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To complete your master\'s degree in physics, your advisor has you design a smal

ID: 1965785 • Letter: T

Question

To complete your master's degree in physics, your advisor has you design a small, linear accelerator capable of emitting protons, each with a kinetic energy of 11.0 keV. (The mass of a single proton is 1.67 10-27 kg.) In addition, 1.00 109 protons per second must reach the target at the end of the 1.40-m-long accelerator.
(a) What the average power must be delivered to the stream of protons?
µW

(b) What force (assumed constant) must be applied to each proton?
N

(c) What speed does each proton attain just before it strikes the target, assuming the protons start from rest?
m/s

Explanation / Answer

(a) The average power deliverd to each proton is ( P)   = nE / t                                                                                         = ( 1*109 )( 11.0 keV ) / (1 s )                                                                                          = ( 1*109 )( 11.0 keV )( 1.6*10-16 J / 1 keV ) / (1 s )                                                                                          = ( 17.6*10-7 W ) ( 1 W / 10 -6 W )                                                                                          = 1.76 W ----------------------------------------------------------------------------------------------- ---------------------------------------------------------------------------------------------- (b) Work done on the proton is equal to the kinetic energy given to the proron . The force delivered to the each proton is ( F ) = W / d                                                                                =  ( 11.0 keV )( 1.6*10-16 J / 1 keV ) / ( 1.40 m)                                                                                = 1.26*10-15 N -------------------------------------------------------------------------------------------------- -------------------------------------------------------------------------------------------------- (c) The velocity of the proton just before it srtikes the target is                                                                    v = sqrt( 2k / m )                                                                        = sqrt ( ( 2) ( ( 11.0 keV )( 1.6*10-16 J / 1 keV )   / ( 1.67*10-27 kg ) )                                                                         = 1.45*106 m/s                                                                                =  ( 11.0 keV )( 1.6*10-16 J / 1 keV ) / ( 1.40 m)                                                                                = 1.26*10-15 N -------------------------------------------------------------------------------------------------- -------------------------------------------------------------------------------------------------- (c) The velocity of the proton just before it srtikes the target is                                                                    v = sqrt( 2k / m )                                                                        = sqrt ( ( 2) ( ( 11.0 keV )( 1.6*10-16 J / 1 keV )   / ( 1.67*10-27 kg ) )                                                                         = 1.45*106 m/s
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