What is the magnitude of the electric force of attraction between an iron nucleus (q = +26e) and its innermost electron if the distance between them is 1.5 × 10−12 m?

Therefore the magnitude of required electric force is  292.864×10⁻⁵ N.

Explanation:

The number of proton of iron is 26.

 C is the charge of each proton.

Total charge of 26 proton is =(26×1.6×10⁻¹⁹) C

                                             =41.6×10⁻¹⁹ C

The charge of innermost electron is = - C

Coulomb's Law:

The attraction or repulsion force of two charges is

Let  be two positive charge and the distance between the charges be r

Then F∝

and  

Therefore,

Here

=   - C   and    =41.6×10⁻¹⁹ C

r = 1.5× 10⁻¹² m

K= 9.9×  10⁹C²N⁻¹m⁻²

Therefore

  N

  = -292.864×10⁻⁵ N

The mins sign shows the direction of the force.

Magnitude of electric force is only the numerical value of the force.

Therefore the magnitude of required electric force is  292.864×10⁻⁵ N.

0.11 m/s and 0.28 m/s should be correct