Question

What is the magnitude of the force on an electron at a distance of 7.00 angstrom from the uranium nucleus?What would the magnitude of the force be if the distance of the electron from the nucleus were doubled?

Answer 1

To solve this problem we will use the concepts given in Coulomb's laws where the electrostatic force of a body is defined. In this case on an electron in a Uranium atom. Recall that in a neutral state the uranium atom has 92 electrons. The magnitude of the force on an electron is

`F = (ke(92e))/(d^2)`

Here,

k = Coulomb's Constant

e = Charge of electron

d = Distance between them

Replacing we have that,

`F = (92(9*10^(9))(1.6*10^(-19))^2)/((7*10^(-10))^2)`

`F = 4.3258*10^(-8)N`

Therefore the force on this electron is
`4.3258*10^(-8)N`

Now the magnitude of the force on an electron is

`F = (ke(86e))/((2d)^2)`

`F = (92(9*10^9)(1.6*10^(-19))2)/(4(7*10^(-10))^2)`

`F = 1.0814*10^(-8)N`

If the distance of the elctron from the nucleus were double the force would be
`1.0814*10^(-8)N`