no effect on any voltage it measures and its output can be passed on with no effect
on the next stages
...
1
...
There is no potential difference between the inputs (infinite gain means
that for any output voltage the corresponding voltage across the
input terminals is small; virtually zero
...
At junction X, no current will flow into the op-amp, so all the input current will flow
through both resistors
...
R1
0V
+VIN
Rf
X
X
0V
IS
IS =
IS
+VIN
R1
both currents are equal
rearranging:
- VOUT
IS =
0 - VOUT - VOUT
=
Rf
Rf
- VOUT +VIN
=
Rf
R1
Gain =
VOUT
=
VIN
Rf
R1
For an ideal op-amp, the gain depends only on the ratio of two resistors
...
7 kW
+
+
V
V
0V
-6V
+10
If we investigate the
characteristics of a
SATURATED
practical inverting mode
amplifier, we obtain a
graph like the one shown
opposite
...
0
=
DVIN
0
...
DVOUT
-0
...
6
-0
...
2
-0
...
4
+0
...
8
+1
+ VIN
-2
-4
SATURATED
-6
- supply voltage
-8
-10
- VOUT
The output voltage of the op-amp is derived from the supply voltage
...
In the situation where the
output voltage has reached this limit, the op-amp is said to be saturated
...
When an a
...
signal is applied to the inputs of a suitable op-amp, a low gain will produce an
an amplified a
...
Increasing the gain will drive the op-amp into saturation an the output
signal will br ‘clipped’ close to the supply voltages upper and lower limits
...
6 V
(b) The input voltage is raised to
2
...
Explain what happens
to the output voltage
...
6
VOUT = - 6
...
0 Volts, the amplifier will go into saturation
...
Higher Physics : Electricity and Electronics
Page 29
Differential Mode
Rf
R1
R4
V1
If
Rf
R1 =
VOUT
R3
V2
R3
R 4 , the following relationship applies
VOUT = ( V 2 - V 1 )
Rf
R1
This arrangement amplifies the difference between the input voltages
...
100 kW
Example
...
6 V
50
...
VOUT = ( V 2 - V 1 )
Rf
R1
= ( 50
...
6 ) x 10
= - 1
...
Higher Physics : Electricity and Electronics
Page 30
Differential mode and the Wheatstone bridge
...
This allows the amplification of the small p
...
s generated by resistive
sensors like strain gauges and resistance thermometers
...
One of the strain gauges is the dummy, the other is the measuring sensor
...
Op-amps used to control external devices
...
However, the output current from the op-amp is enough to
turn a transistor on and off and this can be used to provide a current
...
The
frequency of a wave is equal to the frequency of the source which generated it
...
medium 1
medium 2
v1 = f x l1
v2 = f x l2
The period of a wave is the time
taken to complete one wave cycle
of movement
...
wave direction
particle
movement
wave direction
Transverse Waves : Particles affected
by the wave move at right angles to the
direction of movement of the wave
...
particles are pushed together then pulled apart as wave passes
particle
movement
Longitudinal Waves :Particles affected by longitudinal waves move in the same direction
as the movement of the wave
...
Higher Physics : Radiation and Matter
Page 2
Amplitude
The amplitude of a wave is a measure of the energy carried by the wave
...
The amplitude
represents the position of maximum potential energy
...
One wave interferes with the effect of the other
...
If the sources are coherent, however, it is possible to generate stable wave patterns or
interference patterns
...
We can set up two
coherent waves in a
ripple tank using two
dippers driven at the
same frequency
...
Higher Physics : Radiation and Matter
Wave trains of large amplitude
waves
...
maxima
minima
low amplitude waves in
between the high
amplitude
...
Page 3
Constructive Interference
The troughs and peaks of one wave coincide with the troughs and peaks of the
other
...
maxima
Destructive Interference
The peaks of one wave coincide with the troughs of the other
...
Condition for a maxima (constructive interference)
Path difference = d1 - d2 = nl ( n = 0, 1, 2, 3, 4
...
We can set up two loudspeakers and
drive them from the same signal source
as shown
...
Y
loud
quiet
loud
LS
quiet
signal
gen
...
Maxima are located at A, B
and C
...
Path difference at B = 0 ( straight ahead )
loud
C
3
...
7 m
Path difference at C = l
= 3
...
2
= 0
...
5
-1
(speed of sound in air = 340 ms )
A
f = 680 Hz
Higher Physics : Radiation and Matter
Page 5
10
0
mV
Double slit interference patterns
...
