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
