Question

What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point charge. Express your answer to two significant figures and include the appropriate units.

Answer 1

Step-by-step explanation:

It is known that charge on xenon nucleus is
`q_(1)` equal to +54e. And, charge on the proton is
`q_(2)` equal to +e. So, radius of the nucleus is as follows.

r =
`(6.0)/(2)`

= 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

=
`5.5 * 10^(-15) m` (as 1 fm =
`10^(-15)`)

Therefore, electrostatic repulsive force on proton is calculated as follows.

F =
`(1)/(4 \pi \epsilon_(o)) (q_(1)q_(2))/(d^(2))`

Putting the given values into the above formula as follows.

F =
`(1)/(4 \pi \epsilon_(o)) (q_(1)q_(2))/(d^(2))`

=
`(9 * 10^(9)) (54e * e)/((5.5 * 10^(-15))^(2))`

=
`(9 * 10^(9)) (54 * (1.6 * 10^(-19))^(2))/((5.5 * 10^(-15))^(2))`

= 411.2 N

or, =
`4.1 * 10^(2)` N

Thus, we ca conclude that
`4.1 * 10^(2)` N is the electric force on a proton 2.5 fm from the surface of the nucleus.