Section Three: Light and Matter Topic One: … 1 of 17 Section Three: Light and Matter Topic One:...

of 17

Section Three: Light and Matter Topic One: Electromagnetic Waves electromagnetic waves are transverse waves that have both an electric and a magnetic component. A changing

magnetic field (B) would produce a changing electric field (E) that is at right angles to the magnetic field and that a changing electric field will produce a changing magnetic field. The net result of the interaction of the changing E and B fields is the production of an electromagnetic wave that moves away from the wave source at the speed of light.

electromagnetic spectrum includes radio waves, microwaves, infrared radiation, visible light, x-rays and gamma rays. They differ in frequency f and wavelength λ but all travel at the speed of light.

Electric Fields and Waves Shake the end of a stick back and forth in still water and you will produce waves on the water surface. Similarly shake a charged rod back and forward in empty space and you will produce electromagnetic waves in space. This is because the shaking charge can be considered a varying electric current and a varying magnetic field surrounds such an electric current. No medium is required. The vibrating fields move out from the vibrating source and at any point in space the electric field is perpendicular to the magnetic field and both are perpendicular to the direction of the wave. Light is energy that is emitted by accelerating charges - often electrons in atoms. This energy travels in a wave that is partly electric and partly magnetic. A changing electric field E will produce a changing magnetic field B.. This electromagnetic wave is transverse in nature because both the E and B fields are perpendicular to the direction of propagation of the wave. These waves are waves of fields not matter and can propagate (move) in empty space. Accelerating electric charges give rise to electromagnetic waves.

Reception of EM Wave by an Antenna One kind of antenna consists of one or more conducting rods. The electric field in the EM wave exerts a force on the electrons in the conductor causing them to move back and forth at the frequencies of the wave. Note: A horizontal antenna radiates (or receives) waves that are

polarised in the horizontal plane. A vertical antenna radiates (or receives) waves that are

polarised in the vertical plane.

Polarisation The plane of polarisation of an electromagnetic wave is the plane defined by the direction of travel and the direction of the oscillating electric field.

Unpolarised light oscillates in many directions perpendicular to the direction of travel. These oscillations can be resolved into vertical and horizontal components. Unpolarised light can be passed through a polariser to produce plane polarised light in which the oscillations are confined to a single plane.

A polariser (or polarising filter) can also be used as an analyser to determine the polarisation of light. If the light passes unaltered, the polarising filter is aligned with the polarisation plane of the light. If the light is completely blocked, the polarising filter is at right angles to the plane of polarisation of the light.

APPLICATION: LADS (Laser Airborne Depth Sounder) This device is an Australian system which aims to accurately chart water depth in the regions of sea surrounding Australia – of course it can be used to chart fresh water lakes as well. As a device to measure depths in water it is not unique. This was done previously using sound waves emitted vertically down from slow moving ships near the coast. LADS uses the same physical idea in that it involves the measuring the time difference between emission and reception of a pulse of waves that have been both reflected from the top water surface and it bottom. If the speed of the wave in the water is known then Time difference × speed = depth 2

of 17

LADS uses optical rather than sonar methods. A pulsed beam of laser light is directed down from a plane – from an infra red laser. The pulse width is very short to improve accuracy. This light reflects from the surface to the plane, which is carrying the source. An electronic device doubles the frequency of the laser light so that a beam of green visible light is also produced. The infra red light is almost all reflected from the water surface, the green light penetrates the surface and is reflected from the bottom of the sea, or lake. Both the infra red and green light travel at the same speed in air. The plane flies quite low above the water surface and by measuring the time delay between a pulse being reflected from the surface and one (green) (infra red) from the bottom the time for double the depth for the green light (green) to travel at a known speed, in water, can be calculated. The speeds of green light, in water are about 2.24 × 108 m s-1 – it depends on whether salt water or fresh water is used. In addition problems arise because of light energy being absorbed or scattered by particles in the water so that high-energy lasers are needed. infrared

green Topic Two: Interference of light Key Terms interference patterns are produced when two (or more) coherent sources produce waves of the same frequency

and amplitude which superimpose in the same region of space (ie they overlap). sources are coherent when there is a constant phase relationship between the waves emitted by the sources. crest is the highest point of that portion of

a transverse wave above the equilibrium position.

trough is the lowest point of that portion of a transverse wave below the equilibrium position.

destructive interference occurs if the amplitude of the resultant of two interfering waves is smaller than the amplitude of either wave.

constructive interference occurs if the amplitude of the resultant of two interfering waves is larger than the amplitude of either wave.

diffraction occurs when waves bend and spread out as they pass an obstacle or through a narrow opening. The amount of diffraction depends on the wavelength of the waves relative to the size of the obstacle. In the case of a narrow opening, the amount of bending increases as the size of the opening decreases.

diffraction grating consists of a large number of closely space parallel slits which diffract light incident on the grating. The diffracted light will exhibit a pattern of light and dark spaces due to constructive interference and destructive interference.

polarization is a property of light that indicates that light is a transverse wave phenomenon. A transverse wave is a wave in which the oscillations or vibrations of the wave are at right angles to the direction of motion of the wave. An electromagnetic wave in which the electric vector is vibrating in only one plane is said to be plane polarized.

Speed, Frequency and Wavelength The back and forth vibratory motion (often called oscillatory motion) of a swinging pendulum or mass suspended on a spring produces a sine curve. This curve depicts a wave: Wavelength (λ) is the distance from one crest to another crest, more generally the shortest distance between successive identical parts of the wave that are in phase. λ is measured in metres. Amplitude (A) is the distance of the maximum displacement point of the wave from the mean position. (metres.)

