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Question: The value of charge on the oil droplets experimentally observed were \( - 1.6 \times {10^{ - 19}}\) ...

The value of charge on the oil droplets experimentally observed were −1.6×10−19 - 1.6 \times {10^{ - 19}} and −4×10−19 - 4 \times {10^{ - 19}} coulomb. The value of the electronic charge, indicated by these results is:

A. 1.6×10−19 B. −2.4×10−19 C. −4×10−19 D. −0.8×10−19  A.{\text{ }}1.6 \times {10^{ - 19}} \\\ B.{\text{ }} - 2.4 \times {10^{ - 19}} \\\ C.{\text{ }} - 4 \times {10^{ - 19}} \\\ D.{\text{ }} - 0.8 \times {10^{ - 19}} \\\
Explanation

Solution

In order to solve the given problem we will first understand the relation between the charge on the oil droplet of any sort of charge and the value of electronic charge. We will use the basic formula relating the two of them. On the basis of the charge given in both the cases by the method of estimation and rounding we will find the probable number of electrons in each of the cases and on the basis of the number of electrons given we will find the electronic charge.

Complete step by step answer:
We know that the term charge on the oil droplet or object means the total charge possessed by all of the electrons which is represented as qq . But the term electronic charge means the charge possessed by each electron or one electron which is normally represented as ee .
The relation between these two terms is given by:
q=neq = ne
Here the term “n” denotes the number of electrons. For the given two cases the two cases are having the difference in the charge as −1.6×10−19 - 1.6 \times {10^{ - 19}} and −4×10−19 - 4 \times {10^{ - 19}} .
This difference must have arisen due to the difference in the number of electrons in both the cases.
Let for the first case the number of charges is mm .
Let for the second case the number of charges is m′m' .
Let us modify the formula above as follows:
∵q=ne ⇒e=qn  \because q = ne \\\ \Rightarrow e = \dfrac{q}{n} \\\
As, now we know that for both the cases the electronic charge is same, so we have:
∵e=qn ⇒e=qn=−1.6×10−19m=−4×10−19m′  \because e = \dfrac{q}{n} \\\ \Rightarrow e = \dfrac{q}{n} = \dfrac{{ - 1.6 \times {{10}^{ - 19}}}}{m} = \dfrac{{ - 4 \times {{10}^{ - 19}}}}{{m'}} \\\
Also we know that the number of electrons can only be integer.
So we have the integer value as follows when
m=2,m′=5m = 2,m' = 5
So, for the first case we have 2 electrons and for the second case we have five electrons.
Now let us substitute the value in the above formula to find the value of electric charge “q”.
∵e=qn=−1.6×10−19m=−4×10−19m′ ⇒e=qn=−1.6×10−192=−4×10−195 ⇒e=−0.8×10−19  \because e = \dfrac{q}{n} = \dfrac{{ - 1.6 \times {{10}^{ - 19}}}}{m} = \dfrac{{ - 4 \times {{10}^{ - 19}}}}{{m'}} \\\ \Rightarrow e = \dfrac{q}{n} = \dfrac{{ - 1.6 \times {{10}^{ - 19}}}}{2} = \dfrac{{ - 4 \times {{10}^{ - 19}}}}{5} \\\ \Rightarrow e = - 0.8 \times {10^{ - 19}} \\\
Hence, the value of the electronic charge, indicated by these results is −0.8×10−19 - 0.8 \times {10^{ - 19}} .
So, the correct answer is “Option D”.

Note: In order to solve the given problems students must remember the basic relation between the total charge, number of charged particles and the electronic charge. Mostly the formula is used to find the number of electrons in the case of flow of electrons as current but this formula is valid for all types of charged particles and only difference is that the charge of the particle changes every time for different charged particles.