The gap
behaves as a source of microwaves
...
These are generated from the same source of microwaves so the wave
patterns are coherent and will form an interference pattern
...
A typical
response is shown in the graph
...
He used double slit interference to show that light could form an interference pattern
...
Waves can
interfere and cancel each other out
...
Higher Physics : Radiation and Matter
Page 6
The diffraction grating
...
The slits act as sources of diffracted waves
...
These are ideal generating interference patterns
...
From the digram, this occurs
where:
dm
nl
nl = d sin q
q
direction to
0th order
d is the distance between each slit in metres
q is the angle between the direction of the maxima and the direction of the 0th order
...
3rd order n = 3
Measuring the wavelength of monochromatic light
Monochromatic light is light with a single wavelength
...
2nd order n = 2
1st order n = 1
X
q
laser
0th order n = 0
D
nl = d sin q
2l = d X
D
sin q = tan q
Higher Physics : Radiation and Matter
for small angles
Page 7
Spectrometer
telescope
slit
q
collimator
light
source
0th order
diffraction
grating
turntable
Light from the source passes through a narrow
slit into the collimator
...
The interference ‘fringes’ are viewed using a
telescope (focussed on infinity)
...
The viewing
angle is measured on the turntable
...
The first order maximum is viewed at an angle of 15
...
Calculate the wavelength of the light
...
5
d = 2 x 10 -6 m
q = 15
...
The 0th order spectra is a band of white light
where all wavelengths interfere constructively
...
The blue colours are refracted
through the greatest angle
...
The wavelengths of light in the visible spectrum range from 700 nm at the red end to 400 nm at
the blue end
...
Higher Physics : Radiation and Matter
Page 9
Refraction of Light
medium
A
medium
B
refracted ray
q2
normal
angle of
refraction
q1
angle of
incidence
incident ray
The speed of light depends on the medium through which it is travelling
...
This change of direction at the boundary between two media is called refraction
...
Normally examples will refer to monochromatic light
...
Sources of monochromatic light include lasers
...
The absolute refractive index ( n ) of an optical material is based on monochromatic light
passing from a vacuum into the material
Vacuum
qM
normal
qV
Material
n =
sin q V
sin q M
monochromatic
light
Measurement of refractive index
The speed of light in air is approximately equal to the speed of light in a vacuum so we can
measure the refractive index using light from air into the material and be confident of
accuracy to the fourth significant figure
...
ray from ray
box directed
at centre of
face
A narrow ray of light is
projected at the centre of
the flat face of a semicircular block
...
The normal is drawn
...
sin q 1
n =
sin q 2
refracted ray
emerges
undeviated
q2
q1
semi-circular block
of material
monochromatic
light
Higher Physics : Radiation and Matter
Page 11
Material
Absolute
Refractive
index
Water
1
...
31
Perspex
1
...
50
Diamond
2
...
00
Glycerol
The absolute refractive index is a ratio and
so has no units
...
1
...
33
sin q = 0
...
45 O
n =
glass
30 O
sin 45
sin 30
= 1
...
absolute refractive index =1
...
As the angle of incidence increases
O
the angle of refraction will eventually reach 90
...
The angle of incidence which gives rise to an angle of refraction equal to 90 O is
called the critical angle
...
Refracted ray
with weak reflected
ray,
q equal to the critical
angle
...
Strong reflected ray
...
Total internal
reflection with no
refracted ray
...
We can chose a section of that ray where an
incident wave front is about to enter medium 2 at B and a refracted wave front is just
leaving A
...
Higher Physics : Radiation and Matter
Page 14
Example : Find the speed of monochromatic light in glass with an absolute refractive
index of 1
...
c speed of light in vacuum = 3 x 10 8 ms -1
c
n =
v
8
v =
3 x 10
1
...
46 =
3 x 10
v
8
= 2
...
White light contains light with a range in frequencies
...
The white light is dispersed; spread
out, by refraction
...
light
glass prism
red
blu
e(
(n
n=
=1
...
5
18
5)
)
Absolute refractive indices for optical materials are quoted for different wavelengths of
light
...
The different refractive indices also mean that the speed of light is also dependant on
frequency
...
Carried by white light, the signal would be scrambled as the red signal
raced ahead of the blue signal
...
n = 1
...
5
An optical fibre is made from two types of glass, one surrounded by the other
...
5, the outer glass has an absolute refractive
index of 1
...
Calculate the critical angle for light passing through the inner glass
...
5
for outer glass v =
=
c
1
...
