Download CHAPTER 28 Sources Of Magnetic Field

 September 6, 2012Dr. Afaf AbdelHady Physics c CHAPTER 28
Sources Of Magnetic Field
28.1 Magnetic field of a moving
charge
To find the magnetic field B of a point
charge q moving with constant velocity v
at the point P, let's call the location of the
moving charge at a given instant the
source point and the point P where we
want to find the field the field point.
Experiments show that the magnitude of
B is proportional to q , the particle
velocity v , 1 r 2 and to the sinϕ . But the
direction of B is perpendicular to the plane
containing the line from source point to
field point and the particle velocity v .
B=
B=
µ0 q v sin ϕ
4π
r2
µ0 qv × rˆ
4π r 2
µ0
is a proportionality constant. B
4π
has the greatest magnitude when sinϕ = 1
or ϕ = 90 . If the charge q is negative, the
directions of B are opposite.
Where
The unit of B is tesla (T).
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September 6, 2012Dr. Afaf AbdelHady Physics c 28.2 Magnetic Field of a Current Element
There is a principle of superposition of magnetic fields:
The total magnetic field caused by several moving charges is the vector sum of the fields
caused by the individual charges.
We begin by calculating the magnetic field caused by a short segment dl of a current
carrying a conductor, the total moving charge dQ in
the segment is
dQ = nqAdl
Where n is the number of the moving charged
particles per unit volume, A is the cross sectional
area, and Adl is the volume.
The moving charges in this segment are equivalent
to a single charge dQ , travelling with a velocity
equal to a drift velocity v . The magnitude of the
resulting field dB at any point P is
dB =
µ0 dQ vd sin ϕ µ0 dq nvd Adl sin ϕ
=
, but
4π
r2
4π
r2
n dq vd A = I , then
dB =
µ0 Idl sin ϕ
4π
r2
In the vector form, using the unit vector r̂ , we have
dB =
µ0 Idl × rˆ
4π r 2
This is the Biot and Savart law.
We can use this law to find the total magnetic field
at any point in space due to the current in a
complete circuit.
B=
µ0
4π
∫
Idl × rˆ
r2
2
September 6, 2012Dr. Afaf AbdelHady Physics c 28.3 Magnetic Field of a Straight Current-Carrying Conductor
An important application of the Biot and Savart law is finding the magnetic field produced by
a straight current carrying conductor.
From the figure
r = x 2 + y 2 and sin ϕ = sin(π − ϕ ) = x
µ
B= 0
4π
a
∫ (x
−a
2
x2 + y 2
xdy
+ y 2 )3 2
The final result is
B=
µ0 I
2a
4π x x 2 + a 2
When the length 2a of the conductor is very great in
comparison to its distance x from the point P, we can
consider it to be infinitely long. When a is much
larger than x , x 2 + a 2 is approximately equal to a ,
hence in the limit a → ∞
B=
µ0 I
2π x
Thus, at all points on a circle of radius r around the conductor, the magnitude of B is
B=
µ0 I
2π r
(near a long, straight, current-carrying conductor)
Note:
The total magnetic flux through any closed surface is always zero. This implies that there are
no isolated magnetic charges or magnetic monopoles. Any magnetic field lines that enters a
closed surface must also emerge from that surface.
∫ B.d A = 0
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September 6, 2012Dr. Afaf AbdelHady Physics c 28.4 Force Between Parallel Conductors
The lower conductor produces a B field that, at
the position of the upper conductor, has magnitude
B=
µ0 I
2π r
The force that this field exerts on a length L of the
upper conductor is
F = I'L× B
Where the vector L is in the direction of the
current I ' and has the magnitude L. Since B is
perpendicular to the length of the conductor and
hence to L , the magnitude of this force is
F = I ' LB =
µ 0 II ' L
2π r
And the force per unit length is
F µ 0 II '
=
L 2π r
Applying the right-hand rule to F = I ' L × B shows
that the force on the upper conductor is directed
downward.
Note:
1- the two conductors carrying current in the
same direction attract each other. If the
direction of either current is reversed, the forces also reverse.
2- Parallel conductors carrying currents in opposite directions repel each other.
Defining the Ampere:
The attraction or repulsion between two straight, parallel, current-carrying conductors is the
basis of the official SI definition of the ampere:
One ampere is that unvarying current that, if present in each of two parallel conductors of
infinite length and one meter apart in empty space, causes each conductor to experience a
force of exactly 2 × 10−7 newtons per meter of length.
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September 6, 2012Dr. Afaf AbdelHady Physics c 28.5 Magnetic Field of a Circular
Current Loop
We can use the law of Biot and Savart to find
the magnetic field at a point P on the axis of the
loop, at a distance x from the center. As the
figure shows, dl and r are perpendicular, and
the direction of the field dB lies in the x-y plane.
Since
r 2 = x2 + a2
dB =
dBx = dB cos θ =
µ0 I
dl
2
4π ( x + a 2 )
µ0 I
dl
a
2
2
2
4π ( x + a ) ( x + a 2 )1 2
dB y = dB sin θ =
µ0 I
dl
x
2
2
2
4π ( x + a ) ( x + a 2 )1 2
The situation has rotational symmetry about the x-axis, so there can not be a component of
the total field B perpendicular to this axis. For every element dl there is a corresponding
element on the opposite side of the loop, with opposite direction. These two elements give
equal contributions to the x-component of d B , but opposite components perpendicular to the
x-axis. Thus all the perpendicular components cancel and only the x-component survive.
Bx = ∫
µ0 I
µ 0 Ia
adl
=
dl
2
2 32
4π ( x + a )
4π ( x 2 + a 2 )3 2 ∫
The integral of dl is just the circumference of the circle ∫ dl = 2π a , then
Bx =
µ0 Ia 2
2( x 2 + a 2 )3 2
The direction of the magnetic field is given by a
right-hand rule.
Magnetic field on the axis of a coil
Bx =
µ0 NIa 2
2( x 2 + a 2 )3 2
(on the axis of N circular loops)
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September 6, 2012Dr. Afaf AbdelHady Physics c Bx =
µ0 NI
2a
(at the center of N circular loops)
28.6 Ampere's Law
It states that: the line integral of B around any closed
path equal to µ0 times the net current through the area
enclosed by the path.
∫ B.d l = µ I
0 encl
The positive sense of current is determined by a right-hand
rule
28.7 Applications of Ampere's law
Examples 28.7 to 28.10
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September 6, 2012Dr. Afaf AbdelHady Physics c CHAPTER 29
Electromagnetic Induction
29.1 Electromagnetic Experiments
Here is what we observe:
1- When there is no current in the electromagnet, so that
B = 0 , the galvanometer shows no current.
2- When the electromagnet is turned on, there is a
momentary current through the meter as B increases.
3- When B levels off at a steady value, the current
drops to zero, no matter how large B is.
4- With a coil in a horizontal plane, we squeeze it so as
to decrease the cross-sectional area of the coil. The
meter detects current only during the deformation, not
before or after. When we increase the area to return
the coil to its original shape, there is current in the
opposite direction, but only while the area of the coil is changing.
5- If we rotate the coil a few degrees about the horizontal axis, the meter detects current
during the rotation, in the same direction as when we decrease the area. When we
rotate the coil back, there is a current in the opposite direction during this rotation.
6- If we jerk the coil out of the magnetic field, there is a current during motion, in the
same direction as when we decreased the area.
