Question

Protons carry a charge of +1.6e-19 C. If the repulsive force between two protons in an atomic nucleus equals 80 N, how far apart are the protons, in picometers (that's trillionths of a meter)? Use "E" notation for your answer.

Answer 1

Solution)

We know, according to Coulomb's law

Force F=k*q1*q2/(r^2)

r^2 =k*q1*q2/(F)

So, distance, r = sqrt ((k*q1*q2)/(F))

= sqrt ((k*e^2)/(F))

=sqrt ((9*10^9 N.m^2/C^2*(1.60*10^-19 C)^2)/(80 N))

=1.69*10^-15 m

Distance, r = 1.69*10^-3 picometers=1.69E3

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