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. Part A Part complete What is the electric force on a proton 1.0 fm from the surface of the nucleus? Express your answer in newtons. Fnucleusonproton F n u c l e u s o n p r o t o n = 780 N SubmitPrevious Answers Answer Requested Part B What is the proton's acceleration? Express your answer in meters per second squared.

Answer 1

Answer:

A: $777.6\ N.$

B: $\rm 4.656\times 10^{29}\ m/s^2.$

Explanation:

Given:

• Charge on the 125 Xe nucleus, $\rm q = +54e.$

• Diameter of the 125 Xe nucleus, $\rm d=6.0\ fm = 6.0\times 10^{-15}\ m.$

• Distance of the proton from the surface of the nucleus, $\rm a=1.0\ fm = 1.0\times 10^{-15}\ m.$

Part A:

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

$\rm E=\dfrac{kQ}{r^2}.$

where, k = Coulomb's constant whose value = $9\times 10^9\ \rm Nm^2/C^2.$

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

$\rm E=\dfrac{kq}{r^2}=\dfrac{(9\times 10^9)\times (+54 e)}{r^2}$

• $\rm e$ = elementary charge, having value = \rm 1.6\times 10^{-19}\ C.

• $\rm r$ = distance of the proton from the center of the nucleus =$\rm a+\text{Radius of the nucleus}= a + \dfrac d2 = 1.0+\dfrac{6.0}2=4.0\ fm = 4.0\times 10^{-15}\ m.$

Therefore,

$\rm E=\dfrac{(9\times 10^9)\times (+54 \times 1.6\times 10^{-19})}{(4.0\times 10^{-15})^2}=4.86\times 10^{21}\ N/C.$

The electric force on a charge q due to an electric field is given as  

$\rm F=qE$

For the proton, $\rm q = e =1.6\times 10^{-19}\ C.$

Thus, the electric force on the proton is given by

$\rm F = 1.6\times 10^{-19}\times 4.86\times 10^{21}=777.6\ N.$

Part B:

According to Newton's second law,

$\rm F=ma$

where, a is the acceleration.

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

Therefore, the proton's acceleration is given by

$\rm a = \dfrac{F}{m_p}=\dfrac{777.6}{1.67\times 10^{-27}}=4.656\times 10^{29}\ m/s^2.$