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September 6, 2012Dr. Afaf AbdelHady Physics c 7- If we decrease the number of turns of the coil by unwinding one or more turns, there
is a current during unwinding, in the same direction as when we decrease the area. If
we wind more turns onto the coil, there is a current in the opposite direction during
the winding.
8- When the magnet is turned off, there is a momentary current in the direction opposite
to the current when it was turned on.
9- The faster we carry out any of these changes, the greater the current.
10- If all these experiments are repeated with a coil that has the same shape but different
material and different resistance, the current in each case is inversely proportional to
the total circuit resistance. This shows that the induced emfs that are causing the
current do not depend on the material of the coil but only on its shape and the
magnetic field.
Note: The common element in all these experiments is changing magnetic flux through the
coil connected to the galvanometer.
29.2 Faraday's Law
The common element in all induction effects is changing
magnetic flux through the circuit. Remember that
Φ B = B. A = BA cos ϕ
If B is uniform over a flat area A .
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September 6, 2012Dr. Afaf AbdelHady Physics c Faraday's law of induction states:
The induced emf in a closed loop equals the negative of the time rate of change of
magnetic flux through the loop.
In symbols, Faraday's law is
ε =−
dΦB
dt
If we have a coil with N identical turns, and if the flux varies at the same rate through each
turn, the total rate of change through each turn is
ε = −N
dΦB
dt
29.3 Lenz's Law
Lenz's Law is a convenient method for determining the direction of an induced current or
emf. Lenz's law is not an independent principle: it can be derived from Faraday's law.
Lenz's law states: The direction of any magnetic induction effect is such as to oppose the
cause of the effect.
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September 6, 2012Dr. Afaf AbdelHady Physics c 29.3 Motional Electromotive Force
Figure 19.15a shows the rod moves to the right at a
constant velocity v in a uniform magnetic field B directed
into the page. A charged particle q in the rod experiences
a magnetic force with magnitude F = q vB . This magnetic
force causes the free charges in the rod to move, creating a
positive charge at the upper end and a negative charge at
the lower end. This in turn creates an electric field E
within the rod. There is a downward electric force with
magnitude qE to cancel exactly the upward magnetic
force with magnitude qvB . Then qE = qvB and the
charges are in equilibrium. The magnitude of potential
difference
Vab = Va − Vb = EL = vBL
Now suppose the moving rod slides along a stationary Ushaped conductor, forming a complete circuit. No magnetic force acts in the charges, but the
charge redistributes itself, creating an electric field within it. This field establishes a current
in the direction shown. The moving rod has become a source of electromotive force. We call
this emf a motional electromotive force, denoted by ε .
ε = vBL
Note: The motional emf; length and velocity perpendicular to uniform B .
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September 6, 2012Dr. Afaf AbdelHady Physics c Chapter 33
THE NATURE AND PROPAGATION OF LIGHT
A wave traveling in a certain direction can be represented either by:
aRays, which are straight lines representing the direction in which the wave propagates (travels).
b- Wave fronts, which are parallel lines perpendicular to the rays. The distance between two successive
wave fronts is equal to the wavelength. 1-Two theory of light
A source of light may be considered to give out energy either as a stream of particles called photons, or a
continuous of energy along a ray in a given direction. The first concept gives rise to the so called
corpuscular or particle theory of light, while the second has its origin in the wave theory of light.
The concept that light energy is propagating as wave motion is more convenient as far as propagation of
light through a medium is considered. Phenomena such as reflection, refraction, interference, diffraction,
and polarization found their explanation in the wave theory of light. The particle theory is of importance in
the study of the origin of light spectra and light's interaction with matter.
a-
Corpuscular Theory of Light
Was proposed by I. Newton in 1704. According to this theory a source of light continuously emits tiny,
light, and elastic particles called corpuscular in all directions. According to Newton, they have the
following properties:
1-
Can travel through matter with the very high velocity of light
2Have the property of reflection from a polished surface, or transmission through a transparent
medium.
3-
Produce the sensation of vision when fall on the eye.
Newton applied his laws of mechanics to explain several properties of light such as its rectilinear motion,
reflection, and refraction. Corpuscular theory of light could not explain:
11
September 6, 2012Dr. Afaf AbdelHady Physics c 1.
Observed change in the velocity of light in denser and rarer media.
2.
Phenomenon of double refraction.
3.
Phenomenon of interference, diffraction, and polarization.
b-
Wave Theory of Light
The first statement on the wave nature of light was done by C. Huygens in 1679. According to this theory a
luminous body is a source of disturbance in a hypothetical medium. Huygens' light waves have the
following properties:
1. Propagate from an uncollimated luminous equally in all directions of space with the speed of
light along a ray which at a given instant of time is always perpendicular to the light wave
front at this instant.
2. For a point source, wave fronts at any instant of time from a spherical surface with the centre
at the source.
3. All points on a given wave front serve as point sources of spherical secondary disturbances
called secondary wavelets.
4. Secondary waves travel through space with the same speed as the primary wave, with the
envelope of all the secondary wavelets after any given interval of time giving rise to the
secondary wave front.
5. Secondary wave fronts are formed only in the forward direction.
Points 3-5 constitute what is known as Huygens's
Principle.
Notice: The new wave front at a later time is found by
constructing a surface tangent to the secondary
wavelets or, as it is called, the envelope of the
wavelets.
Light rectilinear propagation
The original wave front AA' is travelling outward from
the source, as indicated by the arrows. Let v be the
speed of propagation of the wave; then in time t it
travels a distance vt . We construct several circles
(traces of spherical wavelets) with radius r = vt ,
centered at points along AA'. The trace of the envelope
of these wavelets, which is the new wave front, is the
curve BB'. The light ray will not bend, and rectilinear
propagation is verified.
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September 6, 2012Dr. Afaf AbdelHady Physics c Reflection and Huygens's principle
We consider a plane wave approaching a plane reflecting
surface. The lines AA', OB', and NC' represent successive
positions of a wave front approaching the surface MM'.
From plane geometry the angle θa between the incident
wave front and the surface is the same as that between
the incident ray and the normal to the surface and
therefore the angle of incidence. Similarly, θ r is the angle
of reflection OP = vt .
Now OB, by construction, is tangent to a circle of radius
vt with centre at A.
The triangles APO and OQA are congruent because they
are right triangles with the side AO in common and
AQ = OP = vt . Therefore,
≺ θa =≺ θr
(law of reflection)
Refraction and Huygens's Principle
The angles θa and θ r between the surface and the incident
and refracted wave fronts are the angles of incidence and the
angle of refraction, respectively. From right triangle AOQ,
sinθa =
v at
AO
And from the right triangle AOB,
sinθb =
v bt
AO
Combining these, we find
sinθa v a
=
sinθ b v b
Since na = c v a and nb = c v b
nb v a
=
na v b
Thus,
13
September 6, 2012Dr. Afaf AbdelHady Physics c So ,
sinθa nb
=
sinθ b na
or na sinθa = nb sinθ b
LIGHT WAVES
Light is a form of energy and has the following properties (see also
the wave):
1.
Light waves are transverse electromagnetic waves which
transfer energy from one place to another.
2.
Light travels in vacuum with the greatest speed in the
universe (c=3×10⁸ m/s).
Wavelengths of light are very short (about 5×10⁻⁷ m), but the
3.
frequencies of light are very high.
4.
Light travels in straight line*, therefore, it can produce
shadows and inverted images
5.
Light can be reflected, refracted and diffracted.
6.