Dis

plac

emen

t

W a ve le n gt h

A m p l itu de

+ v e

- ve

C r e s t

T ro u g h

ocean floor

water surface

of 17

v = fλperiod1frequency =

Frequency (f) is the number of to-and-fro vibrations in time. f is measured in cycles per second or hertz. where Period (T) is the time taken to complete one oscillation. Wavemotion involves a transfer of energy from one point to another, with no net transfer of matter.

Wave Equation here: v = speed or velocity of the wave (in m s-1) f = frequency in hertz (Hz)

λ = wavelength in metres (m). Transverse Waves A wave where the motion of the oscillations of the wave are at right angles (perpendicular) to the direction of travel.

It consists of a series of crests and troughs.

Longitudinal Waves The medium moves back and forth parallel to the direction in which the wave travels. The wave consists of a series of compressions and rarefactions. Note: Transverse waves and longitudinal waves transfer the energy along the medium. The matter does not transfer but moves to and fro about a fixed position. This relationship holds for all kinds of waves whether they are water waves, sound waves, radio waves, or light waves. The speed of light is a universal constant, c = 3.0 × 108 m s-1 in a vacuum. Electromagnetic Waves Light is energy that is emitted by accelerating electric charges – often electrons in atoms. This energy travels in a wave that is partly electric and partly magnetic. Such a wave is an electromagnetic wave. Light is a small portion of the broad family of electromagnetic waves that include radio waves, microwaves and x-rays. The range of the electromagnetic waves is called the electromagnetic spectrum. The range of wavelengths in the visible spectrum is shown below. (1 nm = 1.0 × 10-9 m)

Wave Fronts A wave front is a line on a surface on which all points are in phase at any instant. This means that when one point is at the crest, so are the others. Huygens’ Principle “Every point on a wave front can be considered as a source of tiny wavelets that spread out in the forward direction at the speed of the wave itself. The new wave front is the envelope (ie. the common tangent to) of all the wavelets.”

Note: This principle is useful in predicting the future position of a wave front when an earlier position is known. Suppose that AB is a wave front for waves created from a point source S (ie. the wave is travelling away from S at a speed v). In the wave front AB which is travelling away from a source S with a speed v. To find the wave front after a short time t tiny circles, centred on wavefront AB, radius r = vt represent the Huygens’ (imaginary) wavelets. The tangent to all these wavelets, the curve CD, is the new position of the wavefront. Interference

A monochromatic light source emits a wave of a single colour and therefore has single frequency and wavelength in a vacuum(in air). e.g. Red light has wavelength of about 7.5×10-7m and hence a frequency of about 4×1014Hz. Two sources are coherent if the waves created by them have a constant phase relationship. An interference pattern is observed only when the sources are coherent. If two tiny light bulbs are used as sources, an interference pattern would not be seen. The light emitted by one light globe would have a random phase with respect to the second light bulb. Two such sources, whose output waves bear no fixed relationship to each other, are called incoherent sources. Wave Addition Wave Addition is the superposition of wave forms (interference)

v

R = R a r e f a c t io n C = C o m p r e s s i o n

C C C C R R R R

P a r t ic le V i b r a t io n

λ

C r e s t + C r e s t

+ =

T r o u g h + T r o u g h

+ =

C o n s t r u c t i v e I n t e r f e r e n c e

Crest + Trough

+ = Destructive Interference

of 17

Young’s Double Slit Interference Convincing evidence for the wave nature of light was provided by Thomas Young who was able to measure the wavelength of visible light using an interference effect. If light from a single monochromatic source falls onto a screen containing two closely spaced slits S1 and S2, bright and dark fringes are seen on the screen behind.

Young realised that the bright fringes of light resulted from light waves from both holes (or slits) arriving crest to crest (constructive interference - more light). Similarly, the dark areas resulted from light waves arriving trough to crest (destructive interference - no light). Young’s experiment can done in the laboratory with two closely spaced holes (or slits) with a sodium vapour lamp, light from which is passed through a single slit providing the monochromatic coherent source (a laser is even better because no first single slit is required only a double slit). The arrangement above is equivalent to two rocks being thrown into a lake, or

when sound from two loudspeakers interferes. These two slits have circular waves leaving them because of the diffraction effect.

Using the diagrams we can see how an interference pattern is produced on the screen. The waves of wavelength λ are shown entering the slits S1 and S2 which are a distance apart d. The waves spread out in all directions after passing through the slits. The diagrams show only three different angles. (a) (b) Diagram (a) shows that at the centre of the screen (θ = 0°) the waves from each slit travel the same distance and are in phase. ie. a crest of one wave arrives at the same time as the crest of the other wave. Hence the amplitude of both waves add to form a larger amplitude and this is known as CONSTRUCTIVE INTERFERENCE and there is a bright spot in the centre of the screen. Diagram (b) shows that at an angle θ the lower wave travels an extra distance of one whole wavelength, and the waves are in phase. Note that from the shaded triangle the extra distance equals d sinθ. At the point on the screen constructive interference occurs again since the paths of the two rays differs by one wavelength (or any whole number of wavelengths). Diagram (c) shows that at this angle of θ the lower wave travels an extra distance equal to one-half wavelength, so that the two waves arrive at the screen fully out of phase. At this point on the screen since the extra distance travelled is one half wavelength (or 3/2λ, 5/2λ and so on) the two waves are exactly out of phase when they reach the screen: the crest of one wave arrives at the same time as the trough of the other and so produce zero amplitude. This is DESTRUCTIVE INTERFERENCE and the screen is dark. The pattern on the screen is a series of bright and dark lines (or fringes) as shown below.

