Question

A uranium and iron atom reside a distance R = 58.70 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionized. Calculate the distance r from the uranium atom necessary for an electron to reside in equilibrium. Ignore the insignificant gravitational attraction between the particles.

What is the magnitude of the force on the electron from the uranium ion?

Answer 1

Distance R = 58.7 x 10 ^-9 m

Net charge of Uranium q = 1.6 x 10 ^-19 C

Net charge of iron q ' = 2 x 1.6 x 10 ^-19 C      Since it is doubly ionized.

Charge of electron e = 1.6 x 10 ^-19 C

The electron is in equilibrium ,when the force on electron due to Uranium is equal to force on electron due to iron.

i.e., the force on electron due to Uranium = force on electron due to iron

                     Kq e / x ² = K q ' e / ( R-x) ²

Where x = distance of the electron from Uranium

                    q / x ² = q ' / ( R-x) ²

                   q / q ' = x ² / ( R-x) ²

       x ² / ( R-x) ² = 1 / 2

        x ²   = 0.5 ( R-x) ²

            x = (R-x )

              = 0.7071 ( R -x )

i.e., x = 0.7071 ( R - x)

       x = 0.7071 R - 0.7071 x

1.7071 x = 0.7071 R

          x = 0.414 R

            = 24.31 nm

Or x = - 0.7071 ( R - x)

       x = - 0.7071 R + 0.7071 x

0.2929 x = 0.7071 R

          x = 2.414 R

             = 141.7 nm

Therefore the distance x from the uranium atom necessary for an electron to reside in equilibrium = 24.31 nm

the magnitude of the force on the electron from the uranium ion F = Kq e / x ²

Where   x = 24.31 x 10 ^-9 m

            K = 8.99 x 10 ^-9 N m ² / C ²

Substittue values you get F = 3.89 x 10 ^-13 N