Question

The magnitude of the electrostatic force between two identical ions that are separated by a
distance of 5.0x10-10 m is 3.7x10-9 N. (a) What is the charge of each ion? (b) How many
electrons are "missing" from each ion (thus giving the ion its charge imbalance)?

Answer 1

charge on each ion = 3.2 × 10^-19 C

and 2 electrons are missing from each ion.

Let magnitude of charge on each ion is Q.

given, magnitude of electrostatic force, F = 3.7 × 10^-9 N

seperation between ions , r = 5 × 10^-10 m

using Coulomb's formula,

F = KQ²/r²

⇒Q² = Fr²/K

= {3.7 × 10^-9 × (5 × 10^-10)²}/(9 × 10^9)

= {3.7 × 25 × 10^-29}/{9 × 10^9}

= {3.7 × 25/9} × 10^-38

Q = √{3.7 × 25/9 × 10^-38}

= 3.2 × 10^-19 C

hence, charge on each ion = 3.2 × 10^-2 C

we know charge on each electron = 1.6 × 10^-19 C

so, number of electrons are missing from each ion = charge on each ion/charge on electron

= 3.2 × 10^-19/1.6 × 10^-19

= 2

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Answer 2

$\huge\bold\purple{Answer:-}$

(a) Charge on each ion: q=2e

(b) Two electrons are missing

Explanation:

(a) We have,

r = 5.0*10^{-10}\\ \\F_q = 3.7*10^{-9} \\ \\ F_q=\frac{kq^2}{r^2}\\ \\=>3.7*10^{-9}=\frac{9*10^{9}*q^2}{(5*10^{-10})^2}\\ \\=>q=\pm 3.2*10^{-19} C=2e

(b) Here, Charge can be both negative or positive but we can talking about missing electrons ,then we consider charge to be positive.

Hence,

Ion Becomes A^{2+}

This means Charge loses 2 electrons to obtain stable configuration.

That is, 2 electrons are missing which gave rise to charge imbalance.