Young’s Double Slit Interferometer The coherent sources produce a pattern of uniformly spaced bright (constructive interference) and dark (destructive interference) fringes. The point O is a maximum intensity or reinforcement since the wavelets from the sources S1 and S2 meet at O at the same time. The waves travelling from S1 and S2 to the screen will have a path difference. If the path difference (the extra distance travelled by one wavelet to reach the same point on the screen) is:

of 17

Ldx

Lx

==

=

≈≈=

CSPD

small) (since tansin)difference(path dsinCS

2

2

θθθθ

an odd number of half wavelengths then DESTRUCTIVE INTERFERENCE an even number of half wavelengths then CONSTRUCTIVE INTERFERENCE. A constant phase relationship between the sources is a necessary condition for observable interference to occur, they are COHERENT (implies same λ). To achieve maximum contrast the amplitudes must be equal.

A single source S produces wavelets that meet S1 and S2 at the same time (since SS1=SS2). Therefore the sources S1 and S2 have a constant same phase relationship (in fact the same phase) so they are coherent. From the diagram below

If consider a position of constructive interference then

...)0,1,2,3,4,n (where n

ofnumber wholeCSPD 2

==

===

λ

λLdx

For a particular constructive band (fringe) nCS 12 λ==

Ldx

The path difference for the next band (fringe) is ( ) 1nCS2 λ+==L

dx2

Rearranging these two equations so that x is the subject of the formulae gives:

dLx λn

1 = ( )

dLx λ1n

2+

= ( )

dL

dLxx λλ n1nW 12 −

+=−= so

dLλ

=W

Note: W is the fringe separation or band width. since λ is small we need L to be large

and d small to see W. W is proportional to λ if L and d are

constant. WRED > WVIOLET since λRED>λVIOLET.

The diagram of the intensity pattern of interference fringes produced by the double slit experiment. The arrow marks the central fringe.

APPLICATION: CD (Compact Disc) player The CD player uses a moulded plastic disc for the reproduction of sound. The CD works digitally using a binary code. The CD has a spiral groove engraved on its surface (using a pulsed high-powered laser). The spiral goes from the inside of the disc out. In the grooves there are a succession of pits ie The CD can be duplicated using a process where the engraved moulding is coated with a thin layer of either Aluminium or Silver (both good reflectors of light as well as good conductors of electricity). This metal layer is then coated with a transparent, acrylic plastic. When the disc is placed in the CD player and turned on, the narrow beam of light from the low power, solid state (and hence robust) laser can be alternatively reflected from the upper and lower humps on the disc - the disc is read from underneath;

SCREENDouble Slit

Single Slit

S S1

S2

M O

P

θ θ d

L

x

C

of 17

λ/4 2 × λ/4 = λ/2, thus there is a λ/2 total path difference between light reflected from the two, "depths" on the disc due to destructive interference when path difference = λ/2

annul zero path difference therefore reinforce.

λ/4

The digital signals received are transformed into electrical pulses, which can then be transformed into pressure variations that we detect as sound. The diagram below shows how the system is set up Reading a CD. The fine beam of a laser, focused even more finely with lenses, is directed at the under surface of a rotating compact disc. The beam is reflected back from the areas between the pits but reflects much less than the pits. The reflected light is reflected as shown, reflected by a half-reflecting mirror MS. The strong and weak reflections correspond to the 0s and 1s of the binary code representing the audio or video signal. The advantage of a CD over other methods of sound reproduction is that nothing mechanical (like a needle with a record) actually touches the disc and also light from the pits is such that dust causes no sound distortion. The reflected laser beam is also used to control the spin rate of the disc. The rate varies from 200 revs/minute when near the centre to 500 revs/minute near the circumference. This ensures a constant linear speed. A standard CD stores about 74 minutes of sound. Sample Problems: WAVES: Interference 1. Consider the arrangement for Young’s double slit experiments using monochromatic light. (a) What effect would decreasing the distance between the two slits have on the distance between the centres of the

bright bands?

dLλ

=W If d is decreased then W is increased (inverse proportionality with λ and L constant). 2. What is the distance between the slits for a double slit arrangement which, illuminated with yellow light of

wavelength 6.0×10-7m, produces and interference pattern 1.5m away with bright bands 0.30 cm apart?

L = 1.5 m, W = 0.30 cm = 3×10-3 m, λ = 6.0×10-7 m

3. Monochromatic light passes through two parallel slits 0.03cm apart and then falls on a screen 120cm further away. The first bright band is 0.16cm from the centre of the interference pattern. Find the wavelength of the light.

d = 0.030 cm = 3×10-4 m, L = 1.2 m, W = 0.16×10-2 m

4. Young’s double slit experiment. (a) In a Young’s double slit experiment using sodium light, nine interference fringes were found to occupy 3.6 mm

when viewed through a micrometer eyepiece with an image plane was 240 mm from the slits. Calculate the separation of the slits, assuming the wavelength of sodium light to be 6×10-4 mm.

10 bands occupied 3.6mm (9 fringes) L = 240 mm, d = ?, λ = 6×10-7 m, W = 3.6/10 = 0.36 mm

m103103

5.1106d

d5.1106103

dLW

43

7

73

−−

−

−−

×=×

××=

××=×⇒=

λ

m1042.1

1031016.0103

2.11016.0LW

742

4-2

−−−

−

×=×××

=

××

=×⇒=

λ

λλd

m104

104.2107.6106.3W

4

174

−

−−−

×=

×××=×⇒=

ddd

Lλ

of 17

(b) When the sodium lamp was replaced by a white light source with a green filter, and the eyepiece moved 120 mm further from the slits, five green fringes were found to occupy 2.3 mm in the eyepiece. Find the mean wavelength of the light passed by the filter.

