 # A chemist in an imaginary universe, where electrons have a different c petrusrexcs 2021-12-15 Answered
A chemist in an imaginary universe, where electrons have a different charge than they do in our universe, performs the Millikan oil drop experiment to measure the electrons
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Step 1
The Millikan experiment consisted of finding the minimum possible charge by obtaining the smallest drop possible. He then inferred that the charge of the electron was due to the fact that it was not possible to find a lower charge.

Step 2
We must find a number that when we divide in the four drop # it would give us a quotient that is a whole number. We can do this by doing trial and error.
By trial and error, we found out that the number is $-2.3×{10}^{-19}C$
DROP A: $\frac{-6.9×{10}^{-19}C}{-2.3×{10}^{-19}C}=3$
DROP B: $\frac{-9.2×{10}^{-19}C}{-2.3×{10}^{-19}C}=4$
DROP C: $\frac{-11.5×{10}^{-19}C}{-2.3×{10}^{-19}C}=5$
DROP D: $\frac{-4.6×{10}^{-19}C}{-2.3×{10}^{-19}C}=2$
Answer: $-2.3×{10}^{-19}C$
###### Not exactly what you’re looking for? Kindlein6h
Millikan experiment consisted in finding the minimum possible charge by obtaining the tiniest drop possible.
Then, he inferred that was the charge of the electron, given that it was not possible to find a lower charge.
Given these recorded data:
$-6.9\cdot {10}^{-19}$
$-9.2\cdot {10}^{-19}$
$-11.5\cdot {10}^{-19}$
$-4.6\cdot {10}^{-19}$
You must find a number that divide the four data and yields a whole number.
This number is $-2.3\cdot {10}^{-19}$.
In that way you find:
$-6.9\cdot \frac{{10}^{-19}}{-2.3}\cdot {10}^{-19}=3⇒3$ electrons
$-9.2\cdot \frac{{10}^{-19}}{-2.3}\cdot {10}^{-19}=4⇒4$ electrons
$-11.5\cdot \frac{{10}^{-19}}{-2.3}\cdot {10}^{-19}=5⇒5$ electrons
$-4.6\cdot \frac{{10}^{-19}}{-2.3}\cdot {10}^{-19}=2⇒2$ electrons
And you can conclude that the charge of the electron is $-2.3\cdot {10}^{-19}$
Answer: $-2.3\cdot {10}^{-19}$
###### Not exactly what you’re looking for? nick1337

Step 1
The idea is the number of electrons is a whole number. So to find a charge that a “whole multiple” will satisfy all the charges on the drops.
Step 2
Let X = the charge….A, B, C and D are whole.
$AX=-5.7×{10}^{-19}C.$
$BX=-7.6×{10}^{-19}C.$
$CX=-9.5×{10}^{-19}C.$
$DX=-3.8×{10}^{-19}C.$
Difference between BX and CX is $1.{910}^{-19}$.  That’s the smallest difference.
Step 3
Therefore,
$3×\left(-1.{910}^{-19}\right)=-5.7×{10}^{-19}C$
$4×\left(-1.{910}^{-19}\right)=-7.6×{10}^{-19}C$
$5×\left(-1.{910}^{-19}\right)=-9.5×{10}^{-19}C$
$2×\left(-1.{910}^{-19}\right)=-3.8×{10}^{-19}C.$
So the charge of imaginary electron is $-1.{910}^{-19}C$