Diffraction, EM and particle Waves

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1 Diffraction, EM and particle Waves RESOURCE MATERIAL Web: Text: chapter, 3 pp sections 1-5 andbook: study guide 4 For quiz 4, you should know: 1. Resolution of the eye 2. Electromagnetic (EM) waves: c = λf E = hf 3. Particle waves: E = hf, E = ½mv 2 p = h/λ, p = mv 4. Particle wave functions and probability Re: SG4 in Lab quiz #10, you should know: (There is also Absorption of light) 1. Resolution of the eye 2. y/l = λ/2d (double slit), y/l = λ/a (single slit) 3. y/l = 1.22λ/a (circular aperature) 1

2 Electromagnetic Waves 1. Coupled oscillating electric (E) and magnetic (M) fields Electric field y x Magnetic field z 2. Can propagate in vacuum (unlike sound) 3. Transverse waves: E and M are perpendicular to each other and direction of propagation. c = lf 4. c = m/s (vacuum) SPECTRUM Know ordering Gamma rays X-rays ultra -violet vis infra -red Radio

3 Visible Light Spectrum λ (nm) Diffraction of Waves Bending of waves around the edges of obstacles. Significant when λ size of obstacle [ physical optics ] For λ << size of obstacle ray optics [as in SG3] source emits uniformly in all directions 3

4 Double Slit Interference light d detector screen L Observe alternating bright and dark bands (fringes). Due to constructive and destructive interferenc of waves coming from both slits SEE LAB 4,pg. 1-2 Cylindrical waves from both slits are in phase, i.e. crests leave slits simultaneously. Constructive interference line. Crest meets crest, trough meets trough. Bright fringe incident parallel wave front Destructive interference line. Crest meets trough. Dark fringe 4

5 P d l 1 l 2 L y bright fringe path lengths from both slits are equal screen Point P on screen will be a bright fringe if the difference in path lengths, l 2 l 1, is an integer multiple of the wavelength, i.e. if l 2 l 1 = nλ, where n = 0,1,2,3, P will be a dark fringe if l 2 l 1 = m(λ/2), where m = 1,3,5 (odd multiple of λ/2) Let y = distance from central bright fringe (intensity maximum) to nearest dark fringe (intensity minimum). Can show [lab 4, pg.2] using small angle approx. (adding amplitudes of sources). y = Ll/(2d) need on LQ#10 5

6 Double slit intensity pattern Intensity y distance along detector screen light Single Slit Diffraction y Observe one bright central fringe and progressively less bright side fringes a L a/2 l 1 l 2 L D y Imagine each point in space across slit is a source of cylindrical waves 6

7 To calculate total interference pattern add amplitudes of all sources at point D by grouping point sources into pairs separated by a/2. As before destructive interference results at point D if l 2 l 1 = m(λ/2). Therefore if we let y = distance from central intensity maximum to adjacent minimum Then we find (replace d in previous formula with a/2): y = ll/a Number of point sources in single slit is arbitrary. Different overlapping interference patterns from many sources washes out equal fringes found for 2 slit diffraction and causes broad central maximum. Single slit intensity pattern Intensity 2y distance along detector screen 7

8 Diffraction by a circular aperture Observe one bright central spot and progressively less bright circular fringes light a L Result is similar to single slit diffraction a α L y Detector screen Distance on detector from central max. to first min. y/l» a = 1.22 l/a with n = 1 8

9 Resolution source 1 α Consider the diffraction pattern of two separate light sources source 2 If the angular separation a of the sources is large enough, their diffraction patterns do not overlap and the image of each source can be distinguished. But for small a, the diffraction patterns will overlap. resolved two point sources unresolved 9

10 Rayleigh Criterion The images of two equally bright point sources of light can just be resolved if the central maximum of diffraction pattern coincides with the first minimum of the other. S 1 y S 2 α a L Therefore the minimum angular separation at which the sources S 1 and S 2 can be resolved is y/l = a = λ/a y/l = a = 1.22 λ/a for a slit for circular aperture E.g., for eye (circular aperture), a = pupil diameter NB: α is related to separation, y and distance L by a = y/l 10

