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 fm = 1 femtometer = 1×10−15m.) Hint: Treat the spherical nucleus as a point charge.

A)

What is the electric force on a proton 9.0 fm from the surface of the nucleus?

B)

What is the proton's acceleration?

Answer 1

Part A.

Electrostatic force is given by:

F = k*q1*q2/R^2

q1 = Charge on 125Xe atom = +54e = 54*1.6*10^-19 C

q2 = charge on proton = +e = 1.6*10^-19 C

R = distance between Xe nucleus and proton = 9.0 fm + radius of nucleus = 9.0 fm + 6.0 fm/2 = 12.0 fm

R = 12.0*10^-15 m

So,

F = 9*10^9*54*1.6*10^-19*1.6*10^-19/(12.0*10^-15)^2

F = 86.4 N = Force on proton

Part B.

Using Newton's 2nd law:

F_net = m*a

a = F_net/m

m = mass of proton = 1.67*10^-27 kg

F_net = Force on proton due to nucleus = 86.4 N

So,

a = F_net/m

a = 86.4/(1.67*10^-27)

a = 5.17*10^28 m/s^2 = acceleration of proton

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