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Transcript
Physics 3313 – Review 1 Monday March 8 , 2010 Dr. Andrew Brandt 3/8/2010 3313 Andrew Brandt 1 Time Dilation Example • Muons are essentially heavy electrons (~200 times heavier) • Muons are typically generated in collisions of cosmic rays in upper atmosphere and, unlike electrons, decay ( t 2.2 sec) 0 • For a muon incident on Earth with v=0.998c, an observer on Earth would see what lifetime of the muon? • 2.2 sec? 1 1 2 16 v c2 • t=35 sec • Moving clocks run slow so when an outside observer measures, they see a longer time than the muon itself sees. (It’s who is the observer that matters not who is actually moving) 3/8/2010 3313 Andrew Brandt 2 More about Muons • They are typically produced in atmosphere about 6 km above surface of Earth and frequently have velocities that are a substantial fraction of speed of light, v=.998 c for example 8 m 6 vt0 2.994 x10 • • • • • sec x 2.2 x10 sec 0.66km How do they reach the Earth? Standing on the Earth, we see the muon as moving so it has a longer life (moving clocks run slow) and we see it living t=35 sec not 2.2 sec , so it can travel 16 times further than 0.66 km that it “thinks” it can travel, or about 10 km, so it can easily cover the 6km necessary to reach the ground. But riding on a muon, the trip takes only 2.2 sec, so how do they reach the ground??? Muon-rider sees the ground moving, so the length contracts and is only L0 6 /16 0.38km so muon can go .66km, but reaches the ground in only .38km 3/8/2010 3313 Andrew Brandt 3 Velocity Addition Example • Lance is riding his bike at 0.8c relative to observer. He throws a ball at 0.7c in the direction of his motion. What speed does the observer see? vx' v vx vv ' 1 2x c 3/8/2010 vx .7c .8c 0.962c .7c .8c 1 c2 3313 Andrew Brandt 4 Relativistic Momentum Example • A meteor with mass of 1 kg travels 0.4c • Find its momentum, what if it were going twice as fast? compare with classical case a) v 0.4 c 1.09 v b) 0.8 c 1.67 c) p mv 1.2 108 d ) 2.4 108 3/8/2010 p mv 1.09 1kg 0.4 3 108 m kg m 1.31108 sec sec p 1.67 1 0.8 3 108 4.01 108 kg m sec kg m sec kg m sec 3313 Andrew Brandt 5 Relativistic Energy E mc2 KE mc 2 KE ( 1)mc 2 E (mc 2 ) 2 p 2 c 2 • Ex. 1.6: A stationary bomb (at rest) explodes into two fragments each with 1.0 kg mass that move apart at speeds of 0.6 c relative to original bomb. Find the original mass M. Ei Mc2 KEi Mc2 0 E f 1m1c 2 1m2c2 Mc 2 2m1c 2 1 2 M 2kg / 0.8 2.5kg v c2 • What if the bomb were not stationary? 3/8/2010 3313 Andrew Brandt 6 Properties of Photoelectric Effect • Existence of photoelectric effect was not a surprise, but the details were surprising 1) Very little time (nanoseconds) between arrival of light pulse and emission of electron 2) Electron energy independent of intensity of light 3) At higher frequency get higher energy electrons Minimum frequency (0) required for photoelectric effect depends on material: slope=E/ =h (planck’s constant) 3/8/2010 3313 Andrew Brandt 7 Einstein Explains P.E. Effect • Einstein explained P.E. effect: energy of light not distributed evenly over classical wave but in discrete regions called quanta and later photons 1) EM wave concentrated in photon so no time delay between incident photon and p.e. emission 2) All photons of same frequency have same energy E=h, so changing intensity changes number (I=Nh, where N is rate/area) but not energy 3) Higher frequency gives higher energy h 6.626 1034 J sec • Electrons have maximum KE when all energy of photon given to electron. • is work function or minimum energy required to liberate electron from material ( =h 0 ) KE max h h h 0 3/8/2010 3313 Andrew Brandt 8 Example of PE for Iron • a) Find given 0 1.11015 Hz h 0 15 15 1 4.14 10 eV sec1.110 s 4.5eV • b) If p.e.’s are produced by light with a wavelength of 250 nm, what is stopping potential? KEmax eV0 • • • 6 h 1.24 10 eV m 4.5eV 9 250 10 m =4.96-4.5 =0.46 eV (is this your final answer?) NO! it is V0=0.46V (not eV) x-ray is inverse photo-electric effect, can neglect binding energy, since x-ray is very energetic 3/8/2010 3313 Andrew Brandt 9 Photon Energy Loss h (1 cos ) mc 2me c 2 .511MeV 2 e e 3/8/2010 3313 Andrew Brandt 10 De Broglie E pc h hv h h p p c • Tiger example • He hits a 46 g golf ball with a velocity of 30 m/s (swoosh) . • Find the ; what do you expect? dB • 6.6 x1034 J sec h v c 1 4.8 1034 m mv .046kg 30m / sec • The wavelength is so small relevant to the dimensions of the golf ball