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Question: The electrostatic force of repulsion between two equal positively charged ions is \( 3.7 \times {10^...

The electrostatic force of repulsion between two equal positively charged ions is 3.7×109N3.7 \times {10^{ - 9}}N , when they are separated by a distance of 5A˙5\dot A . What is the number of electrons which are missing from each ion?

Explanation

Solution

Hint
Convert the separation of two charged ions to metre. Then, use the Coulomb’s Law which is expressed by-
F=14πε0q1q2r2F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}
where, q1{q_1} is the charge of first charged ions and q2{q_2} is the second charged ion
rr is the separation between charged ions
It is given that charge of ions is equal
Find the charge and then calculate number of electrons by using formula- q=neq = ne
where, nn is the number of electrons and ee is the charge of an electron.

Complete step by step answer
Let the force of repulsion between two charged ions be FF , separation between them be rr and charge on the first and second ions be q1{q_1} and q2{q_2} respectively.
According to the question, it is given that
F=3.7×109NF = 3.7 \times {10^{ - 9}}N
r=5A˙=5×1010mr = 5\dot A = 5 \times {10^{ - 10}}m
The charge on both the ions is equal to each other then,
q1=q2=q{q_1} = {q_2} = q
Using the expression of Coulomb’s Law which can be expressed as-
F=14πε0q1q2r2F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}
Now, putting the values given in question in their respective places
3.7×109=9×109×q1q2(5×1010)23.7 \times {10^{ - 9}} = 9 \times {10^9} \times \dfrac{{{q_1}{q_2}}}{{{{(5 \times {{10}^{ - 10}})}^2}}}
q1=q2=q\because {q_1} = {q_2} = q
q2=19×109×3.7×109×(5×1010)2q2=92.5×10299×109\therefore {q^2} = \dfrac{1}{{9 \times {{10}^9}}} \times 3.7 \times {10^{ - 9}} \times {(5 \times {10^{ - 10}})^2} {q^2} = \dfrac{{92.5 \times {{10}^{ - 29}}}}{{9 \times {{10}^9}}}
Doing further calculations, we get
q2=10.28×1038q=3.2×1019C{q^2} = 10.28 \times {10^{ - 38}} q = 3.2 \times {10^{ - 19}}C
Now, we got the value of charge for two charged ions
To calculate the number of electrons missing from each ion let nn be the number of electrons missing and ee be the charge of electrons. Use the expression-
q=neq = ne
Putting values in their respective places
3.2×1019=n×1.6×1019n=3.2×10191.6×1019n=23.2 \times {10^{ - 19}} = n \times 1.6 \times {10^{ - 19}} n = \dfrac{{3.2 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} \therefore n = 2
The number of missing electrons from each ion is 2.2.

Note
Coulomb’s Law states that force between two charged particles is directly proportional to product of their charges and is inversely proportional to separation between them. Its S.I unit is N(Newton).N(Newton).