Question

. The atomic radii of Li+ and O2 ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). (b) What is the force of repulsion at this same separation distance?

Answer 1

Answer:

Explanation:

Given that,

Charge on Li, q_1=1q

1

=1

Charge on Oxygen, q_2=-2q

2

=−2

Atomic radii of lithium, r_1=0.068\ nmr

1

=0.068 nm

Atomic radii of oxygen, r_2=0.140\ nmr

2

=0.140 nm

Distance between charges, r=0.068+0.140=0.208\times 10^{-9}\ mr=0.068+0.140=0.208×10

−9

m

(a) The force pf attraction is given by :

F=k\dfrac{q_1q_2}{r^2}F=k

r

2

q

1

q

2

F=9\times 10^9\times \dfrac{1\times (-2)}{(0.208\times 10^{-9})^2}F=9×10

9

×

(0.208×10

−9

)

2

1×(−2)

F=-4.16\times 10^{29}\ NF=−4.16×10

29

N

(b) Let F' is the force of repulsion. We know that when both charges are like, then there is force of repulsion.

F = -F'

F'=4.16\times 10^{29}\ NF

′

=4.16×10

29

N

Hence, this is the required solution.