Light obeys the relation: v = f λ
Since the speed (v) is constant (for a given medium), it follows that the frequency (f) of a wave is inversely
proportional to its wavelength…
Refractive index of a medium n=
speed of light in air (vacuum) c
= speed of light in medium
v
In any material, v = f λ , since f is the same in any material as in vacuum and v is always less than the
wave speed c in vacuum, λ is also correspondingly reduced. Thus the wavelength λ of light in a material
is less than the wavelength λ0 of the same light in vacuum. So, f = c λ0 = v λ . Combining this with the
equation n = c v , we find n = λ0 λ or
λ = λ0 n
(wavelength of light in a material)
Polarization
Polarization is a characteristic of all transverse waves. This chapter is about light, but to introduce some
basic polarization concepts, let's go back to transverse waves on a string. For a string that in equilibrium
lies along the x-axis, the displacement may be along the y-direction. In this case the string always lies in
the xy-plane. But the displacement might instead be along the z-axis, then the string always lies in the xzplane.
When a wave has only y-displacements, we say that it is linearly polarized in the y-direction; a wave with
only z-displacement is linearly polarized in the z-direction. For mechanical waves we can build a
14
September 6, 2012Dr. Afaf AbdelHady Physics c polarizing filter, or polarizer, that permits only waves with a certain polarization direction to pass. This
filter passes waves that are polarized in the y-direction but blocks those that are polarized in the z-direction.
The same language can be applied to electromagnetic waves, which also have polarization. An
electromagnetic wave is a transverse wave; the fluctuating electric and magnetic fields are perpendicular to
each other and to the direction of propagation. We always define the direction of polarization of an
electromagnetic wave to be the direction of the electric-field vector E, not the magnetic field, because
common electromagnetic wave detectors respond to the electric forces on electrons in materials, not the
magnetic forces.
Polarizing Filters
Waves emitted by a radio transmitter are usually linearly polarized. Light from ordinary sources, such as
light bulbs, is not polarized. Such light is called unpolarized light or natural light. To create a polarized
light from unpolarized natural light requires a filter. The most common polarizing filter for visible light is a
material known by the trade name Polaroid, widely used for sunglasses and polarizing filters for camera
lenses. This material incorporates substances that have dichroism, a selective absorption in which one of
the polarized components is absorbed much more strongly than the other. A Polaroid filter transmits 80%
or more of the intensity of a wave that is polarized parallel to a certain axis in the material, called the
polarizing axis. An ideal polarizer passes 100% of the incident light waves that is polarized in the direction
of the filter's polarizing axis but completely blocks all light that is polarized perpendicular to this axis. We
will assume that all polarizing filters are ideal.
15
September 6, 2012Dr. Afaf AbdelHady Physics c Since the intensity of an electromagnetic wave is proportional to the square of the amplitude of the wave
and the ratio of transmitted to incident amplitude is cos ϕ , so the ratio of transmitted to incident intensity is
cos 2 ϕ . Thus the intensity of the light transmitted through the analyzer is
I = I max cos 2 ϕ
(Malus's law, polarized light passing through an analyzer)
Where I max is the maximum intensity of light transmitted ( at ϕ = 0 ) and I is the amount transmitted at
angle ϕ . Malus's law applies only if the incident light passing through the analyzer is already linearly
polarized.
Polarization by reflection
Brewster discovered that when the angle of incidence is equal to the polarizing angle θ p , the reflected ray
and the refracted ray are perpendicular to each other, so θb = 90 − θ p .
16
September 6, 2012Dr. Afaf AbdelHady Physics c From refraction law
n a sinθ p = nb sinθ b
n a sinθ p = nb sin(90-θ p ) = nb cos θ p
tanθ p =
nb
na
(Brewster's law for the
polarizing angle)
17
September 6, 2012Dr. Afaf AbdelHady Physics c Chapter 35 INTERFERENCE Interference refers to any situation in which two or more waves overlap in space. When this occurs, the total wave at any point at any instant of time is governed by the principle of superposition. This principle applies to electromagnetic waves and is the most important principle in all of physical optics. The principle of superposition states: When two or more waves overlap, the resultant displacement at any point and at any instant is found by adding the instantaneous displacements that would be produced at the point by the individual waves if each present alone. In optics, sinusoidal waves are characteristic of monochromatic light (light of a single color). Two monochromatic sources of the same frequency and with any definite, constant phase relationship are said to be coherent. We use the term coherent waves (for light waves) to refer to the waves emitted by two such sources. When waves from two or more sources arrive at a point in phase, the amplitude of the resultant wave is the sum of the amplitudes of the individual waves; individual waves reinforce each other. This called constructive interference. For constructive interference to occur at b, the path difference r2 − r1 sources must be an integral multiple of the λ . r2 − r 1 = m λ
the for the two wavelength
(m=0, ± 1, ± 2,.......) At point c, the path difference is a half‐integral wavelengths. Waves from the two sources arrive at exactly a half‐ cycle out of phase. The resultant is the difference between the two individual If the individual amplitudes are equal, then the total is zero. This cancellation or partial cancellation of the waves is called destructive interference. The for destructive interference is number of point c amplitude amplitudes. amplitude individual condition 18
September 6, 2012Dr. Afaf AbdelHady Physics c 1
r2 − r 1 = (m + λ )
2
(m=0, ± 1, ± 2,.......) Two –source interference of light (Young's experiment) d sin θ = m λ
1
d sin θ = ( m + λ )
2
(m=0, ± 1, ± 2,...) constructive interference (m=0, ± 1, ± 2,...)
destructive interference The pattern on the screen is a succession of bright and dark bands, or interference fringes, parallel to the slits S1 and S2 . The center of the pattern is a bright band corresponding to m=0; this point on the screen is equidistant from the two slits. We can derive an expression for the positions of the centers of the bright bands on the screen. In the figure, y is measured from the center of the pattern. Let ym be the distance from the center of the pattern ( θ =0 ) to the center of the m th bright band. Let θ m be the corresponding value of θ ; then y m = R tan θ In the experiment, the distances y m are often much smaller than the distance R from the slits to the screen. Hence θ m is very small, tan θ m is very nearly equal to sin θ m y m = R sin θ
19
September 6, 2012Dr. Afaf AbdelHady Physics c ym = R
mλ
d
y m = ( m + 1 2)λ
(constructive interference or max. intensity) R
d
(destructive interference or min. intensity) The distance between any two adjacent maxima or minima is known as fringe separation β . It is independent of the order of the fringe m. β = y m +1 − y m = λ
R
d
This experiment provides a direct measurement of λ . Amplitude in Two‐Source Interference E p2 = E 2 + E 2 − 2E 2 cos(π − φ )
= E 2 + E 2 + 2E 2 cos φ
Then, using the identity 1 + cos φ = 2 cos 2 (φ 2) , we obtain E p2 = 2E 2 (1 + cos φ ) = 4E 2 cos 2 (φ 2) E p = 2E cos(φ 2)
(amplitude in two-source
interference)
When the two waves are in phase, φ = 0 and E p = 2E . When they are exactly a half cycle out of phase, φ = π , cos φ 2 = 0 and E p = 0 . Intensity in Two‐source interference We can express the intensity I in terms of the maximum intensity I 0 : I = I 0 cos 2 (φ 2) For some phase angle φ the intensity is I 0 , four times as great as for an individual wave source, but for other phase angle the intensity is zero. If we average over all possible phase differences (average of cos 2 (φ 2) is 1 2) . So, the intensity I 0 is just twice the intensity from each individual source, as we should expect. Interference in Thin Films 20
September 6, 2012Dr. Afaf AbdelHady Physics c You often see bright bands of color when light reflects from a thin layer of oil floating on water or from a soap bubble. These are the results of interference. Light waves are reflected from the front and back surfaces of such thin films, and constructive or destructive interference between the two reflected waves occurs in different places for different wavelengths. Thin Films and phase shifts during reflection Let's look at a simplified situation in which monochromatic light reflects from two nearly parallel surfaces at nearly normal incidence. Figure 35.12 shows two plates of glass separated by a thin wedge, or film, of air. We want to consider interference between the two light waves reflected from the surfaces adjacent to the air wedge. When we carry out the experiment, the bright and dark fringes appear, but they are interchanged. Along the line where the plates are in contact, we find a dark fring, not a bright one. In that case the two waves that are reflected at the line of contact are a half‐cycle out of phase even though they have the same path length. This phase shift can be predicted from Maxwell's equations and electromagnetic nature of light. Suppose a light wave with electric‐field amplitude E i is travelling in an optical material with refractive index n a . It strikes, at normal incidence, an interface with another refractive index nb . The amplitude E r of the reflected wave is proportional to the amplitude E i of the incident wave and is given by Er =
n a − nb
Ei
n a + nb
(normal incidence) Fig(a): When n a
nb , light travels more slowly in the first material than in the second. In this case, E i and E r have the same sign, and the phase shift between the incident and the reflected wave is zero. Fig(b): When n a = nb , the amplitude E r is zero. There is no reflected wave. 21
September 6, 2012Dr. Afaf AbdelHady Physics c Fig(c): When n a ≺ nb , light travels more slowly in the second material than in the first. In this case, E i and E r have opposite signs, and the phase shift between the incident and the reflected wave is π or half cycle. We can summerize this discussion mathematically. The conditions for constructive and destructive interference are 2t=mλ
(m=0,1,2,..)
