Sol9 - UAHuntsville Spring, 2010 PH 113: Homework Solutions...

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UAHuntsville Spring, 2010 PH 113: Homework Solutions No. 9 Solution 52: (a) We use: V stop = hf −Φ e = hc λ 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 We find ( ) ( ) ( ) 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 = = = = × × = × Solution 53: (a) (1 cos ) e h m c θ Δλ = . In this case φ = 180° (so cos = –1), and the change in wavelength for the photon is given by Δλ = 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 Δλ Δλ 1 1 2 1 2 50 0 1 2 50 0 418 2 keV keV) / 0.511MeV keV . (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 ): ( )( ) 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 × = = = × λ × × (b) We set eV K m v p = = 1 2 2 and solve for the voltage: ( ) ( ) 2 27 6 2 4 19 1.6705 10 kg 3.96 10 m/s 8.18 10 V 81.8 kV. 2 2 1.60 10 C p m v
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Sol9 - UAHuntsville Spring, 2010 PH 113: Homework Solutions...

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