What is the magnitude of electric force of attraction between an iron nucleus (q = 26e) and its innermost electron if the distane between them is 1.5 x 10^-12 m?



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Posted on (Answer #1)

This is an example of an electrostatic force that obeys Coulomb's Law. The size of the force will be proportional to the product of the charge of the nucleus and the innermost electron.  It will also be inversely proportional to the square of the distance or separation.

The iron nucleus has 26 protons, each of which has a charge of 1.6x10-19C = 41.6x10^-19C; the innermost electron has a charge of =-1.6x10^-19C

Coulomb's Law predicts the force will be

Fe = kQpQe/d^2

where k is coulomb's constant, Qp is the charge of the nucleus, Qe is the charge of the electron, and d is the average separation.

Fe = 9.9x10^9(41.6x10^-19)(1.6x10^-19)/(1.5x10-12)^2

Fe =293x10^-5N = 2.93x10^-3N

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