The electrostatic force of repulsion between two equal positively charged ions is $ 3.7 \times {10^{ - 9}}N $ , when they are separated by a distance of $ 5\dot A $ . What is the number of electrons which are missing from each ion?
Answer
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Hint
Convert the separation of two charged ions to metre. Then, use the Coulomb’s Law which is expressed by-
$ F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} $
where, $ {q_1} $ is the charge of first charged ions and $ {q_2} $ is the second charged ion
$ r $ 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 = ne $
where, $ n $ is the number of electrons and $ e $ is the charge of an electron.
Complete step by step answer
Let the force of repulsion between two charged ions be $ F $ , separation between them be $ r $ and charge on the first and second ions be $ {q_1} $ and $ {q_2} $ respectively.
According to the question, it is given that
$ F = 3.7 \times {10^{ - 9}}N $
$ r = 5\dot A = 5 \times {10^{ - 10}}m $
The charge on both the ions is equal to each other then,
$ {q_1} = {q_2} = q $
Using the expression of Coulomb’s Law which can be expressed as-
$ F = \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 \times {10^{ - 9}} = 9 \times {10^9} \times \dfrac{{{q_1}{q_2}}}{{{{(5 \times {{10}^{ - 10}})}^2}}} $
$ \because {q_1} = {q_2} = q $
$ \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
$ {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 $ n $ be the number of electrons missing and $ e $ be the charge of electrons. Use the expression-
$ q = ne $
Putting values in their respective places
$ 3.2 \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. $
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). $
Convert the separation of two charged ions to metre. Then, use the Coulomb’s Law which is expressed by-
$ F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} $
where, $ {q_1} $ is the charge of first charged ions and $ {q_2} $ is the second charged ion
$ r $ 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 = ne $
where, $ n $ is the number of electrons and $ e $ is the charge of an electron.
Complete step by step answer
Let the force of repulsion between two charged ions be $ F $ , separation between them be $ r $ and charge on the first and second ions be $ {q_1} $ and $ {q_2} $ respectively.
According to the question, it is given that
$ F = 3.7 \times {10^{ - 9}}N $
$ r = 5\dot A = 5 \times {10^{ - 10}}m $
The charge on both the ions is equal to each other then,
$ {q_1} = {q_2} = q $
Using the expression of Coulomb’s Law which can be expressed as-
$ F = \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 \times {10^{ - 9}} = 9 \times {10^9} \times \dfrac{{{q_1}{q_2}}}{{{{(5 \times {{10}^{ - 10}})}^2}}} $
$ \because {q_1} = {q_2} = q $
$ \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
$ {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 $ n $ be the number of electrons missing and $ e $ be the charge of electrons. Use the expression-
$ q = ne $
Putting values in their respective places
$ 3.2 \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. $
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). $
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