Question

Two ions having charges +e and – e are 4.0 × 10–10 m apart. Calculate the
potential energy of the system

Answer 1

Two ions having charges +e and -e are 4 × 10^-10 m apart.

so, $q_1$ = +e = 1.6 × 10^-19C

$q_2$ = -e = -1.6 × 10^-19C

r = 4 × 10^-10 m

using formula of potential energy,

$U=\frac{kq_1q_2}{r}$

= (9 × 10^9 × 1.6 × 10^-19 × -1.6 × 10^-19)/(4 × 10^-10)

= (-9 × 2.56 × 10^-29)/(4 × 10^-10)

= (-9 × 0.64) × 10^-19

= -5.76 × 10^-19 J

hence, potential energy of the system is -5.76 × 10^-19 J

as we know, 1eV = 1.6 × 10^-19 J

so, potential energy of the system in eV is -5.76 × 10^-19/1.6 × 10^-19 = -3.6eV