(constructive reflection, no relative phase shift)
2t=(m+1 2)λ
(m=0,1,2,..)
(destructive reflection, no relative phase shift)
If one of the two waves has a half‐cycle reflection phase shift, the conditions are: 2t=(m+1 2)λ
2t=mλ
(m=0,1,2,..)
(m=0,1,2,..)
(constructive reflection, half-cycle relative phase shift)
(destructive reflection, half-cycle relative phase shift)
Note that: We must use the wavelength in the material. If the material has a refractive index n, then λ
λ = 0 where λ0 is the wave length in vacuum. n
Newton's Rings Figure 35.17a shows the convex surface of a lens in contact with a plane glass plate. A thin film of air is found between the two surfaces. When you view the setup with monochromatic light, you see circular interference fringes (Fig. 35.17b). These were studied by Newton and are called Newton's rings. The Michelson Interferometer 22
September 6, 2012Dr. Afaf AbdelHady Physics c Suppose the angle between mirror M 2 and the virtual image of M 1 is just large enough that, five or six fringes are present in the field of view. If we now move the mirror M 2 slowly either backward or forward a distance λ 2 , the difference in path length between rays 1 and 2 changes by λ , and each fringe moves to the left or right a distance equal the fringe spacing. If we observe the fringe positions through a telescope with a crosshair eyepiece and m fringes cross the crosshair when we move the mirror a distance y, then y =m
λ
or
2
λ=
2y
m
If m is several thousand, the distance y is large enough that it can be measured with good accuracy, and we can obtain an accurate value for the wavelength λ . Alternatively, if the wavelength is known, a distance y can be measured by simply counting fringes when M 2 is moved by this distance. 23
September 6, 2012Dr. Afaf AbdelHady Physics c Chapter 36 DIFFRACTION Diffraction is described as " the bending of light around an obstacle" (edge). But the process that causes diffraction is present in the propagation of every wave. When part of the wave is cut off by some obstacle, we observe diffraction effects that results from interference of the remaining parts of the wave fronts. Diffraction phenomena are divided into two general classes: 1‐ Fraunhofer diffraction in which the source of light and the screen on which the diffraction pattern is observed are effectively at infinite distances from the aperture causing the diffraction. (It is simpler to treat theoretically) 2‐ Fresnel diffraction in which either the source, or the screen, or both are at finite distances from the aperture or obstacle. (It's theoretical treatment is more complex). 36.1 Diffraction from a single slit The difference in path length to point P is (a / 2) sin θ where a is the slit width and θ is the angle between the perpendicular to the slit and the line from the center of the slit to P. Suppose this path difference happens to be equal to λ / 2 ; then light from these two strips arrives at P with a half cycle phase difference, and cancellation occurs. Similarly, light from two strips immediately below the two in the figure also arrive at P a half cycle out of phase. The result is complete cancelation at P for the combined light from the entire slit, giving a dark fringe in the interference pattern. This occurs when (a / 2) sin θ = ± λ / 2
sin θ =
mλ
a
or sin θ = ± λ / a (m= ± 1, ± 2,.......)
(dark fringes) 24
September 6, 2012Dr. Afaf AbdelHady Physics c The values of θ are often so small that the approximation sin θ ≈ θ where θ is in radians is a very good one. mλ
(m= ± 1, ± 2,.......) (for small anglesθ )
a
If the distance from the slit to screen is x and the vertical distance of the mth dark bands from the center of the pattern is y, then tan θ = y m / x . For small angle tan θ ≈ θ , then θ=
ym = x
mλ
a
(for y m
x) Intensity in the single‐slit pattern The amplitude E p of the resultant electric field at P is equal to the chord AB, which is 2(E 0 / β ) sin( β / 2) , then E p = E0
sin( β / 2)
β /2
The intensity at each point on the screen is proportional to the square of the amplitude. If I 0 is the intensity in the straight‐ ahead direction where θ = 0 and β = 0 , then the intensity I at any point is 25
September 6, 2012Dr. Afaf AbdelHady Physics c 2
⎡ sin( β / 2) ⎤
I = I0 ⎢
⎥ ⎣ β /2 ⎦
The path difference between the ray from the top of the slit and the ray from the middle of the slit is (a / 2) sin θ . The path difference between the ray from the top of the slit and the ray from the bottom of the slit is twice this. The phase difference is 2π / λ . So β=
2π
λ
a sin θ
and 2
⎡ sin(π a sin θ / λ ) ⎤
I = I0 ⎢
⎣ π a sin θ / λ ⎥⎦
We might expect the peaks to occur where the sine function reaches the values ±1 1, where β = ±π , ±3π , ±5π ,... , or in general β ≈ ± (2m + 1)π
(m=0,1,2,....) This is approximately correct, but because of the factor in the denominator, the maxima do not occur precisely at these points. To find the intensities at the side maxima, using the approximate expression for β , then Im ≈
I0
1 2 2
(m + ) π
2
Where I m is the intensity of the mth side maximum and I 0 is the intensity of the central maximum. Width of the single‐slit pattern For small angles the width (angular spread) of the diffraction pattern is inversely proportional to the slit width a or, more precisely, to the ratio of a to the wavelength λ . With the approximation sin θ = θ the position θ1 of the first minimum beside the central maximum, corresponding to β / 2 = π , is θ1 =
26
λ
a
September 6, 2012Dr. Afaf AbdelHady Physics c 36.2 Diffraction From Double Slit The expression for the intensity shown in Fig. 36.12c is proportional to the product of the two‐slit and single‐slit expressions. 2
φ ⎡ sin( β / 2) ⎤
I = I 0 cos ⎢
where
2 ⎣ β / 2 ⎥⎦
2π d
2π a
sin θ
sin θ
φ=
β=
λ
λ
2
Note that in Fig.36.12c, every fourth interference maximum at the sides is missing because these interference maxima ( mi = ±4, ±8,... ) coincide with the diffraction minima ( md = ±1, ±2,... ). Figures 36.12c and 36.12d show that as you move away from the central bright maximum of the two slit pattern, the intensity of the maxima decrease. This a result of the single slit modulated pattern. The narrower the slits, the boarder the single‐slit pattern and the slower the decrease in intensity from one interference maximum to the next. 36.3 Circular Apertures and Resolving Power 27
September 6, 2012Dr. Afaf AbdelHady Physics c The diffraction pattern formed by a circular aperture is of special interest because of its role in limiting how well an optical instrument can resolve fine details. The diffraction pattern formed by a circular aperture consists of a central bright spot surrounded by a series of bright and dark rings. We can describe the pattern in terms of the angle θ , representing the angular radius of each ring. If the aperture diameter is D and the wavelength is λ , the angular radius θ1 of the first dark ring is given by sin θ1 = 1.22
λ
D
The angular radii of the next two dark rings are given by sin θ2 = 2.23
λ
D
sin θ3 = 3.24
λ
D
Between these are bright rings with angular radii given by sin θ = 1.63
λ
D
,
2.68
λ
D
,
3.70
λ
D
And so on. The central bright spot is called the Airy disk. The angular radius of the Airy desk is that of the first dark ring. The intensities in the bright rings drop off quickly with increasing angle. When D is much larger than the wavelength, the usual case for optical instruments, the peak intensity in the first ring is only 1.7% of the value at the center of the Airy disk, and the peak intensity of the second ring is only 0.04%. Most (85%) of the light energy falls within the Airy disk. Diffraction and Image formation A widely used criterion for resolution of two objects, proposed by Rayleigh and called Rayleigh's criterion, is that the objects are just barely resolved (that is, distinguishable) if the center of one diffraction pattern 28
September 6, 2012Dr. Afaf AbdelHady Physics c coincides with the first minimum of the other. In that case the angular separation of the image centers is given by sin θ1 = 1.22
λ
