The correct option is B V=32.82 volt
For proton, u=hmλ
∴u=6.626×10−341.672×10−27×0.005×10−9
=7.93×104 metre sec−1 .....(i)
If accelerating potential is V, then velocity acquired by the changed particle having charge Q and mass m,
(1/2)mu2=Q.V
u=√2Q.Vm=√2×1.602×10−191.672×10−27×V
By Eqs.(i) and (ii)
V=32.82volt