Question

An electron and a proton are a distance r= 7.6 Times 10-9m apart. How much energy is required to increase their separation by a factor of six? ev

Answer 1

Potential energy between 2 charges = kq1q2/r

When r =7.6x10^9m, initial energy = (9x10^9) x (1.6x10^-19) x (-1.6x10^-19) / (7.6*10^-9)
= -3.11x10^-20 J

When r = 6 x (7.6x10^9)m, initial energy = (9x10^9) x (1.6x10^-19) x (-1.6x10^-19) / (6 x 7.6x 10^-9)
= -0.518x10^-20 J

Increase in energy = final value - initial value
= (-0.518x10^-20) - ( -3.11x10^-20)
= 2.592x10^-20 J

1eV = 1.6x10^-19J so
Increase in energy = 2.592x10^-20 / (1.6x10^-19)
= 0.162eV

Therefore, energy required to increase their separation by a factor of six is 0.162eV