Answer with Explanation:
We are given that
[tex]V(x)=Vsin(\frac{2\pi x}{\lambda})[/tex]
[tex]V_0=3800 V,\lambda=1.0 mm[/tex]
a.At x=0
V(0)=0
Maximum potential, [tex]V_{max}=V_0=3800 V[/tex]
The proton must have minimum speed when V(x) is maximum
Initial kinetic energy =[tex]q\Delta V[/tex]
[tex]\frac{1}{2}mv^2=q(V_{max}-V(0))=q(3800-0)=3800q[/tex]
q=[tex]1.6\times 10^{-19} C[/tex]
[tex]m=1.67\times 10^{-27} Kg[/tex]
[tex]v^2=\frac{2\times 3800 q}{m}=\frac{3800\times 2\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}[/tex]
[tex]v=\sqrt{\frac{2\times 3800 q}{m}}=\sqrt{\frac{3800\times 2\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}[/tex]
[tex]v=8.5\times 10^5 m/s[/tex]
b.The maximum speed reached by a proton when V(x) is minimum
[tex]V_{min}=-3800 V[/tex]
[tex]v=\sqrt{\frac{2q(V_{max}-V_{min}}{m}}[/tex]
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}(3800-(-3800))}{1.67\times 10^{-27}}[/tex]
[tex]v=1.21\times 10^6 m/s[/tex]
