Answer:
c i guess, decrease temperature
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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.
NH4NO3 b) C8H8O4 c) O3 d) C3H5(NO3)3