Two protons each of mass 1.67 times 10^-2 kg and charge e are initially at rest

Question

Two protons each of mass 1.67 times 10^-2 kg and charge e are initially at rest separated by 1.0 times 10^-14 m (approximately the radius of a nucleus). When released, the protons fly apart. Determine the change in their electric potential energy when they are 1.0 times 10^-10 m apart (approximately the radius of an atom), e = 1.6 times 10^-19 C, k = 9.0 times 10^9 N m^2/C^2, Express your answer to two significant figures and include the appropriate units. If the electric potential energy change is converted entirely into the kinetic energy of the protons (shared equally), what is the speed of one proton when 1.0 times 10^10 m from the other proton? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Here ,

part A )

as the potential energy due to charges is given as

U = k * q1 * q2/d

d is the distance

change in electric potential energy = k * q1 * q2 (1/d2 - 1/d1)

change in electric potential energy = 9*10^9 * (1.602 *10^-19)^2 * (- 1/(1 *10^-14) + 1/(1 *10^-10))

change in electric potential energy = - 2.31 *10^-14 J

part B)

for the speed of the proton is v

using consrevation of kinetic enegry

2 * 0.5 * m * v^2 = - change in potential energy

1.67 *10^-27 * v^2 = 2.31 *10^-14

solving for v

v = 3.72 *10^6 m/s

the speed of the proton is 3.72 *10^6 m/s

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