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Unformatted text preview: 28 Modern Physics Concepts Chap. 2 12. What are the wavelengthsof electrons with kinetic energiesof (a) 10 eV, (b) 1000eV, and (c) 107 eV? Solution: From Eq. (2.17) p = (1/c)~T2 + 2Tmoc2 and using the de Broglie relation>.. = hip we obtain the de Broglie wavelength as
A= hc . ~T2 + 2Tmoc2 (P2.6) Now apply this equation to the three electron energies. (a) Substitute moc2 = mec2= 0.5110 MeV and T = 10 eV into Eq. (P2.6) to obtain (b) similarly, for T = 103 eV we find
>..= (4.135 X 1015 eV s)(2.998 X 108 mjs) = 3.87 X 1011 m. y106 + 2(103)(0.5110 X 106) eV (c) similarly, for T = 107 eV we find >.= (4.135 X 1015 eV s)(2.998 X 108m/s) = 1.18 X 1013 m. .,.11014 2(107)(0.5110X 106) eV + What is the de Broglie wavelengthof a water molecule moving at a speed of 2400m/s? What is the wavelengthof a 3g bullet moving at 400 m/s? Solution: (a) A water molecule (H2O) hag a rest mags of about m = (18 u){1.661 x 1027 kg/u) = 2.989 X 1026 kg. Its momentum whentraveling at 2400mls is p = mv = (2.989x1026 kg) x (2400 m/s) = 7.18 x 1023 kg m Sl = 7.18 X 1023 J s m1. Thus the de Broglie wavelengthof the water molecule is
6.626 hA== X 1034 J s = 9.23 X 1012 m. p 7.18 X 1023 J s ro (b) A 3g bullet moving at 400 m/s has a momentum p = mv = (0.003 kg) x (400 m/s) = 1.2 kg m s1 = 1.2 J s ml .Its de Broglie wavelength is thus' 6.626 X 10~4 J s h = 5.53 X 1034 m. ),== 1.2 J s ml p July 24, 2002 13. ...
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This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 CADY

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