Deducing the Elementary Charge from Milikan's Oil Drop Experiments

In summary, the researcher Milikan conducted experiments where he observed various charges on a single oil drop. From this data, he attempted to determine the value of the elementary charge. Using a graphical method was not an option and the answer needed to be in 5 significant figures. After subtracting each of the given charges, the researcher found two distinct sets of numbers and took the average of each set. By dividing the average of the two sets, the researcher obtained a value of approximately 2. This led to a smaller number, 1.62 x 10^-19, which was found to be a common factor among all the given charges. From this, the researcher was able to deduce that each charge was a multiple of the elementary
  • #1
imatreyu
82
0

Homework Statement



In an early set of experiments, Milikan observed that the following measure charges, among others, appeared at different times on a single oil drop. What value of elementary charge can be deduced from these data?

1) 6.563 x 10^-19 C
2) 8.204 x 10^-19 C
3) 11.50 x 10^-19 C
4) 13.13 x 10^-19 C
5) 16.48 x 10^-19 C
6) 18.08 x 10^-19 C
7) 19.71 x 10^-19 C
8) 22.89 x 10^-19 C
9) 26.13 x 10^-19 C


Homework Equations



I have no idea. :( Doing this graphically is NOT an option. My answer needs to be 5 significant figures.


The Attempt at a Solution


I subtract each of the answers in the manner... B-A, C-B and so on. From this, I get two distinct sets of numbers. I take the average of each set of data and get 1.62525 and 3.2665.Then I divide the 3.2665 by 1.62525 to get approximately 2. Then I divide 3.2665 by 2 and get 1.63325. . .

I have no idea why that works! What am I doing when I do that? It was explained to me by a friend, but I don't understand why that works or what that is doing. . .
 
Physics news on Phys.org
  • #2


Each charge is a multiply of elementary charge.

That means difference is also always a multiply of elementary charge.

Imagine you are given small bags with identical marbles inside. You can only weight the bags and you are asked to find mass of the single marble. Mass of each bag is a multiply of mass of a single marble, differences are also multiplies of a single marble mass. Lowest possible mass - and lowest possible difference - is that of a single marble.
 
  • #3


Thank you so much for your explanation using the marbles-- it's very informative. Therefore-- yes, I understand that every charge is a multiple of the elementary charge.

However, I'm not sure how that applies to what I wrote above. . .still. . .
 
  • #4


imatreyu said:
Thank you so much for your explanation using the marbles-- it's very informative. Therefore-- yes, I understand that every charge is a multiple of the elementary charge.

However, I'm not sure how that is relevant, as I don't know any information regarding mass.

You do not need to know the mass of the particles in order to find the elementary charge.
 
  • #5


My initial instinct was to use the GCF of all of the values; however none of them are integers and I imagine such an endeavour would be really challenging and hard. . . and maybe not even correct or useful. Ugh. . .
 
  • #6


Using your numbers what is the smallest charge you detect? Are the other measured charges integral amounts of this smallest charge?Could this smallest charge be the elementary charge?
 
  • #7


Well the smallest charge is 6.563 x 10^-19 C. (Right?)
This is obviously not the elementary charge, though. . .
 
  • #8


P.S. It would help if someone could read over my Attempt at a Solution and explain why that's correct/ what that does/ why I need to do that (if that's correct). I'm so sorry about this. I'm just horribly confused here. .
 
  • #9


imatreyu said:
Well the smallest charge is 6.563 x 10^-19 C. (Right?)
This is obviously not the elementary charge, though. . .

6.563? In your first post you identified a charge which was one quarter of this.
 
  • #10


6.563? In your first post you identified a charge which was one quarter of this.

Really? I'm horrible at math; but isn't it that:

6.563 x 10^-19 C < 8.204 x 10^-19 C
6.563 x 10^-19 C < 11.50 x 10^-19 C
6.563 x 10^-19 C < 13.13 x 10^-19 C
6.563 x 10^-19 C <16.48 x 10^-19 C
6.563 x 10^-19 C < 18.08 x 10^-19 C
6.563 x 10^-19 C < 19.71 x 10^-19 C
6.563 x 10^-19 C < 22.89 x 10^-19 C
6.563 x 10^-19 C < 26.13 x 10^-19 C

?
 
  • #11


As I said above you have already identified a number which is smaller than 6.563.What is this number? Does it divide into all of your results an integral number of times? Remember that there are experimental errors and that the numbers are unlikely to be exact integers.
 
  • #12


Ahh.. . I'm sorry. I just don't know what number you're referring to, and I still don't understand. Could you identify what number you're talking about? I don't see anything smaller than 6.563 x 10^-19 ... (Forgive me if this is just totally stupid, it's 2 AM where I am...and I'm a little dead.)

I really need an explanation of how the procedure I described in "Attempt at the Solution" works, if it's correct or if it's totally wrong. . . I would like to understand it. .
 
  • #13


You referred to the number in your first post with your attempt at a solution.The smallest number you found was 1.62 times ten to the minus nineteen approximately.Stick with it a little longer imatreyu and see what happens when you divide all of your results by this smallest number
 
  • #14


You have calculated series of differences, right?

Borek said:
difference is also always a multiply of elementary charge.

Dadface said:
The smallest number you found was 1.62 times ten to the minus nineteen

Do you see a link now?
 

Related to Deducing the Elementary Charge from Milikan's Oil Drop Experiments

1. What is Milikan's Oil Drop Experiment?

Milikan's Oil Drop Experiment was a scientific experiment conducted by Robert A. Milikan in 1909 to determine the elementary charge of an electron, which is the smallest unit of electric charge. The experiment involved observing the motion of tiny oil droplets in an electric field to measure the magnitude of the charge on each droplet.

2. How did Milikan's experiment help determine the elementary charge?

Milikan's experiment involved measuring the motion of oil droplets in an electric field. By adjusting the strength of the electric field and measuring the motion of the droplets, Milikan was able to calculate the charge on each droplet. He then noticed that the charges on the droplets were all multiples of a certain value, which he determined to be the elementary charge of an electron.

3. What was the value of the elementary charge determined by Milikan's experiment?

The value of the elementary charge determined by Milikan's experiment was 1.602 x 10^-19 Coulombs. This value is still accepted by scientists today and is an important constant in the field of physics.

4. How accurate was Milikan's determination of the elementary charge?

Milikan's determination of the elementary charge was accurate to within 1% of the currently accepted value. This level of accuracy was groundbreaking at the time and helped solidify the validity of his experiment and results.

5. Why is determining the elementary charge important?

Determining the elementary charge is important because it is a fundamental constant in physics and is used in many equations and theories. It also helped to establish the concept of the atom and the existence of subatomic particles such as electrons. Additionally, the accurate determination of the elementary charge has led to advancements in technology and has allowed for a better understanding of the behavior of electricity and matter.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
5K
Back
Top