(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
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