Question

Please fully explain

4) A proton is to be accelerated from rest by using an accelerating voltage of 400 volts. a) How fast will the proton be going after the acceleration through 400 volts? b) Should the proton start out at a higher or lower voltage? c) If the proton is to be accelerated with twice the voltage (800 volts), will the final velocity of the 2.77x10' m/s higher proton be twice as large, less than twice as large, or greater than twice as large as in part a? less than twice as large

Answer 1

(a) qV = 1/2mv²

v = sqrt (2qV / m)

where m is mass of proton = 1.67e-27 kg

v = sqrt (2*1.6e-19*400 / 1.67e-27)

v = 2.76e5 m/s

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(b) The voltage will always decrease as proton moves. so it should start at higher potential

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(c) Now, If V = 800 volts

so,

v = sqrt (2*1.6e-19*800 / 1.67e-27)

v = 3.92e5 m/s

as we can see velocity is less than twice as large