Respuesta :
Answer: The magnitude of the electrostatic force between the ions is [tex]1.306\times 10^{28} N[/tex].
Explanation:
Charge on [tex]Na^+,Q_1[/tex]=+1 C
(sodium ion a has 11 protons and 10 electrons, 1 extra proton with positive charge )
Charge on [tex]Cl^-,Q_2[/tex]=-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 = [tex]0.83\times 10^{-9} m[/tex]
[tex]1 nm= 10^{-9} m[/tex]
[tex]F_E=k\frac{Q_1\times Q_2}{r^2}[/tex]
k= [tex]9\times 10^9 N m^2/C^2[/tex]
[tex]F_E=9\times 10^9 N m^2/C^2\times \frac{(+1)(-1)}{(0.83\times 10^{-9} m)^2}[/tex]
[tex]F_E=-1.306\times 10^{28} N[/tex]
[tex]|F_E|=1.306\times 10^{28} N[/tex]
The magnitude of the electrostatic force between the ions is [tex]1.306\times 10^{28} N[/tex].
Answer:
[tex]F=-3.34\times 10^{-10}\ N[/tex]
Explanation:
Given that,
Distance between charges, [tex]d=0.83\ nm=0.83\times 10^{-9}\ m[/tex]
We need to find the electrostatic force between a [tex]Na^+[/tex] and [tex]Cl^{-1}[/tex] ion.
Charge on [tex]Na^+=+1.6\times 10^{-19}\ C[/tex]
Charge on [tex]Cl^{-1}=-1.6\times 10^{-19}\ C[/tex]
The electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{+1.6\times 10^{-19}\times -1.6\times 10^{-19}}{(0.83\times 10^{-9})^2}[/tex]
[tex]F=-3.34\times 10^{-10}\ N[/tex]
So, the magnitude of electrostatic force between two ions is [tex]3.34\times 10^{-10}\ N[/tex]. Hence, this is the required solution.
