Two protons in a nucleus of ²³⁸U are 6.0  10^-15 m apart.The electrostatic potential energy of the pair is approximately A)zero B)4.7  10^-34 J C)4.3  10^-24 J D)3.8  10^-14 J E)2.4  10⁴ J

Question

Quizplus · Accepted Answer

The electrostatic potential energy of two protons, which are positively charged, is given by 
U=k(q1q2/r)
where k = 9 x 10^9 N m^2/C^2 is the Coulomb constant, q1 and q2 are the charges of the protons (equal to +e, where e is the fundamental charge), and r is the distance between them. 
Substituting the given values, we get 
U=(9 x 10^9 N m^2/C^2) [(+e)(+e)/(6.0 x 10^-15 m)]
= 3.8 x 10^-14 J 
Therefore, the answer is D, 3.8 x 10^-14 J.

Two Protons in a Nucleus of 238U Are 6

Two Protons in a Nucleus of ²³⁸U Are 6