Sol9 - 8.18 10 V 81.8 kV. 2 2 1.60 10 C p m v V e-- = = = =...

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UAHuntsville Spring, 2010 PH 113: Homework Solutions No. 9 Solution 52: (a) We use: V stop = hf - F e = hc l - F e = 1240eV×nm 400nm - 1.8eV e =1.3V (b) The speed v of the electron satisfies K max = 1 2 mv 2 = 1 2 mc 2 ´ v 2 c 2 = E photon - F We find ( 29 ( 29 ( 29 photon stop stop 8 2 3 5 2 2 2 2 1.3V 2.998 10 m/s 511 10 eV 6.8 10 m/s. e e e E eV eV e v c m m m c - F = = = = ´ ´ = ´ Solution 53: (a) (1 cos ) e h m c θ ∆λ = - . In this case φ = 180° (so cos = –1), and the change in wavelength for the photon is given by ∆λ = 2 h / m e c . The energy E' of the scattered photon (with initial energy E = hc / λ ) is then E hc E E h m c E hc E E m c e e ' / ( / )( / ) / . ( . . = + = + = + = + = + = λ λ ∆λ ∆λ 1 1 2 1 2 50 0 1 2 50 0 418 2 κες κες 29 / 0.511Μες κες . (b) From conservation of energy the kinetic energy K of the electron is given by K = E – E' = 50.0 keV – 41.8 keV = 8.2 keV.
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Solution 54: (a) We solve v from λ = h / p = h /( m p v ): ( 29 ( 29 34 6 27 12 6.626 10 J s 3.96 10 m/s. 1.6705 10 kg 0.100 10 m p h v m - - - ´ × = = = ´ l ´ ´ (b) We set eV K m v p = = 1 2 2 and solve for the voltage: ( 29 ( 29 ( 29 2 27 6 2 4 19 1.6705 10 kg 3.96 10 m/s
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Unformatted text preview: 8.18 10 V 81.8 kV. 2 2 1.60 10 C p m v V e-- = = = = Solution 55: (a) For the first and second case (labeled 1 and 2) we have eV 01 = hc / 1 , eV 02 = hc / 2 , from which h and can be determined. Thus, ( 29 ( 29 ( 29 ( 29 ( 29 1 2 15 1 1 1 1 17 1 2 1.85eV 0.820eV 4.12 10 eV s. 3.00 10 nm/s 300nm 400nm e V V h c-------= = = l- l - (b) The work function is =--=--= 3 0820 185 300 400 2 27 2 2 1 1 1 2 ( ) ( . ( . . V V l l l l eV)(400 nm) eV)(300 nm) nm nm eV. (c) Let = hc / max to obtain max . = = = hc F 1240 2 27 545 eV nm eV nm. Solution 56: Solution 57: 2 = 1 1-v 2 c 2 v = c 1-1 g 2 p = gmv = gmc 1-1 g 2 = mc g 2- 1 Solution 58: Solution 59:...
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Sol9 - 8.18 10 V 81.8 kV. 2 2 1.60 10 C p m v V e-- = = = =...

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