5 fringes (6W = 23 mm) L = 240 + 120 = 360 mm, λ = ?, W = 2.3/6 = 3.8×10-4m

5. What is meant by the term coherent light sources? Why are they necessary for interference effects to be observed with light and how are they produced? Coherent sources have a constant phase relationship (same λ) and better resolution if same amplitude (same intensity). Light is not normally coherent (except laser light).

6. In a Young’s double-slit interferometer, the separation of the slits is 0.1 mm. What is the band width of the interference pattern produced 50 cm from the slits?

(a) with red light of wavelength 700nm mm35d

WR ⇒=Lλ

(b) with blue light of wavelength 450nm

m102.4104

10360108.3dxW 7

4-

34 −

−− ×=

×××

=×⇒= λλλ so

0.5mm100.1 -3

=×=

Ld

mm5.22WB ⇒=dLλ

I

x

λ R=700nm W R=3.5cm

λ B=450nm W B=2.25cm

of 17

d

Diffraction Gratings Transmission Diffraction Gratings We have discussed diffraction before with regard to water waves as well as light and we now have seen that it refers to the spreading or bending of light around edges. A diffraction pattern exists around any sharp object illuminated by a point source. This resembles the interference fringes of a double slit and is due to the waves being diffracted around the object. Diffraction Grating A large number of equally spaced parallel slits is called a diffraction grating, although it may be more appropriate to use the term interference grating.

The grating can be made by precision machinery by scratching very fine parallel lines on a glass plate. The untouched spaces between the lines serve as the slits. Photographic transparencies of the original grating serve as an inexpensive grating. Transmission Grating: is a glass slab on which a large number of lines have been scratched so that the spacings of scratches is very small (eg. 10,000 scratches per cm). The light is passed through the transparent part thus supplying a many slit source.

Reflection Grating is a mirror surface on which a large number of lines have been scratched.

The light is reflected from the smooth reflecting surface, eg. compact disc. Both operate similarly except that the reflection grating is preferred as less energy from the incident light is absorbed. These are also the most expensive. Reflection gratings are also useful as ultra-violet and infrared light are absorbed by glass. The analysis of the diffraction grating is much like that of Young’s double-slit experiment. We assume parallel rays of light are incident on the grating. Since this light is parallel there is a need for it to be focussed by a convex lense or even the lense in our eye. We also assume that the slits are narrow enough so that diffraction by each of them spreads light over a very wide

angle on a screen and interference can occur with light from all other slits.

Light rays that pass through each slit without deviation (θ=0°) interfere constructively to produce a bright line at the centre of the screen.

Constructive interference also occurs at an angle θ such that the rays from adjacent slits travel an extra distance of Δl = mλ, where m is an integer.

For constructive interference ...) 2, 1, 0,(mmsin ==Δ= whered λθ l

These are the principal maxima and this is the criterion to have a maximum brightness. Note previously in interference: S2C = path difference = d sin θ where m is the order of the pattern eg m = 1 dsinθ=mλ=λ m = 2 dsinθ=mλ=2λ etc. When white light is passed through the grating, each wavelength produces constructive interference at a different angle and so a series of spectra are produced.

(a) a line spectrum with two wavelengths of light, 400 nm and 700 nm.

(b) Diffraction produced by white light. the shorter wavelengths (blue) are diffracted through smaller angles than the longer wavelengths (red), and the 1st order spectrum will show less dispersion (spread) than the 2nd order. the intensity of the 1st order spectrum is greater than the intensity of the 2nd order because of diffraction causing greater dispersion.

of 17

Sample Problems: WAVES: Diffraction

1. Light of wavelength 5.0×10-7m passes through two parallel slits and the falls on a screen 4.0m away. The bright bands of the interference pattern are found to be 2.0cm apart. Find the distance between the slits. The same two slits are now illuminated by light of a different wavelength and it is found that the fifth minimum for this light occurs at the same point on the screen as the fourth minimum for the previous light. Find the wavelength of the second source of light.

m10102

4105.0Wxd

dxW 4

2-

-7−=

×××

==⇒=λλ

metres 1041055.55.4

5.45.5 WsinceW4W5

772

12

121

221

−− ×=××=

=

∝=

λ

λλλ

2. A diffraction grating has 6×105 lines per metre. Find the angular spread in the second order spectrum between red light of wavelength 7.0×10-7m and violet light of wavelength 4.5×10-7m.

°

°==⎟⎟⎠

⎞⎜⎜⎝

⎛×

××==

°==⎟⎟⎠

⎞⎜⎜⎝

⎛×××

==

−

−

24.3 is separationAngular

6.320.539sin1067.1

105.42sinndsinfor violet

56.90.838sin1067.11072sinndsinredfor

1-6

7-1-

1-6

-71-

θλθ

θλθ

so

so

3. A He-Ne gas laser which produces monochromatic light of a known wavelength λ=6.328×10-7m is used to calibrate a grating in a spectroscope. The first order diffraction line is found at an angle of 20° to the incident beam.

How many lines per metre are there on the grating?

elines/metr105.4 sometres1085.1sin20

106.328d106.328dsin20ndsin 56-7

7- ××=°

×=×=°= −soλθ

4. The red light from three separate sources is mixed and passed through a diffraction grating of 3×105 lines per metre. The wavelengths of the three lines are:

Hydrogen 6.56×10-7m Neon 6.50×10-7m Argon 6.97×10-7m (a) Calculate the deviations for the first order diffraction lines of each of these sources.

