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# hw6 - \$34 5-1 2 h 6.626 10 J h = = 7.27 106 m s = mv we...

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5-12 Using p = h " = mv , we find that v = h m " = 6.626 # 10 \$ 34 J % s 9.11 # 10 \$ 31 kg ( ) 1 # 10 \$ 10 m ( ) = 7.27 # 10 6 m s . From the principle of conservation of energy, we get eV = mv 2 2 = 9.11 " 10 # 31 kg ( ) 7.27 " 10 6 m s ( ) 2 2 = 2.41 " 10 # 17 J = 151 eV . Therefore V = 151 V . 5-15 For a free, non-relativistic electron E = m e v 0 2 2 = p 2 2 m e . As the wavenumber and angular frequency of the electron’s de Broglie wave are given by p = h k and E = h " , substituting these results gives the dispersion relation " = h k 2 2 m e . So v g = d " dk = h k m e = p m e = v 0 . 5-17 E 2 = p 2 c 2 + m e c 2 ( ) 2 E = p 2 c 2 + m e c 2 ( ) 2 " # \$ % & 1 2 . As E = h " and p = h k h " = h 2 k 2 c 2 + m e c 2 ( ) 2 # \$ % & ( 1 2 or " k ( ) = k 2 c 2 + m e c 2 ( ) 2 h 2 # \$ % % % & ( ( ( 1 2 v p = " k = k 2 c 2 + m e c 2 h ( ) 2 # \$ % & ( 1 2 k = c 2 + m e c 2 h k ) * + , - . 2 # \$ % % & ( ( 1 2 v g = d " dk k 0 = 1 2 k 2 c 2 + m e c 2 h ) * + , - . 2 # \$ % % & ( ( / 1 2 2 kc 2 = kc 2 k 2 c 2 + m e c 2 h ( ) 2 # \$ % & ( 1 2 v p v g = k 2 c 2 + m e c 2 h ( ) 2 # \$ % & ( 1 2 k 0 1 2 2 3 2 2 4 5 2 2 6 2 2 k 2 c 2 + m e c 2 h ( ) 2 # \$ % & ( 1 2 0 1 2 3 2 4 5 2 6 2 = c 2 Therefore, v g < c if v p > c . 5-18 " x " p # h 2 where " p = m " v = 0.05 kg ( ) 10 # 3 \$ 30 m s ( ) = 1.5 \$ 10 # 3 kg % m s . Therefore, " x = h 2 " p = 6.626 # 10 \$ 34 J % s 4 & 1.5 # 10 \$ 3 kg % m s ( ) = 3.51 # 10 \$ 32 m .

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5-19 K = mv 2 2 = p 2 2 m : 1 " 10 6 eV ( ) 1.6 " 10 # 19 J eV ( ) = p 2 2 1.67 " 10 # 27 kg ( ) \$ p = 2.312 " 10 # 20 kg % m s , " p = 0.05 p = 1.160 # 10 \$ 21 kg % m s and " x " p = h 4 # . Thus " x = 6.63 # 10 \$ 34 J % s 1.16 # 10 \$ 21 kg % m s ( ) 4 & ( ) = 4.56 # 10 \$ 14 m . Note that non-relativistic treatment has been used, which is justified because the kinetic energy is only 1.6 " 10 # 13 ( ) " 100% 1.50 " 10 # 10 = 0.11% of the rest energy.
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