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3 years ago

6

he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee

n 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

1 answer:

Crazy boy [7]3 years ago

6 0

Answer:

a

$F = -1.07 *10^{-8} \ N$

b

$F_r = 1.07 *10^{-8} \ N$

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

$F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}$

Here k is known as the proportionality constant with value $k = 2.31 * 10^ {-28} J \cdot m$

substituting -2 for $Q_{O}$ i.e the charge on oxygen , +1 for $Q_{Li}$ i.e the charge on Lithium and $[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}$ for r

So

$F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }$

$F = -1.07 *10^{-8} \ N$

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

$F_r = 1.07 *10^{-8} \ N$

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