Find the electrostatic force between Na+ ion and a Cl- ion separated by 0.50 nm.?

Question

Can somebody explain to me how to do it... so then i could also appy to a Li+ and a Br-...thank you

Answer 1

The charge on both is only 1 e. ie, i e- is extra in Na= and less in Cl-. Now,

F=q1*q2/r^2.

Therefore, F= e^2/(0.5 * (10^-9))^2

1 nm= 10^-9 m

1 e = charge of e-.

Answer will be identical for the other question.

Answer 2

By coulomb's law

F=k*q1*q2/r^2

k=9x10^9

r=0.50 nanometers = 5.0 × 10-10 meters

q=1.603*10^-19C "charge on electron"

+q=q1= Na+ ion

-q=q2 = Cl- ion

Since the two ions have opposite charge,

the force between them is attractive.

q*-q=

(-1)*q^2

(-1)*(1.603*10^-19)^2

substituting all this into:F=k*(-1)q^2/r^2:

F=9*10^9*(-1)*(1.603*10^-19)^2/( 5.0*10^-10)^2

= -9.251 × 10^-10 Newton "answer"



Do the same with Li+ and Br-