Answer:
[tex]\Delta V=1253V[/tex]
Explanation:
The kinetic energy will be converted into electric potential energy, so we write:
[tex]\frac{mv^2}{2}=q\Delta V[/tex]
We want the potential difference needed for this to happen, which is:
[tex]\Delta V=\frac{mv^2}{2q}[/tex]
Where m and q are the mass and charge of the proton. We then use values:
[tex]\Delta V=\frac{(1.67\times10^{-27}kg)(4.9\times10^5m/s)^2}{2(1.6\times10^{-19}C)}=1253V[/tex]
