Suppose a scientist repeats the Millikan oil-drop experiment but reports the charges on the drops using an unusual (and imaginary) unit called the warmomb (wa). The scientist obtains the following data for four of the drops: $$ \begin{array}{lc} \hline \text { Droplet } & \text { Calculated Charge (wa) } \\ \hline \text { A } & 3.84 \times 10^{-8} \\ \text {B } & 4.80 \times 10^{-8} \\ \text {C } & 2.88 \times 10^{-8} \\ \text {D } & 8.64 \times 10^{-8} \\ \hline \end{array} $$ (a) If all the droplets were the same size, which would fall most slowly through the apparatus? (b) From these data, what is the best choice for the charge of the electron in warmombs? (c) Based on your answer to part (b), how many electrons are there on each of the droplets? (d) What is the conversion factor between warmombs and coulombs?

Step by step solution

01

Identifying the droplet that falls most slowly

To determine which droplet falls most slowly, we need to find out which droplet has the smallest charge. Between droplets A, B, C, and D, we notice that droplet C has the smallest charge, given by: Droplet C: 2.88 x 10^{-8} wa

02

Finding the charge of an electron in warmombs

To find the charge of an electron in warmombs, we need to find the greatest common factor of the charges of the droplets. Let's divide the charges by the smallest charge, i.e., by 2.88 x 10^{-8} wa: Droplet A: \(\frac{3.84 \times 10^{-8}}{2.88 \times 10^{-8}} = 1.333\) Droplet B: \(\frac{4.80 \times 10^{-8}}{2.88 \times 10^{-8}} = 1.67\) Droplet C: \(\frac{2.88 \times 10^{-8}}{2.88 \times 10^{-8}} = 1\) Droplet D: \(\frac{8.64 \times 10^{-8}}{2.88 \times 10^{-8}} = 3\) From these ratios, we can see that the common factor is 1. The charge of the electron in warmombs is thus 2.88 x 10^{-8} wa.

03

Finding the number of electrons on each droplet

As we have already found the ratios in Step 2, we can use these to determine the number of electrons on each droplet: Droplet A: 1.333 electrons - \(1 \frac{1}{3}\) Droplet B: 1.67 electrons - \(1 \frac{2}{3}\) Droplet C: 1 electron Droplet D: 3 electrons

04

Finding the conversion factor between warmombs and coulombs

We know the charge of an electron in coulombs, which is approximately 1.6 x 10^{-19} C. Using the charge of an electron in warmombs, we can find the conversion factor: Conversion factor = (charge of electron in coulombs) / (charge of electron in warmombs) Conversion factor = \(\frac{1.6 \times 10^{-19}}{2.88 \times 10^{-8}}\ wa^{-1}\) Calculating the conversion factor, we get: Conversion factor ≈ 5.56 x 10^{-12} C/wa

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