Question

Decide how the electrons in the following table would be listed, if they were listed in order of increasing electrostatic potential energy. That is, select "1" next to the electron with the lowest electrostatic potential energy, select "2" next to the electron with the second lowest electrostatic potential energy, and so on. electrostatic potential energy electrorn location A 3.4 nm from a cation with a charge of +2(choose one) B 3.4 nm from a cation with a charge of +3(choose one) C2.1 nm from a cation with a charge of +3 (choose one) D 4.8 nm from a cation with a charge of +2 (choose one)

Answer 1

Electrostatic potential energy is given as:

E = $kQq/r^2$ where k = 9*10⁹ N.m²/C²

Charge on electron is -1. So Q = -1.

From above equation we can see that for same radius potential energy is directly proportional to charge. So B<A (Remember Q = -1, so sign is negative)

For same charge, lesser the radius more is potential energy. so C<B and similarly A<D

From above relations we can say that C<B<A<D

For exact values we can calculate as follows,

A: k*-1*2/(3.4)² = -0.173k

B: k*-1*3/(3.4)² = -0.259k

C: k*-1*3/(2.1)² = -0.680k

D: k*-1*2/(4.8)² = -0.086k

From these values also we get the order of potential energy as D>A>B>C