How many electrons make up a charge of 20\[\mu C\]?
\[\begin{align}
& \text{A) 1}\text{.25}\times \text{1}{{\text{0}}^{14}} \\
& \text{B) 2}\text{.23}\times {{10}^{14}} \\
& \text{C) 3}\text{.25}\times {{10}^{14}} \\
& \text{D) 5}\text{.25}\times {{10}^{14}} \\
\end{align}\]
Answer
590.1k+ views
Hint: We need to understand the relation between the charge present in a system and the number of electrons that can constitute the given amount of charge. We can easily solve this problem using these relations to find the number of charges present.
Complete answer:
We know that the charges involved in a system especially in a conductor is either the presence of electrons or their absence such that positive ions are present. The magnitude of charge for both the cases will be the same, given by ‘e’ as \[1.6\times {{10}^{-19}}C\]. The positive ions can represent a positive charge whereas the negative ions or the electrons represent a negative charge.
The charge is a quantised quantity. The value of the charge present in a system can be only the multiples of the single charge ‘e’. There can never be a fractional value for the charge.
In the present case, we are given a charge of 20\[\mu C\]. We can find the number of electrons present or may be absent in this situation by simple mathematical calculation. We can apply the idea of the charge quantisation to solve this. We can divide the given charge by the charge of a single electron to get the number of electrons involved as –
\[\begin{align}
& \text{Number of electrons = }\dfrac{\text{Total charge}}{\text{Charge of one electron}} \\
& \Rightarrow n=\dfrac{20\times {{10}^{-6}}C}{1.6\times {{10}^{-19}}C} \\
& \therefore n=1.25\times {{10}^{14}}electrons \\
\end{align}\]
The number of electrons present in the system of total charge 20\[\mu C\] is \[\text{1}\text{.25}\times \text{1}{{\text{0}}^{14}}\].
From this information we get the verification that the charge is a quantised quantity which will be the whole number always.
The correct answer is option A.
Note:
The quantisation of electric charge is utilised in every aspect where the charges come into play. Even in chemistry, the balancing of chemical equations has the aim of making the electrons involved to be even on both sides so that it doesn’t violate quantisation.
Complete answer:
We know that the charges involved in a system especially in a conductor is either the presence of electrons or their absence such that positive ions are present. The magnitude of charge for both the cases will be the same, given by ‘e’ as \[1.6\times {{10}^{-19}}C\]. The positive ions can represent a positive charge whereas the negative ions or the electrons represent a negative charge.
The charge is a quantised quantity. The value of the charge present in a system can be only the multiples of the single charge ‘e’. There can never be a fractional value for the charge.
In the present case, we are given a charge of 20\[\mu C\]. We can find the number of electrons present or may be absent in this situation by simple mathematical calculation. We can apply the idea of the charge quantisation to solve this. We can divide the given charge by the charge of a single electron to get the number of electrons involved as –
\[\begin{align}
& \text{Number of electrons = }\dfrac{\text{Total charge}}{\text{Charge of one electron}} \\
& \Rightarrow n=\dfrac{20\times {{10}^{-6}}C}{1.6\times {{10}^{-19}}C} \\
& \therefore n=1.25\times {{10}^{14}}electrons \\
\end{align}\]
The number of electrons present in the system of total charge 20\[\mu C\] is \[\text{1}\text{.25}\times \text{1}{{\text{0}}^{14}}\].
From this information we get the verification that the charge is a quantised quantity which will be the whole number always.
The correct answer is option A.
Note:
The quantisation of electric charge is utilised in every aspect where the charges come into play. Even in chemistry, the balancing of chemical equations has the aim of making the electrons involved to be even on both sides so that it doesn’t violate quantisation.
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