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Chapter 3: Problem 14

An ion in a mass spectrometer follows a semicircular path of radius \(15.2 \mathrm{cm} .\) What are (a) the distance it travels and (b) the magnitude of its displacement?

Short Answer

Expert verified
The ion covers a distance of \(47.75 \,cm\) and has a displacement of \(30.4 \,cm\).

Step by step solution

01

Calculate the Distance

The distance covered by the ion along a semicircular path can be calculated by finding the length of half the circumference of a circle, which is given by the formula \(0.5 \times 2\pi r\). Plug in the given radius value \(r = 15.2 \,cm\) into the formula, which yields \(0.5 \times 2\pi \times 15.2 = 47.75 \,cm\). Thus, the distance the ion travels is \(47.75 \,cm\).
02

Calculate the Displacement

The displacement, on the other hand, is the shortest path covered by the ion, which is equal to the diameter of the circular path it took. We can compute this by multiplying the radius by 2. So \(2r = 2 \times 15.2 = 30.4 \,cm\). Hence, the displacement of the ion is \(30.4 \,cm\).

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