4
=
1
...
4
c / 1
...
5
1
...
The intensity of radiation falling on a surface is the radiant energy per second on a square
metre of the surface
...
The cross-sectional area of the beam
at a distance of 1 metre is 4 mm 2
...
P
A
2 x 10 -3
=
4 x 10 -6
= 5 x 10 2
= 500 W m -2
I =
The Inverse Square Law
The intensity of radiation from a point source varies inversely as the square of the
distance from the source
...
double the distance and the same
radiation is spread over four times
the area
...
The rule can be applied to
large sources at large distance from the source where the source appears small to the
observer
...
It also applies
to the force of gravity and the electric forces
...
photodiode
r
Readings are corrected for background light intensity
...
1 / r2
Example
1
...
8 m
I =
0
...
When the lamp was positioned
at a distance of 1
...
35 Wm
...
8 m
...
constant
1
...
35 x 1
...
5
= 0
...
8 m
constant
I =
0
...
79
=
0
...
2 Wm -2
same lamp, same constant
as expected, ½ the distance(approx) 4 times the intensity
...
When certain metal surfaces are exposed to ultra-violet light, electrons are ejected from
the surface
...
uv lamp
uv lamp
zinc
plate
zinc
plate
-
+
negatively charged
electroscope - leaf
collapses as electrons
are lost
...
For a particular metal surface the effect can only take place for light with a frequency
greater than a threshold frequency f O
...
Above this threshold
frequency, the effect takes place, instantly and at all intensities
...
Using apparatus similar to that shown above, different metal surfaces are illuminated by
different frequencies of UV light
...
The electrons then become part of a current which can be
measured using the ammeter
...
Higher Physics : Radiation and Matter
Page 19
Light : Wave or Particle?
The photoelectric effect cannot be explained by wave theory
...
The only factor to be considered would be the intensity of the
light! It should not depend on the frequency of the light
...
Each photon carries a package of energy which depends on the frequency of
the light
...
If the package carries enough
energy, the electron can escape, if not, the electron cannot escape
...
63 x 10 -34 Js
Example Calculate the energy carried by a photon of light with a wavelength of 450 nm
E = hf
=
=
hc
l
6
...
42 x 10 -19 J
Intensity and photons
...
uv light
current
metal plate
IMAX
anode
A
V
-VO
When the stopping potential is plotted
against the frequency of the incident
light it is found to be directly proportional
to the difference between the frequency
of the light and the threshold frequency fO
stopping potential VO / Volts
EK = eVO
2
...
0
0
4
...
VO is termed the stopping potential, the
reverse potential difference required to
stop the emission of electrons
...
The stopping potential indicates the
maximum kinetic energy of the ejected
electrons
...
0
8
...
0
12
...
hfO is the work function of the surface ( F )
...
Higher Physics : Radiation and Matter
Page 21
Example
...
Calculate the maximum kinetic energy of the electrons emitted by the
photoelectric effect ( work function of zinc = 5
...
63 x 10 x 3 x 10
-19
- 5
...
16 x 10 -19 J
-19
Example: Caesium has a work function of 3
...
Find the maximum wavelength
of the light which will cause the photoelectric effect in caesium
...
hc
lO
hc
work function
6
...
21 x 10
= 619 nm
Higher Physics : Radiation and Matter
( orange in the visible spectrum)
Page 22
Emission Spectra
...
When electric current is passed through a gas, the gas glows, emitting light
...
The light is emitted as a series of single wavelengths represented by each line in the
spectrum
...
Each line represents an emitted photon of light
...
Energy levels
...
An electron falling from a higher level to a lower
emits a photon of light with an energy equal to
the difference between the two levels
Line spectra represent all the different possible
ways an electron can fall back down the levels
E = 0 ionisation level
-19
W4
-1
...
42 x 10-19
W2
-5
...
The lowest energy level, W1 ,
is the ground state
...
Only the lowest
energy levels are shown
...
Example:
When an electron falls from level W4 to the ground state
W1 , a photon is emitted with a frequency f
...
63 x 10
-34
-19
Note: Energy
levels are
negative values
-19
x f = -1
...
8 x 10 )
20
...
63 x 10-34
15
= 3
...
Higher Physics : Radiation and Matter
W1
ground state
-21
...
sodium vapour
lamp
...
sodium
atoms
The photons in the light
from the sodium lamp are
absorbed by the sodium
atoms in the flame then
spontaneously re-emitted
...
