Answer:
[tex]v_{0,new} = v0\sqrt{}2[/tex]
Explanation:
Initial work done on the proton is given by, [tex]\DeltaW0 = q V_o[/tex]
we know that, [tex]\DeltaW = \DeltaK.E[/tex]
[tex]qV0 = (1/2) m v_0^2[/tex]
[tex]v_0 = \sqrt{}2 q V_0 / m[/tex] { eq.1 }
If it were accelerated instead through a potential difference of 2V0, then it would gain a speed will be given as :
using the above formula, we have
[tex]v_{0,new} = \sqrt{}2 q (2V0) / m[/tex]
[tex]v_{0,new} = \sqrt{}4 q V0 / m[/tex]
[tex]v_{0,new} = v0\sqrt{}2[/tex]