D The minimum separation of two objects that can just be resolved by an optical instrument is called the limit of resolution of the instrument. The smaller the limit of resolution, the greater the resolution, or resolving power, of the instrument. Rayleigh's criterion shows that resolution (resolving power) improves with larger diameter; it also improves with shorter wavelengths. Ultraviolet microscopes have higher resolution than visible‐light microscopes. Diffraction Grating : section 36.5 from the book 29
September 6, 2012Dr. Afaf AbdelHady Physics c CHAPTER 38 PHOTONS, ELECTRONS, AND ATOMS Maxwell, Hertz and others established firmly that light is an electromagnetic wave. Interference, diffraction, and polarization demonstrate this wave nature of light. When we look more closely at the emission, absorption, and scattering of electromagnetic radiation, we discover a completely different aspects of light. We find that the energy of an electromagnetic wave is quantized; it is emitted and absorbed in particle‐like packages of definite energy, called photons or quanta. The energy of a single photon is proportional to the frequency of the radiation. The internal energy of atoms is also quantized. For a given kind of individual atom the energy can't have just any value; only discrete values called energy levels are possible. 38.1 Photoelectric Effect The photoelectric effect is the emission of electrons when light strikes a surface. This effect has numerous practical applications. To escape from the surface, the electron must absorb enough energy from the incident radiation to overcome the attraction of positive ions in the material of the surface. The minimum amount of energy an individual electron has to gain to escape from a particular surface is called the work function for that surface, denoted by φ . When monochromatic light fell on the cathode, no photoelectrons at all were emitted unless the frequency of the light was greater than some minimum value called the threshold frequency. This minimum frequency depends on the material of the cathode. 30
September 6, 2012Dr. Afaf AbdelHady Physics c When the frequency f is greater than the threshold some electrons are emitted from the cathode with initial speeds. We can determine the maximum kinetic the emitted electrons by making the potential of the relative to the cathode V AC , just negative enough so current stops. This occurs for V AC = −V 0 , where V 0 is frequency, substantial energy of anode that the stopping potential. As the electron moves from the the anode, the potential decreases by V 0 and negative cathode to work −eV 0 is done on the electron; the most energetic electron 1
2
and has cathode with kinetic energy K max − mv max
2
energy at the anode. Using the work‐energy theorem, leaves the K max =
called the zero kinetic we have 1
2
mv max
= eV 0 2
Hence by measuring the stopping potential V 0 , we can determine the maximum kinetic energy with which electrons leave the cathode. If the intensity of light is increased while its frequency is kept the same, the current levels off at a higher value, showing that more electrons are being emitted per time. But the stopping potential V 0 is found to be the same. When the frequency of the incident monochromatic light is increased, the stopping potential V 0 increases. In fact, V 0 turns out to be a linear function of the frequency f . Einstein's photon explanation Einstein postulated that a beam of light consists of small packages of energy called photons or quanta. The energy E of a photon is equal to a constant h times its frequency f. E = hf = h
c
λ
Where h is a universal constant called Planck's constant. A photon arriving at the surface is absorbed by an electron. The electron gets all the photon's energy or none at all. If this energy is greater than the work function φ , the electron may escape from the surface. Greater intensity at a particular frequency means a proportionally greater number of photons per second absorbed, and thus a proportionally greater number of electrons emitted per second and the 31
September 6, 2012Dr. Afaf AbdelHady Physics c proportionally greater current seen in Fig. 38.4. Recall that φ is the minimum energy needed to remove an electron from the surface. Einstein applied conservation of energy to find that the maximum kinetic energy K max =
1
2
= hf − φ mv max
2
Substituting K max = eV 0 , we find eV 0 = hf − φ
(photoelectric effect)
Electron energies and work functions are usually expressed in electron volts (eV). To four significant figures, 1 eV = 1.602 × 10 −19 J Photon Momentum According to the special theory of relativity, every particle that has energy must also have momentum, even if it has no rest mass. Photons have zero rest mass. The momentum of the photon is given by p=
E hf
h
=
= λ
c
c
The direction of the photon's momentum is simply the direction in which the electromagnetic wave is moving. 38.2 Compton scattering When x‐rays strike matter, some of the radiation is scattered. Compton and others discovered that some of the scattered radiation has smaller frequency (longer wavelength) than the incident radiation and that the change in wavelength depends on the angle through which the radiation is scattered. If λ and λ ' are the wavelengths of the incident and scattered radiation, we find that λ '− λ =
h
(1 − cos φ )
mc
(compton scattering) Where φ is the angle between the scattered photon and the incident photon, and m is the electron rest mass. The quantity h
has units of length. mc
32
September 6, 2012Dr. Afaf AbdelHady Physics c We can derive Compton scattering equation from the principles of conservation of energy and momentum. The electron recoil energy may be in the relativistic range, so we have to use the relativistic energy‐ momentum relationships. The incident photon has momentum P and energy Pc , the scattered photon has momentum P ' and energy P 'c . The electron is initially at rest, so its initial momentum is zero and its energy is its rest energy mc 2 . The final electron momentum Pe and the final electron energy is given by E e2 = ( mc 2 ) 2 + ( Pe c ) 2 . Then the energy conservation gives pc + mc 2 = p 'c + E e Rearrange, we find ( pc − p 'c + mc 2 ) 2 = E e2 = ( mc 2 ) 2 + ( pe c ) 2 We can eliminate the electron momentum Pe by using momentum conservation P = P '+ Pe BY using the law of cosines with the vector diagram in Fig.38.27c, we find p = p + p ' − 2 pp 'cos φ 2
e
2
2
Carrying out the necessary algebra leads to this simple result mc mc
−
= 1 − cos φ p'
p
Finally, we substitute p ' = h λ ' and p = h λ , then multiply by h mc to obtain λ '− λ =
h
(1 − cos φ )
mc
(compton scattering) When the wavelength of x‐ray scattered at a certain angle are measured, the curve of intensity per unit wavelength as a 33
September 6, 2012Dr. Afaf AbdelHady Physics c function of wavelength has two peaks (Fig. 38.28). The longer‐wavelength peak represents Compton scattering. The shorter‐wavelength peak, labeled λ0 , is the wavelength of the incident x rays and corresponds to x‐ray scattering from tightly bound electrons. In such scattering processes the entire atom must recoil, so m is t mass of the entire atom rather than a single electron. The resulting wavelength shifts are negligible. 38.3 Continuous Spectra Hot matter in condensed states nearly always emits radiation with a continuous distribution of wavelengths rather than a line spectrum. An ideal surface that absorbs all wavelengths of electromagnetic radiation incident upon it is also the best possible emitter of electromagnetic radiation at any wavelength. Such an ideal surface is called blackbody radiation. By 1900 this radiation has been studied extensively, and two characteristics had been established. 1‐
The total intensity I (average power per area) emitted from the surface of an ideal radiator is proportional to the fourth power of the absolute temperature. This is called Stefan‐Boltzmann law: I =σ T
4
(Stefan-Boltzmann law for a blackbody) Where σ a fundamental physical constant is called the Stefan‐Boltzmann constant. 2‐
