Problem 42 - Chemistry

Question

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. \(2.56 \times 10^{-12}\) zirkombs \(\quad 7.68 \times 10^{-12}\) zirkombs \(3.84 \times 10^{-12}\) zirkombs \(\quad 6.40 \times 10^{-13}\) zirkombs

Step-by-Step Solution

Answer

The charge of the electron in zirkombs is \(6.40 \times 10^{-13}\) zirkombs.

Step 1: Write down all the charges of the oil drops in standard form

Here are the given charges on various oil drops in zirkombs: - \(2.56 \times 10^{-12}\) zirkombs - \(7.68 \times 10^{-12}\) zirkombs - \(3.84 \times 10^{-12}\) zirkombs - \(6.40 \times 10^{-13}\) zirkombs

Step 2: Convert the numbers to their integer values

To find the greatest common divisor of these charges, let's first express them as integers by multiplying them by \(10^{12}\): - \(2.56 \times 10^{-12} \times 10^{12} = 2560\) - \(7.68 \times 10^{-12} \times 10^{12} = 7680\) - \(3.84 \times 10^{-12} \times 10^{12} = 3840\) - \(6.40 \times 10^{-13} \times 10^{12} = 640\) Now the charges are: 2560, 7680, 3840, and 640.

Step 3: Find the greatest common divisor

To find the charge of the electron, we need to find the greatest common divisor (GCD) of these integer charges. The GCD of 2560, 7680, 3840, and 640 is 640.

Step 4: Convert the GCD back to the original units

Now we need to convert the greatest common divisor (GCD) back into zirkombs by dividing it by \(10^{12}\): \(640 \div 10^{12} = 6.40 \times 10^{-13}\)

Step 5: State the charge of the electron

The charge of the electron in zirkombs is \(6.40 \times 10^{-13}\) zirkombs.