Question

The radius of a neon atom is \(69 \mathrm{pm}\), and its mass is \(3.35 \times 10^{-23}\) g. What is the density of the atom in grams per cubic centimeter \(\left(\mathrm{g} / \mathrm{cm}^{3}\right) ?\) Assume the atom is a sphere with \(=\frac{4}{3} \pi r^{3}\)

Short answer

The density of the neon atom is approximately \(2.43 \times 10^4\ \text{g/cm}^{3}\).

Step-by-step solution

Convert the radius to centimeters

First, we should convert the given radius from picometers (pm) to centimeters (cm). We know that 1 meter (m) = \(10^9\) picometers and 1 meter = 100 centimeters, so the conversion factor between picometers and centimeters is \(10^{-11}\). Therefore, we should multiply the given radius by this conversion factor: \(69 \ \text{pm} \cdot 10^{-11}\ \text{cm/pm} = 6.9 \times 10^{-10}\ \text{cm}\)

Calculate the volume of the atom

Now we need to calculate the volume of the neon atom. Since we are assuming that the atom is a sphere, we can use the volume formula for a sphere, which is \(\frac{4}{3} \pi r^3\), where \(r\) represents the radius of the sphere. Using the converted radius from step 1, we can calculate the volume of the neon atom: Volume = \(\frac{4}{3} \pi (6.9 \times 10^{-10}\ \text{cm})^3 ≈ 1.38 \times 10^{-27}\ \text{cm}^{3}\)

Calculate the density of the atom

Now that we have calculated the volume of the neon atom, we can determine its density. The density formula is: Density = \(\frac{\text{mass}}{\text{volume}}\) Using the given mass of the neon atom and the volume calculated in step 2, we can determine the density of the atom: Density = \(\frac{3.35 \times 10^{-23}\ \text{g}}{1.38 \times 10^{-27}\ \text{cm}^{3}} ≈ 2.43 \times 10^4\ \text{g/cm}^{3}\) So, the density of the neon atom is approximately \(2.\ 43 \times 10^4\ \text{g/cm}^{3}\).

Key concepts

Density Formula

Understanding how to calculate density is fundamental in various scientific disciplines. Density is defined as the mass of an object divided by its volume and is expressed using the formula:
\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \].
In the context of atoms, like a neon atom, density provides insights into how tightly packed an atom's mass is within its volume. If we're given the mass of the neon atom and we know the volume it occupies, we can easily determine its density. In our example, after calculating the volume of the neon atom as a perfect sphere, we applied this formula to derive the atom's density. This concept not only applies to tiny particles but also to everyday objects and substances, making it a versatile tool for both theoretical and practical applications in science and engineering. To ensure thorough understanding, one should practice applying the density formula to various problems and become comfortable with different units of mass and volume.

Conversion of Units

When performing atomic density calculations, ensuring correct unit conversion is vital. For our neon atom example, converting the radius from picometers to centimeters was required before we could calculate the volume of the atom. This step is often where students make mistakes, but it is crucial because failing to convert to consistent units will render the density calculation incorrect.
For the conversion, we use the fact that \(1 \text{ meter} = 10^{9} \text{ picometers} \) and \(1 \text{ meter} = 100 \text{ centimeters} \), thus establishing that \(1 \text{ picometer} = 10^{-12} \text{ meters} \) and converting picometers to centimeters with the conversion factor of \(10^{-11} \). It is essential to understand and to be comfortable with the relationship between units, especially when working within the metric system where many units are related by powers of ten.

Volume of a Sphere

In geometry, the volume of a sphere is calculated using the formula \[ \text{Volume} = \frac{4}{3}\pi r^3 \], where \( r \) is the radius of the sphere. The volume of an atom, when assumed to be spherical, can provide a basis for numerous calculations, including the atomic density.
In the case of the neon atom exercise, we used this formula after converting the radius to the appropriate unit of centimeters. The radius was cubed, then multiplied by \(\pi\), and finally multiplied by \(\frac{4}{3}\) to determine the atom’s volume. It's important for students to identify when to use this formula and to understand its derivation from the basic principles of geometry. By mastering the volume calculation of a sphere, one can gain greater insight into various physical and scientific phenomena that deal with spherical objects.