After passing through
the sodium flame, the light
has less photons and is
dimmer than the light which
by-passes the flame
...
This means
that an atom can only absorb photons of light which are identical to the set of photons
which the atom emits as a line spectra
...
This is due to the absorption of photons ant their re-emission in directions away from
the beam
...
stellar absorption spectra
The continuous spectra of stars, like the Sun, contain hundreds of absorption lines
...
The position and strength of the lines indicate the presence and
concentration of various elements
...
Spontaneous emission of radiation is a random process where an atom captures a photon,
holds on to it for a short period of time then re-emits the photon in a random direction
...
When emitted, the photon is in phase with the
stimulating photon and is emitted in the same direction
...
Higher Physics : Radiation and Matter
Page 26
ENERGY
full
mirror
half-silvered
mirror
ENERGY
Lasers are constructed from material which contains atoms which can be excited to the
same state
...
Once in the excited state, any spontaneous emission of a photon will
stimulate the emission of other photons
...
All other photons
escape
...
Some of the photons are absorbed, but as long as more photons are emitted than absorbed,
the intensity of the beam will increase
...
Laser safety
Lasers produce a narrow beam of intense light which can permanently damage the retina
of the eye
...
Example Calculate the intensity of a laser with an output power of 2 mW at a distance
of 1 m when the diameter of the beam is 3 mm
...
5 x 10 -3 ) 2
= 283 Wm -2
How does this compare with the intensity of a 100 W light bulb at 1m?
100 W
1m
The power output of the bulb is distributed
over the surface of a sphere with a radius
of 1 m
...
Higher Physics : Radiation and Matter
Page 27
Energy Levels in Solids
...
When the same atoms are formed into
a solid, the atoms are linked together and the electrons can occupy a whole series of energy
levels grouped into bands
...
Conductors, Insulators and Semiconductors
...
For electrical conduction
to take place there must be electrons in a band and vacant energy levels to move to
...
Electrons in the valence band are part of the interatomic
bonding
...
Conductor
conduction
band
Insulator
Empty
Semiconductor
electron
hole
valence
band
Full
conduction is possible
through the partly
filled conduction band
...
Conduction takes place in both
the conduction band and the
valence band
...
Conductors
...
Insulators
...
Given enough energy, the electrons will become free and the
insulator will conduct
...
Semiconductors
...
As they warm up, thermal energy frees electrons from
bonds leaving holes in the bonding electrons
...
As the electrons move one way, the
hole moves in the opposite direction like a positive charge
...
p and n type semiconductors
...
This process is termed doping
...
This provides extra free
electrons for the conduction
band
...
These create extra holes in
the valence band
...
Doped semiconductor is termed Extrinsic
...
In p-type semiconductor, the current is carried by positive holes in the valence
band
...
5
-1
...
5
0
...
0
pd across diode /V
-20
Forward biassed : Current flows
when pd is greater than 0
...
(in a silicon diode )
p
electrons
n
p
depletion layer
holes
Reverse biassed : No current
flows
n
electrons drift across the junction to fill holes
...
Once enough charge is built up, further electron drift will be prevented
...
This
region is called the depletion layer
...
Depletion layer is reduced in width
...
Higher Physics : Radiation and Matter
p
n
Reverse biassed: p side is made more negative
and n side more positive
...
No
current and depletion layer widens
...
A series resistor is
always included in the circuit to limit the
current flowing through the LED
...
The electrons in the conduction band lose
energy when they combine with holes in
the lower valence band
...
LED ‘s are made which can emit red, orange, green or blue light
...
The Photodiode
...
Photons of light entering a p - n
junction cause the separation of an electron from a hole
...
Photovoltaic Mode:
photon
hole
current
electron
photodiode
RL
When a photon enters the depletion layer, an electron - hole pair is created
...
The electron and the hole recombine when the electron flows to the p - side
through an external circuit
...
Solar cells are basic photodiodes with a large junction area
...
photon
hole
electron
R
V
The current is a result of electron - hole
pair creation and not the applied voltage,
so is independent of the voltage as long
as the voltage is below the breakdown
voltage of the photodiode
...
The current produced is
directly proportional to the rate at which electron - hole pairs are produced which
is directly proportional to the intensity of the incident light
...
light intensity
pin diodes
...
The depletion layer is very thin so that
electron - hole pairs are swept up very
quickly into the circuit
...
These diodes are used to detect the
digital signals passed through optical
fibres
...
A block of p - type semiconductor
(the substrate) is the starting point
...
