Respuesta :
Explanation:
It is known that charge on xenon nucleus is [tex]q_{1}[/tex] equal to +54e. And, charge on the proton is [tex]q_{2}[/tex] equal to +e. So, radius of the nucleus is as follows.
r = [tex]\frac{6.0}{2}[/tex]
= 3.0 fm
Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.
d = r + 2.5
= (3.0 + 2.5) fm
= 5.5 fm
= [tex]5.5 \times 10^{-15} m[/tex] (as 1 fm = [tex]10^{-15}[/tex])
Therefore, electrostatic repulsive force on proton is calculated as follows.
F = [tex]\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}[/tex]
Putting the given values into the above formula as follows.
F = [tex]\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}[/tex]
= [tex](9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}[/tex]
= [tex](9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}[/tex]
= 411.2 N
or, = [tex]4.1 \times 10^{2}[/tex] N
Thus, we ca conclude that [tex]4.1 \times 10^{2}[/tex] N is the electric force on a proton 2.5 fm from the surface of the nucleus.