11 Sample Problem I Two headlights of a car make an angle of 10-3 radians at the eye of an observer. If the person can just distinguish two spots 1 cm apart at a distance of 20 m, will she be able to distinguish the headlights as two distinct objects? Minimum angle of resolution is: a = 1 cm/20 cm = radians Since the angle subtended by the headlights (10-3 rad) is larger than a, the person can distinguish them. Sample Problem II a) A person in a darkened room can just resolve two points of light 1 mm apart at a distance of 5 m. The room lights are now turned on. What happens to the person s s pupils and in what direction must he move to again be able to just resolve the two points of light? 11

12 Darkened room Calculate angle of resolution: α D y/l = 10-3 /5 y = 1 mm α D α D = rad. L = 5 m Lights turned on The person s pupil diameter should decrease. New aperture a L is less than previous value a D. ence, the new angle of resolution is: a L = 1.22 l/a L Since a L > a D, a L > a D Then angle of resolution has increased. But the angle subtended by the point sources is still a D which is less than the angle of resolution with the lights on. Therefore with the lights on, the person must move closer to resolve them. 12

13 b) If the pupil diameter decreases from a D = 4 mm to a L = 2 mm, to what distance must the person move to resolve the points? a D = 1.22 λ/a D a L = 1.22 λ/a L a L /a D = a D /a L = 4/2 = 2 a L = 2 a D = 2( ) = radians Also, a L = y/l L = y/a L = 10-3 /( ) = 2.5 m L =2.5 m y = 1 mm α L α D L = 5 m The text and formula sheet refer also to the angle of resolution inside the eye. α 2 α 1 d n 1 n 2 13

14 By Snell s law n 1 sinα 1 = n 2 sinα 2 n 1 α 1 = n 2 α 2 For small angles α 2 = (n 1 /n 2 )α 1 = α 1 /n 2 For n 1 = 1 (air) α 2 = (1.22λ/a)/n 2 ë á 2 = 1.22 n a 2 = 1.22(λ m /a) λ m = wavelength inside eye Most problems encountered deal with resolution outside eye: use n = 1. 14

15 Particle-like Nature of Light Waves light Exhibited mainly in the interaction of electromagnetic radiation (light) with matter. E.g.. molecule Photoelctric Effect molecule in excited state (structural rearrangement, dissociation, ionization, etc.) Under illumination, the metal plate emits electron with max. Kinetic Energy, Metal plate Collector plate KE max = hf - Φ where Φ = binding energy of electron for that particular metal. Creates Observations show: photocurrent 1) Light exchanges energy with matter in discrete amounts. 2) Energy is proportional to frequency, f, of light 15

16 Einstein (1905): 1. Light consists of particles = photons 2. Energy of one photon: E = hf = hc/λ 3. h = Planck s Constant = J. s 4. A beam of light contains many photons, N, total energy E = Nhf = Nhc/λ Example: A pulse of X-rays of wavelength 0.20 nm has energy J. ow many photons are in the pulse? E = N hc/λ N = Eλ/(hc) N= ( ) ( )/[( ) ( )] N = photons Power and Intensity of Radiation P = E/t = Nhf/t I = P/A 16

17 Wave-like Nature of Particles Matter around is composed of particles: atoms and molecules: air water water: 2 O O chemically bonded atoms Nitrogen: N 2 N N Also can have single atoms: e: e m Atomic Structure: electrons nucleus In fact the nucleus is 10 5 times smaller than the atom. The e- are actually diffuse 17

18 Nuclear Structure composed of: NUCLEONS protons: +ve charge neutron: no charge m eld together by strong nuclear force. Protons and neutrons have similar mass and are much more massive than electrons: m p /m e = # electrons = # protons in neutral atom but number of neutrons may vary: ISOTOPES (SG8). Structure of the Nucleons: quarks 6 types Up Down Charm Strange Top Bottom proton: u d u Neutron u d d 18

19 Mechanic of Atomic and Sub Atomic Particles Classical Mechanics (Newton s laws) vs. Quantum Mechanics Which applies to the Rutherford model of the hydrogen atom? Attractive Coulomb force between negative electron and positive proton: F = kq Analogous to classical planetary solar system and allowed several important predictions but is flawed. The orbiting electron must accelerate to travel in a circle. From electromagnetic theory an accelerating charge (such as the electron) should radiate energy. This loss of energy should occur very quickly (10-6 s) cause the collapse of the atom. If correct, atoms would NOT be COMMON. r 1 q

