Question

E) 6.6 x 10 Q4. Two protons are released from rest when they are 1.00 nm apart. What will be their maximum speed? A) 1.17 x 104 m/s B) 4.22 x 104 m/s C) 1.77 x 10 m/s D) 2.32x 10 m/s E) 3.39 x 10 m/s

Answer 1

Here, the protons will have the maximum velocity, when they have the maximum attainable kinetic energy i.e. when all the potential energy has converted to kinetic energy.

Therefore

U_initial = KE_final

U_initial=Electrostatic potential energy = kq1q2/r

Final kinetic energy of both the protons = 1/2(mv²)+ 1/2(mv²)

=> U_initial = (9 x 10⁹)(1.6 x 10^-19)²/(1 x 10^-9) = 23.04 x 10^-20

=>mv²= 23.04 x 10^-38

=> v = $\sqrt{}$ (23.04 x 10^-20)/(1.67x10^-27) = 1.1745 x 10⁴ m/s

option A is correct