Question

The proton has a radius of approximately \(1.0 \times 10^{-13} \mathrm{~cm}\) and a mass of \(1.7 \times 10^{-24} \mathrm{~g} .\) Determine the density of a proton. For a sphere, \(V=(4 / 3) \pi r^{3}\).

Short answer

The density of a proton is approximately \( \rho = \frac{1.7 \times 10^{-24} \mathrm{~g}}{\frac{4}{3} \pi (1.0 \times 10^{-13} \text{cm})^3} \approx 4.1 \times 10^{14} \text{g/cm}^3\).

Step-by-step solution

Calculate the volume of the proton

Use the formula for the volume of a sphere, which is given by \(V = \frac{4}{3} \pi r^3\), where \(r\) is the radius of the sphere. Substitute the given radius of the proton \(r = 1.0 \times 10^{-13} \text{cm}\) into the formula to find the volume.

Substitute the values into the volume formula

Substitute \(r = 1.0 \times 10^{-13}\text{cm}\) into the volume formula to get \(V = \frac{4}{3} \pi (1.0 \times 10^{-13} \text{cm})^3\). Calculate the numerical value of the volume.

Calculate the density of the proton

Use the formula for density \(\rho = \frac{m}{V}\), where \(m\) is the mass and \(V\) is the volume. Substitute the mass of the proton \(m = 1.7 \times 10^{-24} \text{g}\) and the calculated volume into the formula to determine the density.

Substitute the values into the density formula

Substitute the calculated volume and the given mass into the density formula to find the proton's density. Perform the division to calculate the final density value.

Key concepts

Volume of a Sphere

Understanding the volume of a sphere is fundamental when we are calculating the space it occupies. The volume can be found using the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( V \) is the volume, \( \pi \) is a constant (Pi, approximately 3.14159), and \( r \) is the radius of the sphere. To visualize this, imagine filling a spherical balloon with water; the amount of water it can hold before bursting is analogous to its volume. For a proton, which we treat as a tiny spherical object, you would substitute the radius of the proton in centimeters into this formula to calculate its volume. The concept might seem abstract since a proton is unimaginably small, but it's essentially the same as calculating the volume for any sphere, just on a much smaller scale.

Density Formula

Density is a measure of how much mass is contained in a given volume. The formula to calculate density is expressed as:
\[ \rho = \frac{m}{V} \]
In this formula, \( \rho \) represents the density, \( m \) is the mass, and \( V \) is the volume of the object. To put this into perspective, think about two balls of the same size but made from different materials, say a tennis ball and a lead ball. While both have the same volume, the lead ball is much heavier; thus, it has a higher density. When calculating the density of a proton, we use its mass and the volume we computed from the 'volume of a sphere' formula. This tells us how densely packed the matter is within the proton.

Substitute Values

Substituting values into formulas is a bit like replacing placeholders in a template with specific information. When we are given certain numerical values, such as the radius and mass of a proton, we replace the variables in our formulas with these given numbers. For example, if we use the volume formula of a sphere with a given radius, we literally substitute the place of \( r \) with the proton's radius, maintaining the correct units of measurement. Similarly, when we have the mass of the proton, we replace \( m \) with this value in the density formula. It's essential to substitute values carefully and correctly, as an error in this step can lead to incorrect results in our calculations. Paying attention to details at this stage ensures accuracy in determining the density of objects, such as protons.

Scientific Notation

Scientific notation is a way of expressing very large or very small numbers in a compact form. It uses powers of ten to simplify numbers, which makes it easier to work with in calculations. In scientific notation, a number is written as the product of a number (between 1 and 10) and a power of ten. For instance, the number \(1.0 \times 10^{-13}\) represents the radius of a proton in centimeters. This indicates that the decimal point in the number 1.0 is moved 13 places to the left. Similarly, the proton's mass is given in scientific notation as \(1.7 \times 10^{-24}\) grams. Using scientific notation is especially useful in physics and chemistry, where extremely large or small values, such as those seen in atomic scales, are the norm. It allows for clearer and more efficient calculations without having to write out many zeros.