A coating of insulating silicon oxide is added
...
Conducting metal contacts are added as shown
...
The metal contact on the bottom of the block is an internal
connection
...
gate
source
How it Works
0V
The drain is the positive terminal
...
A reverse biassed p - n junction
is formed at the drain implant
...
This forms
a narrow channel of n - type semiconductor
just below the gate
...
electrons
n
n
p
gate
source
Higher Physics : Radiation and Matter
n
p
source
If the gate voltage is increased further, a wider
channel is created with a lower resistance
...
The flow of current depends on the formation
of an n - channel between the source and the
drain
...
drain
drain
+V
n
n
p
Page 33
MOSFET will not conduct unless
VGS is above the threshold voltage:
usually around 2 V
...
ID
drain
VDS
source
gate
VGS
+V
A MOSFET can be used as a switch in the
same way as a normal transistor
...
When VGS is above the threshold value
the MOSFET is ON
...
The MOSFET can also be used as an amplifier
...
a particles from radioactive Radon gas were directed at the foils
...
This emits a
tiny flash of light when struck by an a particle
...
The number of flashes at different angles q, were noted
...
(2) Some of the a particles were scattered through large angles
...
Rutherfords nuclear
model
positive
solid
electron
‘currants’
nucleus
electrons
Plum pudding
model of atom
Rutherford explained the results by proposing a new model of the atom with almost all of
the mass and all of the positive charge concentrated in a tiny nucleus
...
Most of the a particles would pass through the empty space round the nucleus
...
Higher Physics : Radiation and Matter
Page 35
Radioactivity
A nucleus is made up of protons and neutrons
...
The neutron is similar to the proton but has no charge
...
Atoms
containing the same number of protons but different numbers of neutrons are isotopes of
that element
...
A
X
Z
examples
14
6
C
235
92
U
X = chemical symbol
atomic number Z = number of protons
atomic mass A = number of protons + neutrons
Radioactive atoms contain unstable nuclei
...
The three types of radioactive particle we will consider are
a, b and g particles
...
They consist of two protons and two neutrons
4
a
216
86
Rn
4
He
+2
176
77
Ir
+2
4
He
+2
He
212
84
172
75
Po
a particle decay is confined to heavier
nuclei
...
When a b particle is emitted, one
of the neutrons changes to a proton
...
The atomic
mass is unaffected
...
g particles are usually emitted along with b and a particles
...
Higher Physics : Radiation and Matter
Page 36
Binding Energy
The protons and neutrons making up a nucleus require energy to keep them together
...
This shows up as mass defect; the difference in mass between the nucleus
and the sum of the individual masses of the protons and neutrons which make up the
nucleus
...
Find the binding energy of a helium nuclide
4
2 He
mass 4 He = 6
...
67295 x 10-27 kg
mass neutron = 1
...
69478 x 10-27 kg
mass defect = mass of 2 protons + 2 neutrons - mass of helium nucleus
= 0
...
053 x 10-27 ( 3 x108 )2
= 4
...
When a radioactive particle is emitted from a nucleus, the binding energy per nucleon in
the nucleus is increased
...
The energy carried by the emitted particle and the new nucleus is derived from the reduction
in mass created by the radioactive decay
...
Some large nuclei are at the limit of the force holding the protons and neutrons together:
the strong nuclear force
...
The name for this process is nuclear fission
...
Spontaneous Nuclear Fission The nucleus simply breaks up in a process similar to
radioactivity
...
Induced Nuclear Fission The nucleus which undergoes spontaneous nuclear fission can be
induced to undergo fission if it captures a neutron
...
When the nucleus splits, the new daughter nuclei emit further neutrons which can go on to cause
more fissions
...
In a nuclear reactor the chain reaction is slowed down by controlling the number of
neutrons involved in the reaction
...
1
0
n
+
235
92
3
...
08 x 10
141
56
1
Ba + 3 0 n
3
...
77 x 10 -11 Joules
Higher Physics : Radiation and Matter
Page 38
n
Nuclear Fusion
...
As with fission, the binding energy is increased and the resulting mass defect
appears as energy
...
347 x 10
1H
-27
kg
4
+
2 He
1
0
n
8
...
1 x 10 -29 kg
energy released = 2
...
However, the material involved
is much lower in density than fissionable material, so the energy released per kilogram is
much higher
...
Fusion reactors offer the promise of unlimited energy supplies in the future if the problem
of starting the fusion and containing the enormous temperatures can be solved
...