°==⎟⎟⎠

⎞⎜⎜⎝

⎛××

=

°==⎟⎟⎠

⎞⎜⎜⎝

⎛××

=

°==⎟⎟⎠

⎞⎜⎜⎝

⎛××

=

−

−

−

08.120.209sin1033.3106.97sinargonfor

3.110.195sin1033.3106.50sinneonfor

4.110.197sin1033.3106.56sinhydrogenfor

1-6

7-1-

1-6

7-1-

1-6

-71-

θ

θ

θ

5. Explain why no interference bands are observed when light from two adjacent lamps fall on a screen. Two sources are not coherent. The photons are produced randomly and there is not a constant phase

relationship. 6. A beam of light is polarised by passing it through a piece of ‘polaroid’. In what way does the polarised beam

differ from the original non-polarised beam? The polarised beam has one of its planes of vibrations removed or “filtered” out where as the unpolarised

beam has all planes of vibration present. 7. Describe how you would use a disc of polarising material to show that direct sunlight is not polarised whereas

diffuse light from the blue sky is partially polarised. When a polarising filter is held up to the blue sky and rotated there is a decrease in intensity due to the light from the blue sky being partially polarised.

of 17

8. Calculate the first and second order angles for light of wavelength 400 nm and 700 nm if the grating contains 10,000 line/cm. The grating contain 104lines/cm=106lines/m, which means the separation between slits is d=(1/106)m=1.0×10-6m. In first order (m=1), the angles are:

appear. willordershigher No 1. exceedcannot sin because 700nm

forexist not doesorder second but the 0.53 so

1sin0.800m101.0

104.02msin

order secondIn .0.44 and 6.23 so

0.700sin0.400m101.0

104.01msin

400

7006-

7-

400

700400

7006-

-7

400

θλθ

θλθ

θθ

θλθ

=°=

=×

××==

°=°=

==×

××==

fd

d

9. Spectra Overlap White light containing wavelengths from 400 nm to 750 nm strikes a grating containing 4000 lines/cm. Show that the blue at λ=450 nm of the third order spectrum overlaps the red at 700 nm of the second order. The grating spacing is d = (1/4000) cm = 2.50 ×10-6 m. The blue of the third order occurs at an angle θ given by:

0.540m102.50

104.503msin 6-

-7

=×

××==

dλθ

Red in second order occurs at:

0.560m102.50

10.0072msin 6-

-7

=×

××==

dλθ

which is a greater angle; so the second order overlaps into the beginning of the third-order spectrum. Topic Three: Photons KEY IDEAS photoelectric effect a phenomena which can only be explained if light has particle-like characteristics. The

light particles are called photons. threshold frequency(f0) is the minimum frequency at which electrons are ejected from a surface. work function (W) is the minimum energy required to break an electron free from the attractive forces

which hold the electron to the surface of a metal. de Broglie wavelength is the wavelength associated with a particle of mass m travelling at speed v. The

wavelength is given by λ=h/(mv) where λ is the de Broglie wavelength of the particle. Photoelectric Effect

A metal has outer electrons that are weakly bonded to the atoms and so are free to move throughout the metal.

These electrons may be removed when they absorb a small amount of energy either: when the metal is heated → thermoelectric effect

or when the metal is irradiated with suitable electromagnetic radiation → photoelectric effect.

Experimental Evidence (Observations of the photo-electric effect) 1. Photo-electrons are released only if the frequency of EM radiation is above a

certain minimum frequency. This frequency is called the threshold frequency (ƒ0). 2. If ƒ>ƒ0 the number of photo electrons released per unit time [photoelectric current] is

proportional to the intensity of the radiation. 3. Photoelectron emission is instantaneous and is not dependent on the intensity of

radiation. 4. The maximum kinetic energy of the photoelectrons increases linearly with the frequency

of the incident light and is not dependent on intensity. 5. No electrons are emitted if the frequency is below the threshold frequency no matter

how intense the light. 6. For frequencies above the threshold frequency electrons are emitted with a range of

velocities up to a maximum.

METAL

UV Radiation

Photo Electrons

zinc

UV Radiation

electrons

EK

ff0

f>f0

velocities vary

of 17

CATHODE ANODE

EMRadiation

Wave Theory (Classical Theory) Maxwell’s theory of electromagnetic radiation postulated that light was a continuous wave of fluctuating E and B fields. This theory makes predictions about the photoelectric effect. Electromagnetic (EM) Theory predicts: 1. The energy absorbed by an electron would be dependent on the intensity of radiation i.e. the maximum energy

Kmax of the emitted electrons would increase as the intensity of light increased. (Observed Kmax ∝ (ƒ-ƒ0)) 2. If low intensity radiation were used there would be a time delay before electrons emitted. (Observed instantaneous emission) 3. Electrons will absorb low frequencies more easily than high frequencies. (Observed high frequency releases more photoelectrons) 4. More electrons are emitted if we increase the intensity of the light. (Observed no emission if ƒ>ƒ0 - with no explanation for the threshold frequencyƒ0). Photon Concept of Light A model of light where we view light not as a continuous wave but in discrete bundles or packets. A quantum or a photon is the name given to this discrete packet. A photon is a discrete bundle of energy with E∝f. where h = Planck’s Constant = 6.63 × 10-34 Js These quanta of light fall on the photo-electric surface and one “free” electron completely absorbs one photon. If given enough energy this electron will be emitted. The smallest energy to remove an electron is called the work function (W).

