The nucleus of a 125 Xe 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 −15 m .) Hint: Treat the spherical nucleus as a point charge. Part A What is the electric force on a proton 3.0 fm from the surface of the nucleus? Express your answer in newtons. F nucleusonproton F n u c l e u s o n p r o t o n = nothing N SubmitRequest Answer Part B What is the proton's acceleration? Express your answer in meters per second squared. a proton a p r o t o n = nothing m/ s 2 SubmitRequest Answer Provide Feedback Next

Diameter of the 125 Xe nucleus, [tex]\rm d=6.0\ fm = 6.0\times 10^{-15}\ m.[/tex]
Distance of the proton from the surface of the nucleus, [tex]\rm a=3.0\ fm = 3.0\times 10^{-15}\ m.[/tex]

Part A:

According to Coulomb's law, the electric field due to a charged sphere of charge Q at a point r distance away from its center is given by

[tex]\rm E=\dfrac{kQ}{r^2}.[/tex]

where, k is the Coulomb's constant, having value = [tex]9\times 10^9\ \rm Nm^2/C^2.[/tex]

Therefore, the electric field due to the 125 Xe nucleus at the proton is given by

[tex]\rm E=\dfrac{kq}{r^2}=\dfrac{(9\times 10^9)\times (+54 e)}{r^2}[/tex]

Here,

e is the elementary charge, having value = [tex]\rm 1.6\times 10^{-19}\ C.[/tex]

r is the distance of the proton from the center of the nucleus = [tex]\rm a + \dfrac d2 = 3.0+\dfrac{6.0}2=6.0\ fm = 6.0\times 10^{-15}\ m.[/tex]

Using these values,

[tex]\rm E=\dfrac{(9\times 10^9)\times (+54 \times 1.6\times 10^{-19})}{(6.0\times 10^{-15})^2}=2.16\times 10^{21}\ N/C.[/tex]

Now, the electric force on a charge q due to an electric field is given as

[tex]\rm F=qE[/tex]

For the proton, [tex]\rm q = e =1.6\times 10^{-19}\ C.[/tex]

Thus, the electric force on the proton is given by

[tex]\rm F = 1.6\times 10^{-19}\times 2.16\times 10^{21}=345.6\ N.[/tex]

According to Newton's second law,

[tex]\rm F=ma[/tex]

where, a is the acceleration.

The mass of the proton is [tex]\rm m_p=1.67\times 10^{-27}\ kg.[/tex]

Therefore, the proton's acceleration is given by

[tex]\rm a = \dfrac{F}{m_p}=\dfrac{345.6}{1.67\times 10^{-27}}=2.069\times 10^{29}\ m/s^2.[/tex]

onyebuchinnaji onyebuchinnaji · Answer 2 · 2022-04-07T12:43:23+02:00

(a) The electric force of the proton is 2.56 x 10⁻²⁹ N.

(b) The acceleration of the proton is 0.0153 m/s².

The electric force of the proton is calculated using Coulomb's law of electrostatic force.

F = kq²/r²

F = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹)/(3)²

F = 2.56 x 10⁻²⁹ N

The acceleration of the proton is determined by applying Newton's second law of motion;

F = ma

a = F/m

a = (2.56 x 10⁻²⁹ )/(1.67 x 10⁻²⁷)

a = 0.0153 m/s²

Learn more about Coulomb's force here: https://brainly.com/question/24743340