The intensity is not uniformly distributed over all wavelengths. Its distribution can be measured and described by the intensity per wavelength interval I (λ ) called the spectral emittance. Thus I (λ )d λ is the intensity corresponding to wavelengths in the interval from λ to λ + d λ . The total intensity I is given by ∞
I = ∫ I (λ )d λ
Measured spectral emittances I (λ ) for three different temperatures are shown in Fig. 38.31. Each has a 0
peak wavelength λm at which the emitted intensity per wavelength interval is largest. Experiment shows that λm is inversely proportional to T, so their product is constant. This result is called Wien displacement law. λmT = 2.90 × 10-3 m.K
(Wien displacement law) As the temperature rises, the peak of I (λ ) becomes higher and shifts to shorter wavelengths. Expeiments show that the shape of the distribution function is the same for all temperatures. 34
September 6, 2012Dr. Afaf AbdelHady Physics c Caution: The spectral emittance I (λ ) is not the same thing as intensity I . Intensity is power per unit area, with units W m 2 ; spectral emittance is power per unit area per unit wavelength interval, with units W m 3 . Rayleigh considered light enclosed in a rectangular box with perfectly reflecting sides. He computes the intensity distribution of the radiation emerging from a small hole in the box. It is I (λ ) =
2π ckT
λ4
At large wavelengths this formula agrees quite well with the experimental results shown in Fig. 38.31, but there is serious disagreement at small wavelengths. Planck proposed a formula that fitted the experimental data perfectly at all wavelengths and for all temperatures. I (λ ) =
2π hc 2
λ 5 (e hc λ kT − 1)
(Planck radiation law) Where h is the Planck's constant, c is the speed of light, k is the Boltzmann's constant, T is the absolute temperature and λ is the wavelength. 38.4 Atomic Line Spectra and Energy Levels The origin of line spectra can be understood in general terms on the basis of two central ideas. One is the photon concept; the other is the concept of energy levels of atoms. These two idea were combined by Bohr in 1913. Photon emission by atoms Bohr said, each atom must be able to exist with only certain specific values of internal energy. Each atom has a set of possible energy levels. An atom can have an amount of internal energy equal to any one of these levels, but it cannot have an energy intermediate between two levels. All isolated atoms of a given element have the same set of energy levels, but atoms of different elements have different sets. According to Bohr, an atom can make a transition from one energy level to a lower by emitting a photon with energy equal to the energy difference between the initial and final levels. If E i is the initial energy before transition, E f is the final energy after transition, and the photon's energy is hf , then conservation of energy gives hf =
hc
λ
= Ei − Ef
(energy of emitted photon) 35
September 6, 2012Dr. Afaf AbdelHady Physics c The Hydrogen Spectrum The spectrum of hydrogen had been studied intensively. In an electric discharge tube, atomic hydrogen emits the series of lines. Balmer found by trial and error a formula that gives the wavelengths of these lines which are now called the Balmer series. We may write Balmer's formula as 1
λ
= R(
1
1
− 2)
2
2 n
(n=3,4,5,.....) Where λ is the wavelength, R is a constant called the Rydberg constant. Balmer's formula has a very direct relationship to Bohr's hypothesis about energy levels. E =
hc
λ
= hcR (
1
1
hcR hcR
− 2)= 2 − 2 2
2 n
2
n
hcR
hcR
as the initial energy E i and − 2 as the final 2
2
n
energy E f . The energies of the levels are negative because we choose the potential energy to be zero The above equation agreed if we identify −
when the electron and nucleus are infinitely far apart. The Balmer (and other) series suggest that the hydrogen atom has a series of energy levels, which we may call E n , given by En = −
hcR
n2
(n = 1, 2,3,.....) Other spectral series for hydrogen have been discovered. These are known as 1
1 1
= R( 2 − 2)
λ
1 n
Lyman series Paschen series Brackett series Pfund series 1
λ
1
λ
1
λ
= R(
= R(
= R(
1 1
− )
32 n 2
1
1
− 2)
2
4 n
1
1
− 2)
2
5 n
(n=2,3,4,.....) (n=4,5,6,.....) (n=5,6,7,.....) (n=6,7,8,.....) 36
September 6, 2012Dr. Afaf AbdelHady Physics c Every atom has a lower energy level that includes the minimum internal energy state that the atom can have. This is called the ground‐state level, or ground level, and all higher levels are called excited levels. A photon corresponding to a particular spectrum line is emitted when an atom makes a transition from a state in an excited level to a state in a lower excited level or the ground state. A photon can also be absorbed by a similar atom that is initially I the lower level. 38.5 The Bohr Model Rutherford's discovery of the atomic nucleus raised a serious question. What kept the negatively charged electrons at relatively large distances away from the very small, positively charged nucleus despite their electrostatic attraction? Rutherford suggested that perhaps the electrons revolve in orbits about the nucleus, just as planets revolved around the sun. To solve this problem, Bohr postulated that an electron in an atom can move around the nucleus in certain circular stable orbits without emitting radiation. According to Bohr, there is a definite energy associated with each stable orbit, and an atom radiates energy only when it makes a transition from one of these orbit to another. The energy is radiated in the form of a photon with energy and frequency given by 37
September 6, 2012Dr. Afaf AbdelHady Physics c hf = E i − E f Bohr found that the magnitude of the electron's angular momentum is quantized. As we know that the angular momentum is L = mvr for a particle with mass m moving with speed v in a circle of radius r. So Bohr's argument led to h
2π
where n=1,2,3,…… is called the principal quantum number for the orbit. Each value of n corresponding to a permitted value of the orbit radius. L = mvr = n
L n = mv n rn = n
h
2π
(quantization of angular momentum)
Let's consider a model of the hydrogen atom. This atom consists of a single electron revolving around a single proton. The inward acceleration is v n2 rn . According to Newton's second law, a radially inward net force with magnitude F = m v n2 rn is needed to cause this acceleration. From Coulomb's law F=
1 e2
4πε 0 rn2
Then 1 e 2 mv n2
=
4πε 0 rn2
rn
Solving for rn and v n , we get n 2h 2
rn = ε 0
π me 2
vn =
1 e2
ε 0 2nh
(orbital radii in the Bohr model) (orbital speeds in the Bohr model) The smallest orbit radius corresponds to n = 1 . We will denote this minimum radius, called the Bohr radius, as a0 : 38
September 6, 2012Dr. Afaf AbdelHady Physics c h2
a0 = ε0
π me 2
rn = n 2a0 The permitted orbits have radii a0 , 4a0 ,9a0 , and so on. The kinetic energy and potential energy when the electron is in the orbit with quantum number n: Kn =
Un = −
1
1 me 4
mv n2 = 2 2 2 ε 0 8n h
2
1 e2
1 me 4
=− 2
ε 0 4n 2 h 2
4πε 0 rn
The total energy En = K n +U n = −
1 me 4
ε 02 8n 2 h 2
The potential energy has a negative sign because we have taken the potential energy to be zero when the electron is infinitely far from the nucleus. We interested only in energy differences, so the reference position does not matter. The energy of the atom is least when n=1 and E n has its most negative value. This is the ground level of the atom; it is the level with the smallest orbit, with radius a 0 . For n=2,3,.... the absolute value of E n is smaller and the energy is progressively larger. The ionization energy of the hydrogen atom is the energy required to remove the electron completely. Ionization corresponds to a transition from the ground level (n=1) to an infinitely large orbit radius (n=∞ ) and thus equals -E1 . 39
September 6, 2012Dr. Afaf AbdelHady Physics c CHAPTER 43
Nuclear Physics
THE STRUCTURE OF AN ATOM
1. Plum Pudding Model: An atom was assumed to be a sphere of positive charge with
negatively charged electrons spread through it rather like currants in a pudding (or seeds in
a a watermelon).
