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Unformatted text preview: PHYSICS 2D PROF. HIRSCH Formulas: QUIZ 4 SPRING QUARTER 2005 MAY 6 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
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† 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
† 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 4 SPRING QUARTER 2005 MAY 6 2005 Justify all your answers to all problems Problem 1 (15 points) (a) An electron is described by the wavepacket y ( x, t ) = Ú dk a( k ) e i( kx-w ( k )t ) , with a(k)=C for k between 0.95A-1 and 1.05A-1, and a(k)=0 otherwise. C is a nonzero constant. Estimate the uncertainty in the position of this electron, in A. (b) For the electron described by the wavepacket given in part (a), estimate its speed (group † velocity). Give your answer as v/c. (c) This part is unrelated to parts (a) and (b). Find the de Broglie wavelengths for electrons traveling at speed v=0.01c and for electrons traveling at speed v=0.99c. Give your answers in A. Problem 2 (15 points) A new particle has been discovered that has the same mass as the proton but interacts with electrons in a different way than protons. The potential energy of an electron at distance r from this particle is a U ( r) = r with a=14.4 eV A1/2. (a) Using the uncertainty principle find an expression for the average kinetic energy of an electron in an orbit of radius r around this particle. Justify your answer. † (b) By minimization of the total energy find an expression for the size (radius) of the oneelectron 'atom' where the nucleus is this particle rather than a proton. (c) Find numerical values for the radius of this atom (in A) and for the kinetic, potential and total energy of this atom (in eV). Use h 2 / m = 7.62 eV A 2 Problem 3 ( 15 points) † Y(x) 0 10A 40A x An electron is described by the wavefunction shown in the figure, it has a constant nonzero value between x=10A and x=40A and is zero everywhere else. (a) What is the probability that this electron will be found at a distance 20A or less from the origin, x=0? (b) Estimate the uncertainty in the momentum of this electron. Express your answer in units eV/c. (c) Estimate the kinetic energy of this electron, in eV. Justify all your answers to all problems ...
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