Can someone help me with this
1 answer:
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Explanation:
It is known that charge on xenon nucleus is equal to +54e. And, charge on the proton is
equal to +e. So, radius of the nucleus is as follows.
r =
= 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
= (as 1 fm =
)
Therefore, electrostatic repulsive force on proton is calculated as follows.
F =
Putting the given values into the above formula as follows.
F =
=
=
= 411.2 N
or, = N
Thus, we ca conclude that N is the electric force on a proton 2.5 fm from the surface of the nucleus.