2. Rutherford Experiment (Scattering of Alpha Particles): Alpha particles are tiny particles
which are positively charged and are relatively heavy (on an atomic scale). When alpha
particles were directed towards a very thin metal foil (10-3 mm thick), the number of
alphas scattered at different angles were counted. It was found that:
a. Most of the alpha particles passed straight through the foil without deflection. This
indicates that the atom consists mainly of empty space.
b. Few of the alpha particles were deflected sideways. This indicates that they were
repelled by the positive nucleus.
c. Very few alpha particles were deflected with a large angle (backscattered), indicating
that they collided with a very heavy central nucleus.
d. Because the atom is neutral, the electrons revolving about the nucleus should carry a
negative charge which is equal to the positive charge of the nucleus.
40
September 6, 2012Dr. Afaf AbdelHady Physics c Representation of the structure of an atom
The Nucleus of an atom is a positively charged entity at the atom's center. Its radius is about
10-15 m, which is about 10-5 as large as the radius of the atom. All nuclei are composed of two
types of particles: protons and neutrons, collectively called nucleons. The only exception to
this is the ordinary hydrogen nucleus, which is a single proton. Although the positively
charged protons repel each other, the much stronger, short-range nuclear force holds the
nucleus together. The nuclear attractive force between nucleons decreases rapidly with
particle separation and is essentially zero for nucleons more than 5×10-15 m apart.
The Atom consists of a nucleus containing the protons and neutrons surrounded by
electrons revolving in different levels.
Atomic Number (Z): it is the number of protons in the nucleus. Each proton within the
nucleus carries a charge +e, whereas the neutron carries no charge. If there are Z protons in a
nucleus, then the charge on the nucleus is +Ze. The atomic number is sometimes called the
charge number.
Neutron Number (N): it is the number of neutrons in the nucleus.
The Mass Number (A) of an atom is equal to the number of nucleons (neutrons plus
protons) in the nucleus of the atom.
A= N +Z
Because normal atoms are electrically neutral, the neutral atom has Z electrons outside the
nucleus. These Z electrons determine the chemical behavior of the atom. As a result, all
atoms of the same chemical element have the same value for Z. For example, all hydrogen
atoms have Z=1, while all carbon atoms have Z=6.
In keeping with this notation, we designated the nucleus having a mass number A and an
atomic number Z by the symbolism
A
Z
( CHEMICAL SYMBOL )
For example:
56
26
Fe (iron) has a mass number of 56 and an atomic number of 26; therefore,
it contains 26 protons and 30 neutrons.
41
September 6, 2012Dr. Afaf AbdelHady Physics c Main particles of an atom
Particle (location)
Symbol
Mass (kg)
Charge
Proton (in the nucleus)
p , 11 H
1.67262×10-27 +e
Neutron (in the nucleus)
n , 01 n
1.67493×10-27 0
Electron (outside the nucleus)
e− , β − , −01 e
9.10939×10-31 -e
Isotopes: Atoms of the same element with the same Z, but different N and A values. Isotopes
have the same chemical properties, but differ in some physical properties.
For example, ordinary oxygen consists of three isotopes that have mass number 16, 17, and
18. Each of the isotopes has Z = 8. Hence these isotopes have the following numbers of
neutrons in their nuclei: 16 – 8 = 8, and 17- 8 = 9, and 18 – 8 = 10. It is customary to
represent the isotopes in the following way:
16
8
or simply as
16
O,
17
O,
18
O,
17
8
O,
18
8
O,
O , where it is understood that oxygen always has Z = 8.
Carbon's isotopes are:
11
6
C, 126 C, 136C, 146 C
and hydrogen isotopes are:
1
1
H
ordinary hydrogen
2
1
,
H ,
deuterium
3
1
H
tritium
Ionization:
A positive ion is formed when a neutral atom loses one electron or more.
A negative ion is formed when a neutral atom gains one electron or more.
Stable and Unstable Nuclei: A stable nucleus contains the right balance of protons and
neutrons, and remains unchanged. Having too many or too few neutrons makes it unstable
and likely to change (decay). The unstable nucleus decays by any process which brings it
nearer to the correct number of neutrons for stability.
Radioactivity: It is a spontaneous disintegration (decay) of unstable nuclei of atoms of
heavy elements.
42
September 6, 2012Dr. Afaf AbdelHady Physics c Radioactive decay is a random process. It can not be controlled by heat, catalysts or
chemicals because it is a nuclear reaction, not a chemical one.
Types of radiation:
During disintegration of the nucleus, certain particles and radiation are given out and a
nucleus (and thus a new element) is formed. There are three-types of radiation
1- Gamma radiation (γ): it is electromagnetic waves of very high energy or very short
wavelength. It carries no charges, so it is not deflected by electric or magnetic fields. It
has a high penetrating power, but a weak ionizing power. It has wave properties
(interference and diffraction).
A
Z
→
X*
parent nucleus
A
Z
+γ
X
daughter nucleus
For example:
12
6
→
C*
12
6
C+γ
where * indicates that a nucleus in an excited state.
2- Beta particles ( β − ): they are negatively charged electrons ( β − = −01 e ). They have a
less penetrating power, but are highly absorbed. They are deflected in magnetic or electric
fields as negative charged particles.
→
X
A
Z
parent nucleus
Y +β −
A
Z+1
daughter nucleus
For examples:
14
6
C → 147 N+ β −
Notice that in beta decay, a neutron changes to a proton and electron. It is also important to
note that the electron in these decays is not present before in the nucleus, but is created at the
moment of decay from the rest energy of the decaying nucleus.