0hfW = where f0 = threshold frequency The photoelectrons (photoemission) occurs when a photon (hf) is absorbed by a high energy electron. If f<f0 then no emission.

0

2(max)2

1(max)

hfhf

-WhfmvK

−=

== (for the fastest moving electrons this is just a statement of energy conservation).

0

-ve

W

0

-ve

W

ΔE=hf0

The Photoelectric Cell

The photoelectric cell has a curved metal cathode (cadmium or many other metals) and a thin pole anode sealed in an evacuated glass tube. When EM radiation strikes the cathode electrons are released and may move to the anode, so allowing an electric current to flow through the photo-cell. Photo cells are often used as switches where the photo cell is only “on” while light strikes the cathode, it is “off” at all other times, lift doors, self opening doors.

The photocell may also be used to measure the energy of photoelectrons. In this application the potential difference between the cathode and anode is reversed so that the photoelectrons must do work (lose energy) in order to move to the anode. As the reverse potential is increased, fewer electrons reach the anode and so the photoelectric current decreases (indicating the photo electrons are released with a range of energy). The smallest reverse potential which stops all photo-electrons from reaching the anode, (null current I = 0 amps) is called the stopping voltage VS. It can be seen that Energy of photoelectrons ≤ eVS (ie maximum energy of photoelectrons Kmax = eVS. WhfeVS −= Experiments show that the Kmax is dependent on the frequency of the nature of the metal

surface.

hfE =

0

-ve

W

ΔE=hf

K max eVs

f f0

metal1 metal2 metal3

grad h

of 17

EXAMPLE : Photon Energy

1. Calculate the energy of a photon of blue light, λ=450nm. Since λc=f , we have

( )( ) JhhfE 197

834

104.4105.4

100.31063.6c −−

−

×=×

××===

λ or eV7.2

106.1104.4

19

19

=××

−

−

Photons from a Light Bulb 2. Estimate how many visible light photons a 100W light bulb emits per second. Let's assume an average wavelength in the middle of the visible spectrum, λ≈500nm. The energy emitted in one second (=100 J) is E = n hf where n the number of photons emitted per second and f = c/λ.

Hence ( )( )

( )( )20

834

9

105.2100.31063.6

10500100×=

×××

=== −

−

hcE

hfEn λ

This is an overestimate since much of the 100J of electric energy input is transformed into heat rather than light. If the efficiency is between 1% and 10%, then the number of photons emitted is on the order of 1019, still an enormous number. Photoelectron speed and energy 3. What is the maximum kinetic energy and speed of an electron ejected from a sodium surface whose work function W0 = 2.28 eV when illuminated by light of wavelength (a) 410 nm; (b) 55 nm. (a) For λ = 410 nm.

( )( )

eVorJ

hhfE

03.31085.4101.4

100.31063.6c

19

7

834

−

−

−

×=×

××===

λ JoreVeveV-WhfK(max)19102.175.028.203.3 −×=−==

Since 15max

31

101.52

101.9

−

−

×==

×==

msmKv

kgmwheremvK 2(max)2

1(max)

(b) For λ = 550 nm. hf = 3.6 × 10-19 J = 2.25eV

Since this photon energy is less than the work function, no electrons are ejected. The photoelectric effect, besides playing an important historical role in confirming the photon theory of light, also has many practical applications. eg. burglar alarms, automatic doors, smoke detectors, photographic light meters and film sound track. X-rays X-rays: incident high energy electrons produce high frequency electromagnetic radiation.

Filament F produces a current to heat the wire and emit electrons. Hood H used to focus electrons Anode A electrons accelerated away from filament with a large acceleration due to high PD 40kV to 400kV

between the Anode and the Cathode. Target T tungsten target on which the electrons impinge. The process of X-ray production is very inefficient, about 99% of the energy being lost as heat.

electrons accelerated by large p.d.

X rays

metal

of 17

+ electron path

X ray photon

proton

½mv22

½mv12

X-rays and Photons If the accelerating potential is V the electrons from the Cathode reach the target T with an energy Ve

(Work = potential difference × charge). ie. VemvK == 221

A few electrons penetrate the atom and interact with the electric field with the electron being slowed down.

ie. 2

121

1 mvK = incident 2

221

2 mvK = final

The energy available becomes 2

2212

121

21 mvmvKKK −=−=

Note: energy lost by one electron hfmvmv =− 222

1212

1 where hf is the energy of one X-Ray photon.

The process of X-ray production is such that most of the energy of the electrons heats the target and must be removed from the anode by a cooling liquid usually a silicon oil.

The diagram shows a bremsstrahlung photon produced by an electron decelerated by interaction with a target atom. Because there will be a range of final velocities due to the degree of interaction (slowing down) we have a range of X-ray photons produced. This gives us a continuous range of X-ray photons. However this range has a maximum value which is determined when all of the available electron energy goes into the photon when the electron has a final velocity v2 of zero.

0222

12 == mvK so hfmvmvKK =−=− 2

2212

121

21 becomes max2

121max hfmvK ==

where max

min fc

=λ .

Considering the Intensity Curves X-Ray Quality If the tube voltage is increased we increase the accelerating potential and so the incident electrons have more energy. The maximum energy is higher and also the number of photons increases so a higher intensity of X-rays is produced. X-Ray photons with a higher frequency and a lower wavelength have a greater penetrating power. A beam of higher frequency X-ray photons is said to be “harder” or of higher quality than those where f is less. Lower frequency X-ray photons are said to be "softer".

X-Ray Quantity If the filament current is increased then more electrons are released and there are more electrons available for photon production and so a higher intensity of X-Ray photons is produced.

