Question

(a) find the magnitude of the electrostatic force between a na+ ion and a cl− ion separated by 0.83 nm.

Answer 1

Answer: The magnitude of the electrostatic force between the ions is
`1.306* 10^(28) N`.

Step-by-step explanation:

Charge on
`Na^+,Q_1`=+1 C

(sodium ion a has 11 protons and 10 electrons, 1 extra proton with positive charge )

Charge on
`Cl^-,Q_2`=-1 C

(Chlorine ion a has 17 protons and 18 electrons, 1 extra electron with negative charge)

Distance between the ions,r = 0.83 nm =
`0.83* 10^(-9) m`

`1 nm= 10^(-9) m`

`F_E=k(Q_1* Q_2)/(r^2)`

k=
`9* 10^9 N m^2/C^2`

`F_E=9* 10^9 N m^2/C^2* ((+1)(-1))/((0.83* 10^(-9) m)^2)`

`F_E=-1.306* 10^(28) N`

`|F_E|=1.306* 10^(28) N`

The magnitude of the electrostatic force between the ions is
`1.306* 10^(28) N`.

Answer 2

`F=-3.34* 10^(-10)\ N`

Step-by-step explanation:

Given that,

Distance between charges,
`d=0.83\ nm=0.83* 10^(-9)\ m`

We need to find the electrostatic force between a
`Na^+` and
`Cl^(-1)` ion.

Charge on
`Na^+=+1.6* 10^(-19)\ C`

Charge on
`Cl^(-1)=-1.6* 10^(-19)\ C`

The electrostatic force is given by :

`F=k(q_1q_2)/(d^2)`

`F=9* 10^9* (+1.6* 10^(-19)* -1.6* 10^(-19))/((0.83* 10^(-9))^2)`

`F=-3.34* 10^(-10)\ N`

So, the magnitude of electrostatic force between two ions is
`3.34* 10^(-10)\ N`. Hence, this is the required solution.