Alpha particle (α): it is a helium nucleus 42 He . It has a much less penetrating power, but is
highly absorbed (could be stopped by a sheet of paper), and very strongly ionized. It is
deflected in magnetic or electric fields as positive charged particles
X
A
Z
parent nucleus
→
A−4
Z−2
Y + 42He
daughter nucleus
43
September 6, 2012Dr. Afaf AbdelHady Physics c For examples:
238
92
U
226
88
Ra
→
→
Th + 24 He,
234
90
222
86
Rn + 42 He
Note that the mass number of the daughter nucleus is 4 less than that of the parent nucleus.
Likewise, the atomic number is reduced by
Section 43.3: From the book
TABLE (1)
Properties
α -particles
β -Particles
γ -Rays
Nature
Positively charged
helium nucleus
Negatively charged
electrons
Electromagnetic
waves
Penetration
Absorbed by about
6 cm of air, and by
thin paper
Absorbed by about
100 cm of air, and
by about 3 mm of
aluminum
½ km of air or 4 cm
of lead reduces the
intensity to about
tenth
Ionization
Strong, because they weak
are massive charged
particles
Deflection by
electric and
magnetic field
Very slight, because
they have a large
mass
Large, because they
have a small mass
No deflection,
because they are
uncharged
Detectors
Photographic film
Photographic film
Photographic film
Cloud chamber
Cloud chamber
Cloud chamber
G-M tube
G-M tube
G-M tube
Very weak
44
September 6, 2012Dr. Afaf AbdelHady Physics c Particles and Rays in Magnetic Fields
Alpha particles (α): are heavy and quite slow, thus they need a very strong field
to
deflect them (deflection is small).
Beta particles (β): are much smaller and faster. They are deflected more easily.
Gamma rays (γ): are not particles, and have no charge, so they are not deflected at all.
Background radiations: It is the count rate which is always present even when there is no
artificial radioactive source in the area. The background radiations are present due to the
following:
1. Radioactive materials in the ground and around us, and
2. "Cosmic radiations" which come from the outer space and fall on earth.
The background count-rate should be recorded and subtracted from the count rates obtained
from a radioactive source.
Half-life: The half-life of a radioactive source is “the time it takes for the activity of the
source to fall to half its original value” (irrespective of what this value may be).
OR: It is the time taken for half the nuclei present in any given sample to decay.
OR: It is the time taken for half the atoms of a radioactive substance to decay.
Section 43.4: From the book
45
Example: The half-life of a
radioactive substance is 10 hours.
The original activity of a sample is
measured and found to be 1200
counts per minutes. What count will
be after 40 hours?
Activity (number of atoms surviving)
September 6, 2012Dr. Afaf AbdelHady Physics c 1200
600
300
10
20
time
Answer:
After 10 hours (one half-life) the count rate will be 600 counts per minute.
After 20 hours (two half-life) the count rate will be 300 counts per minute.
After 30 hours (three half-life) the count rate will be 150 counts per minute.
After 40 hours (four half-life) the count rate will be 75 counts per minute.
Danger and Safety: When radiation is absorbed by the cells of the human body it can
cause diseases, such as Leukemia (blood's cancer), cataract (causing blindness), cancer and
sterility. Radiations can also alter the nucleus of cells which causes mutations or handicaps in
newly born babies.
Handling radioactive materials and safety:
1.
2.
3.
4.
5.
6.
7.
Gloves should be worn and the sources held by forceps.
Keep sources as far as possible from a body by using very long tweezers.
Never point the sources towards the human body.
Food should be kept away from the radioactive sources to avoid possible contamination.
Hands should be washed after handling the radioactive materials.
Ensure that the exposure time to any radiation is kept to a minimum.
When not in use, laboratory sources
are always stored in their special
boxes lined with lead, with the
symbol shown for radioactive
material on them.
Note: the time of exposure and the strength of a source determine the extent of its danger to
the body.
Radioactive Waste: Waste of radioactive products is either buried deep into the ground at
special remote sites, or in an ocean bed.
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September 6, 2012Dr. Afaf AbdelHady Physics c Uses of radioactive isotopes
Radioactive isotopes can be made by bombarding substances with particles α , β and γ - rays. These
radioactive isotopes can then be used, with care, in many different ways:
1- Radioactive tracers:
a. Radioactive fertilizers can be fed to plants and then traced through the plant using a G-M
Counter. This is used to develop better fertilizers and insecticides.
b. Similarly, in hospitals, an isotope with a short half-life may be injected into a patient’s body. A
G-M counter is used to trace the movement of the chemical through the body, to check that
certain glands are functioning properly. Tumors can be located because they take up more
radioactive isotopes than other parts of the body.
c. If radioactive pistons are fitted to a
car, then a G-M counter can be used to test the oil for tracers of wear from the piston.
d. Leaks from a pipeline carrying oil or gas can be traced by injecting a radioactive isotope into
the pipeline. This saves digging up
the whole pipeline.
2- Sterilising:
a. γ -rays can be used to kill bacteria in foods or in hospital blankets without making them
radioactive.
b. Hospital equipment can be sterilised after being sealed in plastic bags.
3- Cancer Treatment:
Cancer cells in a patient’s body may be killed by careful use of γ -radiation.
4- Checking welds:
If a γ -source is placed on one side of weld metal and a photographic film is placed on the
other side, weak points will show up when the film is developed.
Or If a source is placed on one side of the weld metal and a G-M counter on the other side,
the thickness of the weld can be controlled by measuring how much γ -radiation passes
through it.
5- Thickness control:
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September 6, 2012Dr. Afaf AbdelHady Physics c In paper mills the thickness of paper can be controlled by measuring how much β -radiation pass
through the paper to a G-M counter. The counter controls the pressure of the rollers to give the
correct thickness.
6- γ -rays kill the microbes that cause fruit and vegetables to decay. Food irradiated by γ ray stays fresh on shelves for longer periods of time.
7- Radioactive dating:
a- carbon dating: It is the process by which radioactivity is used to find the age of anything
that was once living material, like an old piece of wood or charcoal, plants or shells preserved
in rock, ancient manuscripts and bones. When a plant or animal dies, it stops taking in carbon.
But C-14 atoms continue to decay. By comparing the radiation from them with that from
living things, and knowing the half life of C-14, the age of the remains can be worked out.
Example: The activity of a living plant is 18 counts/second for each gram of carbon, and the
activity of a piece of old wood is 3 count/second. How old is the wood if the half-life of C-14 is
5600 years?
Solution: The activity dropped
From 18 count/second ----- 9 count/second in one half-life.
From 9 count/second-----3 count/second in the second half-life.
There are 2 half-lives, so the age of the wood = 2 (5600) = 11200 years
Rock dating: Many rocks contain potassium compounds. When the potassium-40 in these
compounds decays, it forms a stable argon which stays trapped in the rock. By measuring their
amounts, scientists can work out the age of the rock.
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September 6, 2012Dr. Afaf AbdelHady Physics c Assignments (text Book, University Physics)
Chapter 26:
4,8,11,21,40,42,45,50,51,52.
Chapter 28:
1, 3, 9, 15, 22, 25, 30, 36, 39.
Chapter 29:
1, 6, 8, 11, 15, 19, 21, 23.
Chapter 33:
1, 3, 5, 9, 25, 30,
Chapter 35:
1, 8, 9, 12, 19, 37.
Chapter 36:
1,4,8,14,23,28,32,42.
Chapter 38:
3,6,15,24,36,37,38,43,44.
Chapter 43:
1,13, 18,20, 21, 22,26, 28.
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