Question

In a typical Van de Graaff linear accelerator, protons are accelerated through a potential difference of 20 MV. What is their kinetic energy if they started from rest? Give your answer in (a) eV, (b) keV,(c) MeV, (d) GeV, and (e) joules.

Answer 1

Answer:

a) 2 x10^7 eV

b) 2 x10^4 keV

c) 20 MeV

d) 0.02 Gev

e) 3.2 x 10^-12J

Explanation:

The potential difference = 20 x 10^6 V

The charge on the proton = 1.6 x10^-19

The work done to move the proton will be basically the proton will acquire if it accelerates.

Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19

                                                 =3.2 x 10^-12J or 2 x10^7 eV

2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev

Answer 2

Explanation:

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