SolnsChapter3

SolnsChapter3 - 2 mʬÊ=ÊdÊsinÊœ ´ m ¬ = d sinÊ35‚ = 0.574 d ¬/d = 0.574/mÊ “Only” mÊ=Ê1 So ʬ/dÊ= 0.574 3 ¬ = h/p =

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Unformatted text preview: 2. mʬÊ=ÊdÊsinÊœ ´ m ¬ = d sinÊ35‚ = 0.574 d. ¬/d = 0.574/mÊ. “Only” ‡ mÊ=Ê1. So ʬ/dÊ= 0.574 3. ¬ = h/p = h/mÊc = 6.63¿10-28 kg˛m/sÊ/(9.11¿10-31 kg)(3¿10 8 m/s) = 2.43¿10-12 m . 4. qÊV = ªÊmÊu¤ = (mÊu)¤ 2m = p¤ 2m = (h/¬) 2 2m . Solve: ¬ = h √ 2ÊmÊqÊV 5. p = h/¬ = 6.63¿10-34 J˛s 10-6 m = 6.63¿10-28 kg˛m/s. u = p m = 6.63¿10-28 kg˛m/s 9.11¿10-31 m/s = 728m/s . This is actually quite slow as free-electron speeds go. 8. KEÊ= p¤ 2m Ê´ 1.6¿10-13 J = p¤ 2Ê(1.67¿10-27 kg) ‡ÊpÊ=Ê 2.31¿10-20 kg˛m/s.Ê 6.63¿10-34 J˛s 2.31¿10-20 kg˛m/s Ê= 2.87¿10-14 m . 20Ê¿1.6¿10-19 J = p¤ 2Ê(1.67¿10-27 kg) ‡ p = 1.03¿10-22 kg˛m/s. 6.63¿10-34 J˛s 1.03¿10-22 kg˛m/s = 6.41¿10-12 m . 10. p = h/¬ = 6.63¿10-34 J˛s 10-9 m = 6.63¿10-25 kg˛m/s. u = p m = 6.63¿10-25 kg˛m/s 9.11¿10-31 m/s = 7.28¿10 5 m/s . (b) Using result of exercise 4, ¬ = h √ 2ÊmÊqÊV ´ 10-9 m = 6.63¿10-34 J˛s √...
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This note was uploaded on 10/01/2010 for the course PHY 9D taught by Professor Harris during the Fall '09 term at UC Davis.

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