Note: there will be no increase in energy of individual photons so no increase of fmax.

of 17

Problem set example: Photons 1. Calculate the energy of a photon of (a) red light (f=5×1014Hz), (b) violet light (f=7×1014Hz).

Joules103.3105106.63hfE

19-

14-34

×=

×××==

Joules106.4107106.63hfE

19-

14-34

×=

×××==

2. Find the EK of the most energetic electron ejected from a metal by a radiation of frequency 7×1014Hz if the work function of the metal is 2V.

joules1044.1102.31067.4102.3107106.63W-EEJ102.3101.622VW

19-1919-191434-iK(max)

-19-19

×=×−×=×−×××==

×=××==−−

3. What is the threshold frequency of a metal that has a work function of 3.3volts?

Hz107.97106.63105.28

hWfhfWJ1028.5101.63.3V3.3W 14

34-

-19

0019-19- ×=

××

===×=××== so

4. Will light of frequency 4×1014Hz eject electrons from a metal whose work function is 5V? Explain your answer.

emission no so than W less is EJ102.65104106.63hfEJ108101.65V5W

i19-1434-

i

-19-19

×=×××==

×=××==

5. Light of a frequency above the threshold frequency falls on a given metal. As the intensity of the light is increased: (a) what happens to the number of photoelectrons ejected from the metal? More electrons are emitted since more photons are falling on the metal. (b) what effect does it have on the maximum kinetic energy of the electrons? None. Only a more copious emission of electrons. 5. The energy needed to release an electron from sodium metal is 2.3eV. (a) What is the energy in joules? Joules103.682.3eVW 19−×== (b) Calculate the threshold frequency of sodium and hence determine whether or not orange light of wavelength 6.0×10-7m will be able to release electrons from the sodium surface.

Hz105106103cf som106Hz105.55

hWf 14

7

87-14

0 ×=××

==×=×== −λλthen

6. The stopping voltage of a photoelectric cell is 1.0volts when light of wavelength 5.5×10-7m falls on the cathode surface.

Joules1061.3105.5

10310625.6hcEJoules101.61.0VoltsV 197

83419-

S−

−

−

×=×

×××==×== so

λ

(a) What is the maximum kinetic energy of the emitted electrons? Joules106.1E 19

max−×=

(b) What is the work function of the cathode? Joules1001.2W 19−×= (c) What is the threshold frequency of the metal of the cathode? The work function of a metal is 6×10-19J.

5. In a photoelectric experiment using a clean sodium surface, the maximum energy of the emitted photons was measured for a number of different incident frequencies, with the following results.

Frequency (1014Hz) Energy (eV) 11.8 2.60 10.6 2.11 9.9 1.81 9.1 1.47 8.2 1.10 6.9 0.57

Plot the graph of these results and hence find:

Hz1003.3hWf 14

0 ×==

of 17

A

C A

(a) Planck’s constant and then the threshold frequency of sodium.

6. A photosensitive surface C is illuminated with monochromatic light of successively different wavelengths. The diagram below illustrates that potential differences have to be applied between C and A to prevent electrons released by the light from crossing to the collector A and registering as a current on the ammeter G. The values of potential

difference which just stop electrons reaching A are as follows:

Plot a suitable graph and hence determine: (a) the threshold frequency, (b) the work function of the surface A, (c) the value this experiment finds for Planck’s constant, (d) the maximum kinetic energy (in joules) with which electrons are ejected from the surface by the light of wavelength 3.66×10-7m. Wavelength(metres) Frequency(hertz) Energy(joules) Stopping Voltage (volts) 3.66E-07 8.20E+14 2.368E-19 1.48 4.05E-07 7.41E+14 1.84E-19 1.15 4.36E-07 6.88E+14 1.488E-19 0.93 4.92E-07 6.10E+14 9.92E-20 0.62 5.46E-07 5.49E+14 5.76E-20 0.36 5.79E-07 5.18E+14 3.84E-20 0.24

Wavelength (10-7m) Potential Difference (V) 3.66 1.48 4.05 1.15 4.36 0.93 4.92 0.62 5.46 0.36 5.79 0.24

P h o t o - e l e c t r i c E f f e c ty = 7 E - 3 4 x - 4 E - 1 9

- 4 E - 1 9- 3 E - 1 9- 2 E - 1 9- 1 E - 1 9

01 E - 1 92 E - 1 93 E - 1 94 E - 1 95 E - 1 9

0 4 E + 1 4 8 E + 1 4 1 E + 1 5 2 E + 1 5

F r e q u e n c y ( h e r t z )

Ener

gy(m

ax) j

oule

s

S to p p in g V o lta g e y = 7 E -3 4 x - 3 E -1 9

-3 E -1 9

-2 E -1 9

-1 E -1 9

0

1 E -1 9

2 E -1 9

3 E -1 9

0 2 E + 1 4 4 E + 1 4 6 E + 1 4 8 E + 1 4

Fr e q u e n c y(h e r tz )

Ener

gym

ax(jo

ules

)

of 17

7. A potential difference of 10000V is applied across an x-ray tube. Calculate the minimum wavelength for the x-rays produced.

Joules101.6VeavailableEnergy -15×== If all this kinetic energy is converted into a photon what is its wavelength.

metres1024.1hchf101.6E 10min

15- −×===×= λλ

so

8. What potential difference must be applied across an x-ray tube to produce x-rays with a minimum wavelength of 10-11m?

volts101.24V10

103106.6hc101.6V 511

83419- ×=

×××==×× −

−

soλ

Electrons are accelerated in an x-ray tube by a potential difference of 50000V. (a) What energy do they have on reaching the target?

joules108VeEnergy -15×== (b) What is the shortest wavelength of the continuous x-ray spectrum produced?

metres1048.2108

103106.6hcE 1115

834−

−

−

×=×

×××== λ

λso

(c) If the electrons are stopped in a distance of 10-5m, calculate their average acceleration while being stopped.

