exam3_spring2009

exam3_spring2009 - Physics 251 exam 3, May 18, 2009. You...

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Unformatted text preview: 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 = 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. Whats the work function of the metal? Answer: The photon energy hc/ goes towards overcoming the work function 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, = hc E k = 1240 eV nm 150 nm (3 . eV ) = 8 . 27 3 . 0 = 3 . 27 eV 2. An electron is accelerated through a 300 kV potential. Whats 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 = ( 1) mc 2 or = 1 + qV mc 2 = 1 + 300 10 3 eV 511 10 3 eV = 1 . 58 and its velocity can be found from = 1 / radicalBig 1 2 1 2 = 1 / 2 = 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 = h/p but since p = m v and v = c we have = h p = h mc = hc mc 2 = 1240 eV nm 1 . 58 511 10 3 eV . 774 = 0 . 00198 nm ....
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This note was uploaded on 05/28/2011 for the course PHY 251 taught by Professor Rijssenbeek during the Fall '01 term at SUNY Stony Brook.

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exam3_spring2009 - Physics 251 exam 3, May 18, 2009. You...

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