To solve this problem we will start from the conservation of energy. Here we will have that the kinetic energy must be equivalent to the potential energy. That is
[tex]KE = \frac{1}{2} mv^2[/tex]
Here,
m = mass
v = Velocity
[tex]PE = \frac{kq^2}{r}[/tex]
[tex]k = 8.99*10^9 N\codt m^2\cdot C^{-2} \rightarrow[/tex] Coulomb constant
[tex]q = 1.602 x 10^{-19} Coulombs \rightarrow[/tex] Charge Proton
r = Distance
Equation both expression we have that
[tex]2(\frac{1}{2}) mv^2 = \frac{kq^2}{r}[/tex]
The kinetic energy is multiplied by two, by the existence of the two protons.
[tex]r = \frac{kq^2}{mv^2}[/tex]
Replacing the values we have that
[tex]r = \frac{( 8.99*10^9)(1.602 x 10^{-19})^2}{(1.67*10^{-27})(1.5*10^5)^2}[/tex]
[tex]r = 6.1402*10^{-12}m[/tex]
Therefore the distance of closest approach is [tex]6.1402*10^{-12}m[/tex]
