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

$F = \frac{ke(92e)}{d^2}$

Here,

k = Coulomb's Constant

e = Charge of electron

d = Distance between them

Replacing we have that,

$F = \frac{92(9*10^{9})(1.6*10^{-19})^2}{(7*10^{-10})^2}$

$F = 4.3258*10^{-8}N$

Therefore the force on this electron is $4.3258*10^{-8}N$

Now the magnitude of the force on an electron is

$F = \frac{ke(86e)}{(2d)^2}$

$F = \frac{92(9*10^9)(1.6*10^{-19})2}{4(7*10^{-10})^2}$

$F = 1.0814*10^{-8}N$

If the distance of the elctron from the nucleus were double the force would be $1.0814*10^{-8}N$