Question

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 1.5×10−12m?

Answer 1

Answer:

The magnitude of the force on an electron is 0.069 N.

Explanation:

Given that,

Distance between 1.70 A from the plutonium nucleus, $d=1.5\times 10^{-12}\ m$

The number of electron in iron nucleus is +26e.

To find,

The magnitude of the force between an iron nucleus.

Solution,

Total charge in the plutonium nucleus is, $q=26\times 1.6\times 10^{-19}=4.16\times 10^{-18}\ C$. The electric force between charges is given by :

$F=\dfrac{kq^2}{d^2}$

$F=\dfrac{9\times 10^9\times (4.16\times 10^{-18})^2}{(1.5\times 10^{-12})^2}$

F = 0.069 N

So, the magnitude of the force on an electron is 0.069 N. Hence, this is the required solution.