# quiz6 - PHYSICS 2D PROF HIRSCH Formulas QUIZ 6 SPRING...

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Unformatted text preview: PHYSICS 2D PROF. HIRSCH Formulas: QUIZ 6 SPRING QUARTER 2005 MAY 20 2005 † † † † † Lp 1 ;g= ; c = 3 ¥ 10 8 m / s 2 2 g 1 - v /c Lorentz transformation : x ' = g ( x - vt ) ; y ' = y ; z' = z ; t ' = g ( t - vx / c 2 ) ; inverse : v Æ -v uy ux - v Velocity transformation : ux ' = ; uy ' = ; inverse : v Æ -v 2 1 - ux v / c g (1 - ux v / c 2 ) Relativistic Doppler shift : f obs = f source 1 + v / c / 1 - v / c Time dilation/length contraction : Dt = g t ; L = Momentum, energy (total, kinetic, rest) : p = g m u; E = g mc 2 ; K = (g - 1)mc 2 ; E 0 = mc 2 ; E = p 2c 2 + m 2c 4 Electron : me = 0.511 MeV / c 2 Proton : mp = 938.26 MeV / c 2 Neutron : mn = 939.55 MeV / c 2 Æ Æ Electron : me = 9.109 ¥ 10-31 kg Proton : mp = 1.673 ¥ 10-27 kg Neutron : mn = 1.675 ¥ 10-27 kg Atomic mass unit : 1 u = 931.5 MeV / c 2 electron charge = -e, proton charge = e, e = 1.6 ¥ 10 -19 C ; r r r rrr Force on charge q in E and B fields : F = q( E + v ¥ B) ; centripetal acceleration = v 2 / R • Stefan' s law : R = sT 4 , R = power/unit area ; s = 5.67 ¥ 10-8 W / m 2K 4 ; R = cU / 4 , U = energy density = † Ú u(l)dl 0 † † † † † † 8p hc / l hc Planck' s law : u( l, T ) = n ( l) ¥ e ( l, T ) = 4 ¥ hc / lkB T ; Wien' s law : lm T = l e -1 4.96 k B 1 Photoelectric effect : eV0 = ( mv 2 ) max = hf - f , f ≡ work function 2 Photons : E = hf = pc ; f = c / l ; Quantum oscillator : en = nhf ; probability P (en ) µ e-e / k T n B Compton scattering : l' - l = h (1 - cos q ) mec Rutherford scattering : DN = C sin (q / 2) 4 Electrostatics : F = kq1q2 kq (force) ; U = q0 V (potential energy) ; V = (potential) 2 r r 1 1 1 = R( 2 - 2 ) l m n ; R = 1.097 ¥ 10 7 m-1 = 1 911.6 A Hydrogen spectrum : Bohr atom : E n = - ke 2 Z Z 2E ke 2 mk 2e 4 = - 2 0 ; E0 = = = 13.6eV ; E n = E kin + E pot , E kin = - E pot / 2 = - E n 2 rn n 2 a0 2h2 † † hf = E i - E f ; rn = r0 n 2 ; r0 = a0 Z ; a0 = h2 = 0.529 A ; L = mvr = nh angular momentum mke 2 † † † † † X - ray spectra : f 1 / 2 = An ( Z - b) ; K : b = 1, L : b = 7.4 Constants : h = 4.136 ¥ 10-15 eV ⋅ s ; hc = 12, 400 eV A ; k B = 1 /11, 600 eV/K ; ke 2 = 14.4 eV A hc = 1973 eV A ; e = 1.6 ¥ 10-19 C ; N A = 6.02 ¥ 10 23 Conversions : 1eV = 1.6 ¥ 10 -19 joules ; 1A = 10 -10 m = 0.1nm ; 1MeV = 10 6 eV Double slit interference : d sin q = nl (maxima) , d sinq = (n + 1 / 2)l (minima) h E 2p p2 de Broglie : l = ; f = ; w = 2pf ; k = ; E = hw ; p = hk ; E = p h l 2m i( k j x -w j t ) i( kx -w ( k )t ) wave packets : y ( x, t ) = Ú dk a( k ) e , or y ( x, t ) = Â a j e ; DkDx ~ 1 ; DwDt ~ 1 j † † † dw group and phase velocity : v g = dk ; w vp = k ; Heisenberg : DxDp ~ h ; DtDE ~ h Wave function Y( x, t ) =| Y( x, t ) | e iq ( x,t ) ; P ( x, t ) dx =| Y( x, t ) |2 dx = probability † PHYSICS 2D PROF. HIRSCH QUIZ 6 SPRING QUARTER 2005 MAY 20 2005 E -i t h2 ∂ 2Y ∂Y Schrodinger equation : + V(x, t) Y(x, t) = ih ; Y(x, t) = Y(x)e h 2 2m ∂x ∂t • h2 ∂ 2Y Time - independent Schrodinger equation : + V(x)Y(x) = EY(x) ; Ú dx Y* Y = 1 2m ∂x 2 -• † † † Square well : E n = p 2h2n 2 2 npx ; Yn ( x ) = sin( ) 2 2 mL L L mw 2 x 2h ; x op = x , pop = h∂ ; < A >= i ∂x • -• Ú dxY A * op Y Eigenvalues and eigenfunctions : Aop Y = a Y ; Harmonic oscillator : Yn ( x ) = Cn H n ( x )e - uncertainty : ; E= DA = < A 2 > - < A > 2 p2 1 + mw 2 x 2 ; Dn = ±1 2m 2 ; 1 E n = ( n + ) hw 2 † † Step potential : R = Tunneling : ( k1 - k2 ) 2 , ( k1 + k2 ) 2 ; T = 1- R ; x2 k= -2 a ( x )dx x1 2m (E - V ) h2 ; y ~ e -ax T ~ e -2aDx ; T~e Ú a ( x) = † † Justify all your answers to all problems 2 m[V ( x ) - E ] h2 Problem 1 (15 points) An electron is in the ground state of a harmonic oscillator potential, with ground state energy 0.3eV. (a) What is the classical angular frequency of oscillation w, in s-1? Use h = 6.58 ¥ 10-16 eV s . (b) What is the classical amplitude of oscillation, in A (angstroms)? (amplitude of a classical oscillator with the same energy). (c) How much more likely is it to find this electron at position x=0 than at position x=5A? (d) Estimate the speed of this electron. Give your answer as v/c. † Problem 2 (15 points) 10,000 electrons of kinetic energy 8eV propagating in the positive x direction encounter a potential step of height 6eV at x=0. (a) Find the wavenumber of the electrons for x<0 and for x>0, in A-1. Use h 2 / 2 m = 3.81 eV A 2 (b) Estimate how many electrons are reflected and how many are transmitted. (c) If instead of electrons these were protons with the same kinetic energy, would the number reflected be larger, equal or smaller? Justify your answer. † Problem 3 ( 15 points) V0 (i) V0 (ii) (iii) 2V0 1A 3A 1A For the potential barrier on the left (labeled (i)) of width 1A and height V0 , and particles of incident kinetic energy V0 /2, the transmission probability is 10-2. (a) What is the transmission probability for the same particles with the same incident kinetic energy V0 /2 for the potential barrier (ii)? (b) What is the transmission probability for the same particles with the same incident kinetic energy V0 /2 for the potential barrier (iii)? (c) What is the transmission probability for the potential barrier (iii) if the incident particles are twice as heavy and their kinetic energy is V0 ? Hint: use that (ax ) y =ax y ...
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## This note was uploaded on 12/07/2009 for the course PHYS phys 2d taught by Professor Hirsch during the Spring '08 term at UCSD.

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