Question

What is the magnitude of the force on an electron at a distance of 1.70 angstrom from the plutonium nucleus?

Answer 1

Force,
`F=7.04* 10^(-5)\ N`

Step-by-step explanation:

Given that,

Distance between 1.70 A from the plutonium nucleus,
`d=1.7* 10^(-10)\ m`

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,
`q=94* 1.6* 10^(-19)=1.504* 10^(-17)\ C`. The electric force between charges is given by :

`F=(kq^2)/(d^2)`

`F=(9* 10^9* (1.504* 10^(-17))^2)/((1.7* 10^(-10))^2)`

`F=7.04* 10^(-5)\ N`

So, the magnitude of the force on an electron is
`7.04* 10^(-5)\ N`. Hence, this is the required solution.