Respuesta :
Answer:
A: [tex]777.6\ N.[/tex]
B: [tex]\rm 4.656\times 10^{29}\ m/s^2.[/tex]
Explanation:
Given:
- Charge on the 125 Xe nucleus, [tex]\rm q = +54e.[/tex]
- 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=1.0\ fm = 1.0\times 10^{-15}\ m.[/tex]
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
[tex]\rm E=\dfrac{kQ}{r^2}.[/tex]
where, k = Coulomb's constant whose value = [tex]9\times 10^9\ \rm Nm^2/C^2.[/tex]
Therefore, the electric field due to the 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]
- [tex]\rm e[/tex] = elementary charge, having value = \rm 1.6\times 10^{-19}\ C.
- [tex]\rm r[/tex] = distance of the proton from the center of the nucleus =[tex]\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.[/tex]
Therefore,
[tex]\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.[/tex]
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 4.86\times 10^{21}=777.6\ N.[/tex]
Part B:
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{777.6}{1.67\times 10^{-27}}=4.656\times 10^{29}\ m/s^2.[/tex]
