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Question: Two positive ions, each carrying a charge \(q\) are separated by a distance \(d\). If \(F\) is the f...

Two positive ions, each carrying a charge qq are separated by a distance dd. If FF is the force of repulsion between the ions, the number of electrons missing from each ion will be (c being the charge of an electron).
A. Fx3×103×e\dfrac{{\sqrt F x}}{{3 \times {{10}^3} \times e}}
B. x×109F9e2\dfrac{{x \times {{10}^9}\sqrt F }}{{9{e^2}}}
C. x×104F9e2\dfrac{{x \times {{10}^{ - 4}}\sqrt F }}{{9{e^2}}}
D. x×105Fe29\dfrac{{x \times {{10}^5}\sqrt {F{e^2}} }}{9}

Explanation

Solution

we will use the force of repulsion using coulomb’s law, in which the distance between the charges is taken as xx. Also, we know that every ion charge is an integral multiple of an electron, therefore, q=neq = ne. Here, nn is the number of electrons. We will substitute this value of charge in the formula of force of repulsion to get the desired answer.

Formula used:
The formula of force of repulsion using coulomb’s law is given below
F=kq1q2x2F = k\dfrac{{{q_1}{q_2}}}{{{x^2}}}
Here, FF is the force of repulsion, kk is constant of proportionality, q1{q_1} and q2{q_2} are the charges and xx is the distance between the charges.
Also, we know that the charge is an integral multiple of an electron which is given below
q=neq = ne
Here, qq is the charge, nn is the number of electrons and ee is the electron in the charge.

Complete step by step answer:
We know that, according to Coulomb’s law, the force of repulsion is given by
F=kq1q2x2F = k\dfrac{{{q_1}{q_2}}}{{{x^2}}}
Also, we know that the charge is an integral multiple of an electron which is given below
q=neq = ne
Therefore, value of both the charges is given below
q1=ne{q_1} = ne
q2=ne{q_2} = ne
Now, using both the equations to calculate the number of electrons missing from the ion, as show below
F=k(ne)(ne)x2F = k\dfrac{{\left( {ne} \right)\left( {ne} \right)}}{{{x^2}}}
F=k(ne)2x2\Rightarrow \,F = k\dfrac{{{{\left( {ne} \right)}^2}}}{{{x^2}}}
F=9×109(ne)2x2\Rightarrow \,F = 9 \times {10^9}\dfrac{{{{\left( {ne} \right)}^2}}}{{{x^2}}}
Fx2=9×109(ne)2\Rightarrow \,F{x^2} = 9 \times {10^9}{\left( {ne} \right)^2}
(ne)2=Fx29×109\Rightarrow \,{\left( {ne} \right)^2} = \dfrac{{F{x^2}}}{{9 \times {{10}^9}}}
n2=Fx29×109×e2\Rightarrow \,{n^2} = \dfrac{{F{x^2}}}{{9 \times {{10}^9} \times {e^2}}}
n=Fx3×103×e\therefore \,n = \dfrac{{\sqrt F x}}{{3 \times {{10}^3} \times e}}
Therefore, the number of electrons missing from each ion will be Fx3×103×e\dfrac{{\sqrt F x}}{{3 \times {{10}^3} \times e}}.

Hence, option A is the correct option.

Note: Here, the value of constant of proportionality is k=14πε0=9×109k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}. Here, we have not taken 4πε04\pi {\varepsilon _0} because it will not cancel out in the solution. Always remember that any charge in this world can be represented as an integral multiple of charge of an electron.