We can solve the problem by using conservation of energy.
In fact, the electric potential energy lost by the charge when moving through the potential difference is equal to the kinetic energy acquired:
[tex] \Delta U=\Delta K =K_f -K_i =K_f[/tex]
where [tex] K_f [/tex] is the final kinetic energy, and [tex] K_i [/tex] is the initial kinetic energy, which is zero since the particle starts from rest.
We can rewrite the equation above as:
[tex] q \Delta V=\frac{1}{2}mv^2 [/tex]
where
[tex] q=e=1.6 \cdot 10^{-19}C [/tex] is the charge of the ion
[tex] \Delta V [/tex] is the potential difference
[tex] m=4u=4 (1.67 \cdot 10^{-27} kg)=6.7 \cdot 10^{-27} kg [/tex] is the mass of the ion
[tex] v=1.5 \cdot 10^6 m/s [/tex] is the final speed of the ion
Substituting the numbers and rearranging the equation, we can find the potential difference needed:
[tex] \Delta V=\frac{mv^2}{2q}=\frac{(6.7 \cdot 10^{-27}kg)(1.5 \cdot 10^6 m/s)^2}{2(1.6 \cdot 10^{-19}C)}=4.7 \cdot 10^4 V=47 kV [/tex]
