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Suppose that one of millikan's oil drops has a charge of −4.8×10−19

c. how many excess electrons does the drop contain?.

2 Answers

4 votes

The best and correct option is c

User JonH
by
8.1k points
1 vote

Answer: The number of excess electrons that the drop contains are 3.

Step-by-step explanation:

We are given:

Charge on millikan's oil drops =
4.8* 10^(-19)C

To calculate the number of excess electrons, we divide the charge on millikan's oil drop by the change on 1 electron.


\text{Excess electrons}=\frac{\text{Charge on millikan's oil drop}}{\text{Charge on 1 electron}}

We know that:

Charge on 1 electron =
1.60* 10^(-19)C

Putting values in above equation, we get:


\text{Excess electrons}=(4.8* 10^(-19)C)/(1.60* 10^(-19)C)\\\\\text{Excess electrons}=3

Hence, the number of excess electrons that the drop contains are 3.

User Joostschouten
by
7.5k points
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