The electrostatics force of repulsion between two positively charged ions carrying equal charge is `3.7xx10^(-9)N` when these are separated by a distance of `5Å`. How many electrons are missing from each ion?

Answer

2

Explanation

Here, `q_(1) = q_(2) = q = ?`
`F = 3.7xx10^(-9) N, r = 5 Å = 5xx10^(-10)m`
`n = ?`
As `F = (k q_(1) q_(2))/(r^(2)) = (kq^(2))/(r^(2))`
`:. q = sqrt(r^(2) F//k)`
`= sqrt((1)/(9xx10^(9)) (5xx10^(-10))^(2) xx3.7xx10^(-9))`
`= sqrt((25xx3.7)/(9)) xx10^(-19) C = 3.2xx10^(-19) C`
From `q =n e`
`n = (q)/(e) = (3.2xx10^(-19))/(1.6xx10^(-19)) = 2`