Plz solve it

Plz solve it 2 17 A proton moves with a speed of 7.45 x m/s directly towards a free proton originally Find the distance or closest approach for the two protons. Given : 47t 9 X 109 = 1.67 x 10-27 kg ande= 1.6 x 10-19 C

Dear Student,
Let A proton of mass m moves with speed 7.45 × 10⁵ m/s directly towards another proton which is intially at rest . 
we know, proton - proton repulsive force acts . both are repelling . Let r is the distance of closest approach .
Then, K.E energy of moving proton = electrostatic energy between protons 
1/2mv² = Ke²/r 
⇒ r = 2Ke²/mv² = 2 × 9 × 10⁹ × ( 1.6 × 10⁻¹⁹)²/1.675 × 10⁻²⁷ × (7.45 × 10⁵)we
[ Here mass of proton = 1.675 × 10⁻²⁷ Kg, e = 1.6 × 10⁻¹⁹C ] 
∴ r = 5 × 10⁻¹³ m 

Hence, distance of closest approach is 5 × 10⁻¹³ m


Regards