An oil drop has $ 6.39 \times {10^{ - 19}} $ charge. What will be the number of electrons in this drop?
A) 2
B) 4
C) 8
D) 16
Answer
565.2k+ views
Hint: In order to this question, to know the number of electrons present in one oil drop whose charge is given. Firstly, we should write the charge on one electron and then go through the formula $ q = n \times e $ .
Complete step by step answer:
The charge on one oil drop is $ 6.39 \times {10^{ - 19}} $ .
As we know, the charge on one electron is $ 1.60 \times {10^{ - 19}}C $ .
So, from the formula- $ q = n \times e $ , we can find the number of electrons in that one oil drop:
The number of electrons:
$ \because n = \dfrac{q}{e} = \dfrac{{6.39 \times {{10}^{ - 19}}}}{{1.60 \times {{10}^{ - 19}}}} = 4 $
Therefore, the number of electrons in the drop is 4.
Hence, the correct option is (B.) 4
Note:
Millikan's oil-drop experiment was performed by Robert Millikan and Harvey Fletcher in 1909. It determined a precise value for the electric charge of the electron. The electron's charge is the fundamental unit of electric charge, because all electric charges are made up of groups (or the absence of groups) of electrons.
Complete step by step answer:
The charge on one oil drop is $ 6.39 \times {10^{ - 19}} $ .
As we know, the charge on one electron is $ 1.60 \times {10^{ - 19}}C $ .
So, from the formula- $ q = n \times e $ , we can find the number of electrons in that one oil drop:
The number of electrons:
$ \because n = \dfrac{q}{e} = \dfrac{{6.39 \times {{10}^{ - 19}}}}{{1.60 \times {{10}^{ - 19}}}} = 4 $
Therefore, the number of electrons in the drop is 4.
Hence, the correct option is (B.) 4
Note:
Millikan's oil-drop experiment was performed by Robert Millikan and Harvey Fletcher in 1909. It determined a precise value for the electric charge of the electron. The electron's charge is the fundamental unit of electric charge, because all electric charges are made up of groups (or the absence of groups) of electrons.
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