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9 June, 15:14

What is the speed of a proton that has been accelerated from rest through a potential difference of 4.0 kv?

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  1. 9 June, 15:21
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    Answer: v = 1.19*10^-8 m/s

    Explanation: according to the work-energy theorem, the work done in accelerating the proton equals the kinetic energy that it gains.

    Mathematically, we have that

    qV = 1/2mv²

    Where

    q = magnitude of proton = 1.609*10^-16c

    V = 4kv = 4000v

    m = mass of proton = 9.11*10^-31 kg

    v = velocity of proton = ?

    By substituting the parameters, we have that

    1.609*10^-19 * 4000 = 1/2 * 9.11*10^-31 * v²

    1.609*10^-19 * 4000 * 2 = 9.11*10^-31 * v²

    v² = (1.609*10^-19 * 4000 * 2) / 9.11*10^-31

    v² = 12872*10^-19/9.11*10^-31

    v² = 1412.95*10^-19

    v² = 1.41*10^-16

    v = √1.41*10^-16

    v = 1.19*10^-8 m/s
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