20 Introduction to Quantum Mechanics 1. eisenberg Uncertainty Principle The position, x, and velocity, v of a particle cannot be simultaneously measured with high precision. The expression relating the uncertainty in position x and in velocity v is x v = h/m h = Planck s const. and m is the particle s mass 2. Probability a) The probability that a particle is at a given position x can be defined exactly without influencing the velocity v and vice versa. b) Probability behaves like a wave with de Broglie wavelength λ: l = h/p = h/mv where p = mv = momentum 20

21 Double Slit Interference with Particles Expected image with classical particles (e.g. bullets) detector screen electron gun But with electrons when d = h/mv, we see alternating bands of high and low intensity just like double slit interference for waves d This image is produced even if the gun s intensity (particles/area/time) is reduced so that only one electron hits the slits at a time. So it is not the beam that has a wavelike nature, but rather each electron. An electron may land anywhere, but is more likely to end up on a high intensity band. 21

22 NB: When the particles hit the detector, their position is well defined as one would expect for a particle but only the direction of its momentum is known. At the slits, the probability of being in a given position behaves in a wave like manner with wavelength = de Broglie wavelength. Therefore wavelike diffraction applies to particles with mass (electrons) and without mass (photons). massive particles photons p = mv E = ½mv 2 = p 2 /2m E = hc/λ λ = h/p Therefore for photons λ = hc/e = h/(e/c) p = E/c Example: Compare the kinetic energies of an electron and a photon both having wavelength λ = 0.1 nm. E photon = hc/λ = ( )( )/( ) = J E electron = p 2 /(2m) = (h/λ) 2 /(2m) = ½( / ) 2 ( ) -1 = J 22

23 p electron Probability Waves Simplest case for calculating electronic wave function. Applicable to absorption of visible and u.v. light by many biological molecules. Consider Butadiene: C 4 6 C C C C Simple organic molecule (C,, O, ) Atoms held together by chemical bonds electron sharing by nuclei of different atoms. Single bonds: : σ bond electrons are strongly localized between the nuclei of the bond.. Double bonds: == : σ bond + π bond π electrons are NOT strongly localized by bonding nuclei they are delocalized along conjugated sequence. 23

24 The molecules we will describe contain a series of alternating double and single bonds. Such a series is called a conjugated series. E.g.: decene C C C C C C C C C C or more accurately C C C C C C C C C C Another e.g. Vitamin A (C O): NB: Me = C 3 Notice that in reality bonds are bent 24

25 Retinal: Conjugated bond sequence: Only atoms joined by alternating single and double bonds are part of conjugated system. Sequence BEGINS and ENDS with a double bond Nitrogen,Oxygen and Carbon atoms contribute Each N, O or C atom contributes 1 electron to π- bonded network. π electrons in the cloud can be anywhere within the cloud, we can only calculate the probability they are at a given position. We assume electrons are de-correlated (the prob. of finding electron at given position doesn t influence the other electrons), and that they move very rapidly compared to the nuclei. 25

26 The Simple Model We Use Butadiene: C 4 6 actually looks like: C C C C We will; Approximate conjugated chain of m carbon atoms by a straight line of length, l = (m-1)l CC. l CC = length of carbon-carbon π bond = 0.15 nm. Calculate only the probability at position, x, along this line for the other coordinates we will assume that they lie within a box. Ignore other atoms and sigma bonds. For butadiene looks like: C C C C x = 0 x = l = (4-1)0.15 = 0.45 nm 26

27 The probability for finding each electron is described by a WAVE FUNCTION = Y n (x). We ll use a very simple form for the wave function: Y n (x) = 2/l sin(npx/l) (0 x l) Like a standing wave with no time dependence: describes stationary state of electron. π electrons not found outside box so Ψ n (x) = 0 for x = 0 and x = l Wave functions must be continuous (smooth) for x = 0 and x = l: lim Ψn ( x) = 0 and lim Ψn ( x ) = 0. x 0 x l lim Ψn ( x) = 0 means Ψ n (x) sin(kx) [not cos(kx)] x 0 lim Ψn ( x) = 0 gives condition: sin(kl) = 0. That is, x l kl = nπ n = 1, 2, 3,, n is the Quantum Number So far, we have, Ψ n (x) = Csin(nπx/l), C = Constant 27

