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exam3_spring2009

# exam3_spring2009 - Physics 251 exam 3 You may use a...

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Physics 251 exam 3, May 18, 2009. You may use a calculator, and the equation sheet that I have provided. Order of atoms in the periodic table: H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg ... And here are some isotopic masses you might find to be of use: 4 2 He: 4.002 603 142 54 Xe: 141.929 701 143 55 Cs: 142.927 330 144 55 Cs: 143.932 030 145 55 Cs: 144.935 390 143 56 Ba: 142.920 617 144 56 Ba: 143.922 940 145 56 Ba: 144.926 920 143 57 La: 142.916 059 144 57 La: 143.919 590 145 57 La: 144.921 641 235 92 U: 235.043 923 236 92 U: 236.045 562 237 92 U: 237.048 724 238 92 U: 238.050 783 239 92 U: 239.054 288 237 93 Np: 237.048 167 238 93 Np: 238.050 941 239 93 Np: 239.052 931 239 94 Pu: 239.052 157 238 95 Am: 235.048 032 239 95 Am: 239.053 018 240 95 Am: 240.055 288 239 96 Cm: 239.054 951 240 96 Cm: 240.055 519 241 96 Cm: 241.057 647 1. Ultraviolet light with λ = 150 nm is shined on a metal, and a stopping potential of 3.0 Volts is sufficient to make the detected current go to zero in a photoelectric effect experiment. What’s the work function of the metal? Answer: The photon energy hc/λ goes towards overcoming the work function ϕ of the metal and giving the electron a kinetic energy E k which is equal to the stopping potential times electron charge or qV . That is, ϕ = hc λ E k = 1240 eV · nm 150 nm (3 . 0 eV ) = 8 . 27 3 . 0 = 3 . 27 eV 2. An electron is accelerated through a 300 kV potential. What’s its velocity? Its de Broglie wavelength? What strength of magnetic field is required to make it curve with a radius of 0.5 meters? Answer: The electron acquires a kinetic energy of qV = 300 keV, so we must account for relativity as its rest mass is only 511 keV/ c 2 . Therefore we have qV = ( γ 1) mc 2 or γ = 1 + qV mc 2 = 1 + 300 × 10 3 eV 511 × 10 3 eV = 1 . 58 and its velocity can be found from γ = 1 / radicalBig 1 β 2 1 β 2 = 1 2 β = radicalBig 1 1 2 = radicalBig 1 1 / (1 . 58 2 ) = 0 . 774 or v = 0 . 774 · 3 × 10 8 m/sec or 2 . 3 × 10 8 m/sec. Next, the de Broglie wavelength is λ = h/p but since p = γm 0 v and v = βc we have λ = h p = h γmβc = hc γmc 2 β = 1240 eV · nm 1 . 58 · 511 × 10 3 eV · 0 . 774 = 0 . 00198 nm . 1

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Finally, the radius of curvature can be found from qvB = γmv 2 /r giving B = γmv 2 qvr = γmcβ qr = 1 . 58 · (9 . 11 × 10 31 kg ) · (3 . 00 × 10 8 m/s ) · (0 . 774) (1 . 602 × 10 19 C ) · (0 . 5 m ) or B = 0 . 0042 Tesla. 3. A tritium nucleus (hydrogen plus two neutrons; assume that a neutron mass equals a proton mass) captures a muon (mass 105.7 MeV/c 2 ) to briefly form an exotic atom. Calculate the energy and Bohr model orbital radius of the first excited state (not the ground state).
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