A proton (charge e = + 1.60´10 -19 C, mass m =1.67´10 -27 kg) accelerates from r

Question

A proton (charge e = + 1.60´10-19 C, mass m =1.67´10-27 kg) accelerates from rest by crossing apotential difference V = 3.00 V

a). Find the velocity v0 it acquires.

Suppose this proton, having initial velocity v0 frompart (a), moves from very far away directly towards a stationaryheavy ion which is singly charged, i.e. having the same charge asthe proton, +e.

b) Find the distance rmin between the proton and theion at the moment of closest approach.

c) Find velocity of a proton when it is at a distance r =2rmin from the ion.

Explanation / Answer

charge e = + 1.60´10-19 C
mass m = 1.67´10-27 kg Potential difference V = 3.00 V a). the velocity v0 it acquires v =[ 2Ve / m ]                                                =2.397 * 10 ^ 4 m / s (b). At closest approach kinetic energy = potentialenergy          ( 1/ 2) m v ^2 = K q q ' / r from this required distance   r = 2K* e*e / m v ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 plug the values weget    r = 2.398 * 10 ^-10 m

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