( ) 2205-2821

22

12

1-831-

-15K

12

21

ms108.79a10a2101.33-02asv-v0msv

ms101.33109.11082

m2EvmvEnergy

−− ×=××=×==

×=×××

==∴=

thenso

Topic Four: Wave Behaviour of Particles. DeBroglie Matter Waves When light is being transmitted it exhibits wave-like properties. When light is interacting with matter it has particle-like characteristics.

hfE = λh

chf

==p or mvh

ph

==λ

Waves associated with moving matter are known as de Broglie waves or matter waves.

e g (1) mass = 50 kg, velocity = 10 ms-1 mph 36

34

1033.1500

106.6 −×=×

==λ

e g (2) m = 9.1×10-31 kg (mass of an electron),

K = 90 eV, v = 5.6×106 ms-1 mph 10

631

34

103.1106.5101.9

106.6 −×=×××

×==λ

This in is the X-Ray region of the electromagnetic spectrum.

Electron Diffraction If consider the intensity pattern obtained from wave interference eg. X-Rays diffracted by crystal lattice then electron, matter waves could have wavelength in the X-Ray ranges and a crystal could be used as a diffraction grating to produce a diffraction pattern. Davisson and Germer produced electron diffraction in 1927 and the relationship

λθ nd =sin was satisfied using ph

=λ

of 17

APPLICATION: THE ELECTRON MICROSCOPE. Sight is one of the most important human skills, which enables us too observe the structure of materials – natural and artificial because they reflect light. With the eye alone it is impossible to distinguish between two objects less than about 0.1 mm apart. First magnifying glasses and then microscopes have been developed to enable us to distinguish between (ie RESOLVE) objects that are even closer together. Light has, however, a wave nature and diffracts at sharp edges of objects or small gaps between them. The shortest visible wavelength (violet) is of wavelength of 400 nm or 4 × 10-7 m and with the best microscopes the best possible resolving power is only the approximate value of this wavelength. Shorter wavelength radiations eg X-rays (0.10 nm or less) would help – there are problems, however, because no possible ordinary lens has been yet devised and we could not see them with our eyes. De Broglie, postulated, in 1924, that moving electrons had an associated wavelength λ = h / p (p= momentum). This gave physicists the idea that perhaps electron beams could be focused using either or both electric (E) and magnetic (B) fields and thus perhaps an electron microscope could be constructed. The effective wavelength when the electron is accelerated through of potential of 40 kV is: W = Ve = 4 × 104 × 1.6 ×10-19 = 6.4 × 10-15 J. Energy = 1/2 m v2 = 6.4 × 10-15 J v = 1.19 × 108 ms-1

λ = h / p = 6.63 × 10-34 / ( 3.1 × 10-31 × 1.19 × 108) = 6.12 × 10-12 m. with an equivalent increase in resolving power.

The basic structure of the electron microscope is similar to that of a compound light microscope. The electron microscope can produce images that have a much greater magnification than a light microscope and also enables a far greater resolution to be obtained than that possible with a light microscope. This is because the diffraction pattern for high-energy electrons has a much lower width than that for any visible light waves. Hence instead of being restricted to magnifications of less than 2 000, magnifications of many hundreds of thousands are feasible enabling such things as actual structures of crystal lattices to be observed. In the Transmission Electron Microscope produced by a two dimensional image the objective and eyepiece lenses are magnet fields that exert forces on the electron (motion of charged particles in electric and magnetic fields) to bring them into effective focus. Carefully designed current-carrying coils of wire produce these fields. Problem set example: 1.What is the momentum of a photon of x-radiation of wavelength 10-11m?

secN106.610

106.6hp 23-11

-34

×=×

== −λ 2. Calculate the momentum and energy of a photon of yellow light of wavelength 6.00×10-7m?

secN101.11000.6

106.6hp 27-7

-34

×=×

×== −λ

J103.31000.6

103106.6hcE 19-7

8-34

×=×

×××== −λ

3. Calculate the deBroglie wavelength of: (a) a 1000kg car moving at 10ms-1. metres106.6

101000106.6

ph 38-

-34

×=×

×==λ

(b) an electron accelerated through a potential difference of 100V.

metres1023.1109.5101.9

106.6phms109.5

101.9106.12

m2E

vJ106.1106.1100E 10-631

-3416

31

17K1719 ×=

××××

==×=×

××==×=××= −

−−

−−− λso

4. At what speed does an electron have a deBroglie wavelength of 2.0×10-8m?

1-4831

-34

ms106.3100.2101.9

106.6mhv

mvh

ph

×=×××

×==== −−λ

λ so

5. An electron has an energy of 60000eV. Find:

(a) its velocity 1-831

-15K15- ms101.45

101.9109.62

m2EvJ109.660000eVE ×=

×××

==×== −

(b) the electron’s momentum secN1032.1ms101.45109.1mvp 22-18-31 −×=×××== (c) its wavelength.

metres1099.41032.1106.6

ph 12

22

-34−

− ×=××

==λ

Section Three: Light and Matter Topic One: … 1 of 17 Section Three: Light and Matter Topic One:...

Documents

Transcript of Section Three: Light and Matter Topic One: … 1 of 17 Section Three: Light and Matter Topic One:...