28 What is the Amplitude, C? To normalise the wave function (and find C) we need to relate it to the probability. Divide box into thin slices of thickness x. Probability that an electron is inside varies for each slice. Total probability, (sum over all slices) must = 1. Clearly the probability of finding the electron somewhere within two slices is greater than that of finding it in either single slice. Therefore, the probability of finding electron within an interval of thickness x, P x, is roughly proportional to x. P x P x x Px = probability density This approximation is exact for infinitesimal x. P dx = P x dx 28

29 Therefore, the probability that an electron is within a thin slice, x is given by integrating: x+dx P x = x P x dx = Area under P x curve between x and x+ x ow is the Probability Density Px, related to the Wave Function, Y n (x) P x = [Ψ n (x)] 2 Why is it squared?? Probability must be positive Analogous to Intensity of EM radiation where I E 2 + B 2 or sound I = 2π 2 ρf 2 y 0 2 v Since total probability over box = 1 then: l 1 = Ψ ( x) dx = 0 2 n l 0 Csin n ðx/l 2 dx This gives: C = 2/l. Therefore Ψ n (x) has dimensions of (length) -½. 29

30 Properties of Wave Functions Ψ n (x) = 2/l sin(nπx/l) n = 1,2,3, Wave function for n th electronic energy level (More on Energy Levels in SG 8). Ψ 1 x Ψ 1 2 x 0 L 0 L Ψ 2 L/2 x Ψ L L/2 0 L Ψ 3 x Ψ L L/3 2L/3 0 L/3 2L/3 L NB: n = number of ½ wavelengths l e = 2L/n in box 30

31 Sample Problem II Decene C C C C C C C C C C C-C bond length, l CC = 0.15 nm, therefore total length, L = 9 l CC = 1.35 nm a) What is the maximum probability density for the n = 3 electron? P x = [Ψ n (x)] 2 = (2/L)sin 2 (nπx/l) The maximum value of sin 2 ( ) = 1 Therefore max. P x = 2/L = 2/1.35 = 1.48 nm -1 Note: INDEPENDENT of n!! b) At what positions in the molecule does this maximum occur? Solve for sin(3πx/l) = 1 3πx/L = π/2 x = L/6 for 1 st value, there are 3 values for n = 3. Add L/3 and 2L/3 for others 31

32 c) What are the positions of minimum probability density? (Electron spends least time here) Minimum value of P x = 0. Solve for sin(3πx/l) = 0 3πx/L = 0 x = 0 for 1 st value (ALWAYS), there are 3 values for n = 3. Add L/3 and 2L/3 for others. d) What is the probability density for the n = 3 electron at the 5 th carbon atom? C C C C C C C C C C 1 st Carbon, x = 0 5 th Carbon x = 4 l CC = = 0.60 nm 4 bonds from end Therefore: P x = [Ψ 3 (x)] 2 = (2/L) sin 2 (3πx/L) = (2/1.35)sin 2 (3π(0.60)/1.35) = 1.10 nm -1 32

33 e) What is the probability of finding the n = 3 electron in a region of length 0.05 nm centred at the 5 th Carbon atom? P x = P x x = (1.10 nm -1 )(0.05 nm) = (dimensionless) 0.05 nm Ψ 3 2 x C 1 C 2 C 3 C 4 C 5 C 6 C 7 C 8 C 9 C 10 f) What is the probability that the n = 3 electron lies anywhere between the 1 st and 4 th Carbon atom? See Diagram above: Prob. elec. between C 1 and C 4 = Prob.: 0 < x < L/3 = Area under [Ψ 3 (x)] 2 for 0 < x < L/3 = 1/3 (total area under P x curve) = 1/3 33

WAVE NATURE OF LIGHT

WAVE NATURE OF LIGHT Light is electromagnetic radiation, a type of energy composed of oscillating electric and magnetic fields. The fields oscillate perpendicular to each other. In vacuum, these waves