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.

1. What is the electric force on a proton 1.8 fm from the surface of the nucleus?
2. Express your answer to two significant figures and include the appropriate units.
3. What is the proton's acceleration? Express your answer to two significant figures and include the appropriate units.

Answer 1

Electric force is given by:
F = (kQ₁Q₂)/r²
where k is Coloumb's constant of value 9 x 10⁹, Q₁ and Q₂ are charges and r is the separation between them.
Let the charge of the nucleus be located at its center.
The separation of the proton is equal to:
Radius of nucleus + distance from surface
= 3 + 1.8
= 4.8 fm
Charge of Nucleus = 54e
Charge of proton = e
F = (9 x 10⁹ x 54 x 1.60 x 10⁻¹⁹ x 1.60 x 10⁻¹⁹)/(4.8 x 10⁻¹⁵)²
F = 540 N

Part 3)
The mass of the proton = 1.67 x 10⁻²⁹ kg
acceleration = 540/1.67 x 10⁻²⁹
= 3.20 x 10³¹ m/s²