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 = $\frac{6.0}{2}$

= 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

= $5.5 \times 10^{-15} m$ (as 1 fm = $10^{-15}$ )

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

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}$

Putting the given values into the above formula as follows.

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}$

= $(9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}$

= $(9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}$

= 411.2 N

or, = $4.1 \times 10^{2}$ N

Thus, we ca conclude that $4.1 \times 10^{2}$ N is the electric force on a proton 2.5 fm from the surface of the nucleus.

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4 years ago