Answer:
Force, [tex]F=7.04\times 10^{-5}\ N[/tex]
Explanation:
Given that,
Distance between 1.70 A from the plutonium nucleus, [tex]d=1.7\times 10^{-10}\ m[/tex]
The number of electron in plutonium is 94.
To find,
The magnitude of the force on an electron.
Solution,
Total charge in the plutonium nucleus is, [tex]q=94\times 1.6\times 10^{-19}=1.504\times 10^{-17}\ C[/tex]. The electric force between charges is given by :
[tex]F=\dfrac{kq^2}{d^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times (1.504\times 10^{-17})^2}{(1.7\times 10^{-10})^2}[/tex]
[tex]F=7.04\times 10^{-5}\ N[/tex]
So, the magnitude of the force on an electron is [tex]7.04\times 10^{-5}\ N[/tex]. Hence, this is the required solution.