Activity is measured in becquerels
...
Absorbed Dose
...
The absorbed dose, D, is the energy transferred per kilogram of the absorbing material
...
Biological Effect
...
the absorbed dose,
2
...
3
...
Quality Factor
...
The more ionising the radiation, the greater
the quality factor
...
3
fast neutrons 10
Higher Physics : Radiation and Matter
Page 40
Dose Equivalent
...
H = DxQ
Dose equivalent is measured in sieverts (Sv)
The effective dose equivalent takes account of the particular organs and tissues exposed
to the radiation
...
3
nuclear power 0
...
01
occupational 0
...
01
Man Made
0
...
25
0
...
8
natural radioactivity
in air (indoors)
food and
drink
The chart shows the annual average dose equivalent of radiation received by someone
living in the UK
...
A small proportion come from
man made sources: x-rays, residue from nuclear tests, air travel, nuclear power
...
The maximum
recommended dose for the average citizen is 5 mSv / Year
...
Higher Physics : Radiation and Matter
Page 41
Handling Radioactive Materials
...
The thickness of
material which will reduce the intensity of radiation to half of its value is termed the
half-value thickness
...
Example
...
The
half-value thickness of lead for this source is 12 mm
...
-1
thickness / mm Dose equivalent rate / Sv h
0
20
12
10
The dose equivalent rate behind 48 mm
-1
24
5
of lead = 1
...
5
48
1
...
Gamma radiation is a purely random process
...
The intensity of
gamma radiation obeys the inverse square law
...
A gamma ray source has a dose equivalent rate of 50 Sv h -1 at a distance
of 1 metre from the source
...
constant
1
constant = 50
50 =
1 =
1 =
constant
r2
50
r2
r = 7
...
1 metres from the source
Radioactive sources are kept in shielded containers when not in use
...
Everyone working with radioactive sources is fitted with a personal dosimeter to monitor
their accumulated dose equivalent
...
If anyone has absorbed
more than their allowed dose they are excluded from handling sources until their average
accumulated dose falls back to a safe level
...
We can never be sure that the measurement we have
just taken is the true measured value
...
Results are quoted
12
...
02 mV
-
estimated true value
estimated uncertainty
Types of Error
...
This type of error causes readings to be consistently higher or lower than the true value
...
0
...
1
0
1
2 Force/ N
systematic error
20
Parallax Errors
Scales should be read with the eyes
level with whatever is being read
...
19
18
17
16
15
14
13
Higher Physics : Errors and Uncertainties
mirror
Page 1
Random Errors
Random errors apply to repeated measurements
...
The differences occur because
it is impossible to recreate the exact same conditions for each attempt at measurement
...
In Higher,
however, we usually estimate
this by working out the mean
of the range in measurements
...
If we draw a
distribution, we find that it is symmetrical about the true value
mean value
of distribution
measurement
true measurement
uncertainty =
maximum measurement - minimum measurement
number of measurements
Example: Find the mean and random error in the following set of measurements:
5
...
64, 5
...
59, 5
...
67, 5
...
71, 5
...
60
...
44
10
= 5
...
71 - 5
...
12
10
= 0
...
02 ( rounded up )
measurement = 5
...
02
Higher Physics : Errors and Uncertainties
Page 2
Reading Errors
...
Two types of error have to be taken into account
...
This is
an estimate provided by the instrument manufacturer on the guaranteed accuracy of the
instrument
...
This error applies
to all the readings
...
The scale shown opposite can be
read to an accuracy of + half a scale
division
...
Suppose we have a reading of 1
...
reading error = + 0
...
02
-
1
...
05
-
Suppose we have a reading of 4
...
05 4
...
08
calibration error = + 0
...
24
the last significant figure is
an estimate
At bottom of scale, reading error is larger
than calibration error
...
The larger of the two errors is used when
quoting a measurement
...
Digital meters
can be read to an accuracy of + last digit
...
985
+
7
...
001
A digital meter is potentially more accurate than an analogue meter for the same
calibration error as it can be read to more decimal places
...
0
...
3 + 0
...
3
= 5%
Higher Physics : Errors and Uncertainties
Page 3
Error in Calculated Quantities
...
Example: In an experiment to measure the specific heat capacity of water, the following
results were obtained
...
55 - 0
...
1 - 0
...
505 x 3
...
4
energy
C
% error
mass
c =
0
...
1 - 0
Title: all summary notes
Description: these are all summary notes for physics all type of classes
Description: these are all summary notes for physics all type of classes