What is the electric force and acceleration of a proton??

Question

The nucleus of a 125Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q=54e. What is the electric force on a proton 1.20 fm from the surface of the nucleus? What is the proton's acceleration?

Answer 1

USE THESE 2 LAWS:

ELECTRIC FORCE= [COLOUMBS CONTANT* CHARGE1*CHARGE 2] / DISTANCE ^ 2

AND

FORCE= MASS (MASS OF PROTON) * ACCELERATION

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F = kq1q2 / r²

= ((8.99 x 10^9 Nm²/C²) x (8.64 x 10^-18 C) x (1.60 x 10^-19 C)) / (1.20 x 10^-15 m)²

= 8,630 N

The acceleration is:

a = F / m

= 8,630 N / (1.67 x 10^-27 kg)

= 5.17 x 10^30 m/s²

Answer 2

The force on the proton is:

F = kq1q2 / r²

= ((8.99 x 10^9 Nm²/C²) x (8.64 x 10^-18 C) x (1.60 x 10^-19 C)) / (1.20 x 10^-15 m)²

= 8,630 N

The acceleration is:

a = F / m

= 8,630 N / (1.67 x 10^-27 kg)

= 5.17 x 10^30 m/s²