25: Problem 25.30

INTRO:
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.

Part A: 
What is the electric force on a proton 1.4 fm from the surface of the nucleus? 
Hint: Treat the spherical nucleus as a point charge.
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
We know the formula for the force between two charged particles is F = K |q₁| |q₂| r^-2
We simply have to determine which values to plug into this equation and we will get the answer that we're looking for. 
Begin by taking the radius of the nucleus, which is half the diameter, or 3 fm
the added distance of the proton is 1.4 fm, so the total distance of the proton from the center of the nucleus is 3 + 1.4 = 4.4 fm = 4.4⋅10^-15 m

so r = 4.4⋅10^-15 m​

Next we need to determine the two charges. 
We're given that the nucleus has a charge of q = +54e. 
Knowing that e=1.6⋅10^-19 C , q_nucleus = 8.64⋅10^-18 C

q₁ = 8.64⋅10^-18 C​

lastly, we know that the charge of ONE proton is just e, so

q₂ = 1.6⋅10^-19 C​

Now we have all of our values to plug in and can solve for the force
(remember K is the electrostatic constant and equal to 9.0⋅10⁹ Nm²/C²)

∴F = (9.0⋅10⁹ Nm²/C²)(8.64⋅10^-18 C)(1.6⋅10^-19 C)(4.4⋅10^-15 m)^-2​

→F = 642.645 N​

F = 643 N

Part B:
What is the proton's acceleration?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
We know that F=ma, so if we divide the force by the given mass...  a = 3.8×10²⁹ m/s²