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Chapter 2: Problem 52

A chemist in a galaxy far, far away performed the Millikan oil drop experiment and got the following results for the charges on various drops. Use these data to calculate the charge of the electron in zirkombs. $\begin{array}{ll}{2.56 \times 10^{-12} \text { zirkombs }} & {7.68 \times 10^{-12} \text { zirkombs }} \\ {3.84 \times 10^{-12} \text { zirkombs }} & {6.40 \times 10^{-13} \text { zirkombs }}\end{array}$

Short Answer

Expert verified
The charge of an electron in zirkombs is \(1.28 \times 10^{-10}\) zirkombs.

Step by step solution

01

Write down the charges

We have the following charges (in zirkombs) from the Millikan oil drop experiment: \(2.56 \times 10^{-12}\), \(7.68 \times 10^{-12}\), \(3.84 \times 10^{-12}\), \(6.40 \times 10^{-13}\).
02

Express charges in a form without scientific notation

To make it easier to calculate the GCD, we will multiply all charges by \(10^{12}\) to get rid of the scientific notation. Thus, the charges are now: 2560, 7680, 3840, 640.
03

Calculate the greatest common divisor (GCD) of the charges

In this step, we need to find the GCD of the four charges, which is the largest number that divides all of them without leaving a remainder. You can use the prime factorization method or the Euclidean algorithm to find the GCD. In this case, it's easier to inspect the numbers visually and quickly recognize that they are all divisible by 128.
04

Convert the GCD value back to zirkombs

Now that we have the GCD (128) of the charges expressed without scientific notation, we need to convert it back to its original form. To do this, divide the GCD by \(10^{12}\): \[\frac{128}{10^{12}} = 1.28 \times 10^{-10}\] So, the charge of an electron in zirkombs is \(1.28 \times 10^{-10}\) zirkombs.

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Most popular questions from this chapter

Consider 100.0 -g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 \(\mathrm{g}\) of carbon, and the other has 42.9 \(\mathrm{g}\) of carbon. How can these data support the law of multiple proportions if 42.9 is not a multiple of 27.2\(?\) Show that these data support the law of multiple proportions.

Reaction of 2.0 \(\mathrm{L}\) of hydrogen gas with 1.0 \(\mathrm{L}\) of oxygen gas yields 2.0 \(\mathrm{L}\) of water vapor. All gases are at the same temperature and pressure. Show how these data support the idea that oxygen gas is a diatomic molecule. Must we consider hydrogen to be a diatomic molecule to explain these results?

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