that it has no wave like properties • What about an electron with v 107 m / sec ? • This large velocity is still not relativistic so h 6.63 1034 J s 11 7.3 10 m mv 9.11031 kg 107 m/s 11 • Now, the radius of a hydrogen atom (proton + electron) is 5 10 m • Thus, the wave character of the electron is the key to understanding atomic structure and behavior 3/8/2010 3313 Andrew Brandt 11 De Broglie Example 3.2 • • What is the kinetic energy of a proton with a 1fm wavelength Rule of thumb, need relativistic calculation unless pc pc hc mp c2 .938GeV 4.136 1015 ev s 3 108 m / s 1.24GeV 11015 m • Is pc = energy? Units are right, but pc .ne. energy • Since • This implies a relativistic calculation is necessary so can’t use KE=p2/2m pc mp c2 need relativistic calculation E mc 2 p 2c 2 (.938) 2 (1.24) 2 1.55GeV KE E mc 2 1.55 0.938 0.617GeV 617MeV 3/8/2010 3313 Andrew Brandt 12 Wave Velocities • • • • Definition of phase velocity: v p 2 / 2 k d Definition of group velocity: vg k dk v v c For light waves : g p d For de Broglie waves vg dk 3/8/2010 3313 Andrew Brandt 13 Particle in a Box Still 2 2 1 p h 2 • General expression for non-rel Kinetic Energy: KE mv 2 2 2 m 2 m • With no potential energy in this model, and applying constraint on wavelength gives: h2 2 2 2 En n h / 8 mL 2m(2 L / n)2 • Each permitted E is an energy level, and n is the quantum number • General Conclusions: 1) Trapped particle cannot have arbitrary energy like a free particle—only specific energies allowed depending on mass and size of box 2) Zero energy not allowed! v=0 implies infinite wavelength, which means particle is not trapped 3) h is very small so quantization only noticeable when m and L are also very small 3/8/2010 3313 Andrew Brandt 14 Example 3.5 • 10 g marble in 10 cm box, find energy levels En n 2 h 2 / 8mL2 • • • • n2 (6.63 1034 J s)2 64 2 5.5 10 n J 2 1 2 8 10 kg (10 m) For n=1 Emin 5.5 1064 J and v 3.3 1031 m / s Looks suspiciously like a stationary marble!! At reasonable speeds n=1030 Quantum effects not noticeable for classical phenomena 3/8/2010 3313 Andrew Brandt 15 Uncertainty Principle a) For a narrow wave group the position is accurately measured, but wavelength and thus momentum cannot be precisely determined b) Conversely for extended wave group it is easy to measure wavelength, but position uncertainty is large • Werner Heisenberg 1927, it is impossible to know exact momentum and position of an object at the same time x p 3/8/2010 3313 Andrew Brandt 2 E t 2 16 Rutherford Scattering K N ( ) sin 4 2 KE 2 • The actual result was very different— although most events had small angle scattering, many wide angle scatters were observed • “It was almost as incredible as if you fired a 15 inch shell at a piece of tissue paper and it came back at you” • Implied the existence of the nucleus. • We perform similar experiments at Fermilab and CERN to look for fundamental structure 3/8/2010 3313 Andrew Brandt 17 Spectral Lines • For Hydrogen Atom (experimental observation): 1 1 1 R( 2 2 ) nf ni • where nf and ni are final and initial quantum states • R=Rydberg Constant 1.097 107 m1 0.01097nm 1 • Balmer Series nf = 2 and ni=3,4,5 visible wavelengths in Hydrogen spectrum 656.3, 486.3,…364.6 (limit as n) 3/8/2010 3313 Andrew Brandt 18 Bohr Energy Levels E e 2 En 21 8 0 r m n How much energy is required to raise an electron from the ground state of a Hydrogen atom to the n=3 state? E E f Ei E1 E1 1 1 8 2 E1 ( 2 ) 13.6( ) 12.1eV 2 nf ni 3 1 9 Rydberg atom has r=0.01 mm what is n? E? rn n r1 2 n rn 1105 435 a0 5.3 1011 En E1 7.19 105 eV 2 n Bohr atom explains energy levels E E f Ei E1 E1 h hc / n f 2 ni 2 1 E1 1 1 ( 2 2) ch n f ni Correspondence principle: For large n Quantum Mechanics Classical Mechanics 3/8/2010 3313 Andrew Brandt 19 Expectation Values • Complicated QM way of saying average • After solving Schrodinger Eq. for particle under particular condition contains all info on a particle permitted by uncertainty principle in the form of probabilities. • This is simply the value of x weighted by its probabilities and summed over all possible values of x, we generally write this as x *( x, t ) x( x, t )dx 3/8/2010 3313 Andrew Brandt 20 Eigenvalue Examples d2 • If operator is G 2 and wave function is dx d 2 2x d d 2x d 2x G e e 2 e • 2 dx dx dx dx • • • • • So eigenvalue is 4 How about sin(kx)? eigenvalue is –k2 How about x4? Not an eigenvector since 3/8/2010 e2 x find eigenvalue 4e 2 x G 4 d2 4 G 2 x 12 x 2 cx 4 dx 3313 Andrew Brandt 21