Problem 6
Question
Calculate the density of the \(^{24}_{12} \mathrm{Mg}\) nucleus in g/mL, assuming that it has the typical nuclear diameter of \(1 \times\) \(10^{-13} \mathrm{cm}\) and is spherical in shape.
Step-by-Step Solution
Verified Answer
Density of the nucleus is calculated using the mass of 24 amu and the volume of a sphere with a radius of \(5 \times 10^{-14} \mathrm{cm}\).
1Step 1: Calculate the volume of the nucleus
Assuming the nucleus is a sphere, calculate its volume using the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \), where \( r \) is the radius of the sphere. Given the diameter is \(1 \times 10^{-13} \mathrm{cm}\), the radius \( r \) is half of that: \(5 \times 10^{-14} \mathrm{cm}\).
2Step 2: Convert the volume to cubic centimeters
After calculating the volume in cubic centimeters (cc), make sure it is in the correct unit for density calculation (g/mL), by acknowledging that 1 cc is equivalent to 1 mL.
3Step 3: Determine the mass of the nucleus
The mass of the nucleus can be determined using the atomic mass unit (amu). One amu is approximately \(1.6605 \times 10^{-24} \mathrm{g}\). For a \(^{24}_{12} \mathrm{Mg}\) nucleus, which has an atomic mass of 24 amu, the mass is \(24 \times 1.6605 \times 10^{-24} \mathrm{g}\).
4Step 4: Calculate the density of the nucleus
Density is mass divided by volume. Using the mass from Step 3 and volume from Step 2, calculate the density \( \rho = \frac{m}{V} \).
Key Concepts
Volume of a SphereAtomic Mass UnitDensity Formula
Volume of a Sphere
When considering the volume of a sphere, we are looking at how much space inside the sphere is actually occupied. This is crucial when we're dealing with objects like atomic nuclei that are assumed to be spherical in shape. The formula to find the volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \), where \( V \) represents the volume and \( r \) stands for the radius of the sphere. As radius is half of the diameter, it's important to divide the given diameter by two before plugging the value into the formula. Remember that if the diameter is not already in centimeters, you'll need to convert it since the standard unit for volume in the context of this formula is cubic centimeters (cc), which directly equates to milliliters (mL) in terms of density calculation.
To illustrate, with a nucleus having a diameter of \(1 \times 10^{-13}\) cm, the radius would be \(5 \times 10^{-14}\) cm. Calculating the volume then simply follows the formula, and any student can perform this computation using a standard calculator or computational software that supports scientific notation and the value of pi.
To illustrate, with a nucleus having a diameter of \(1 \times 10^{-13}\) cm, the radius would be \(5 \times 10^{-14}\) cm. Calculating the volume then simply follows the formula, and any student can perform this computation using a standard calculator or computational software that supports scientific notation and the value of pi.
Atomic Mass Unit
Delving into the atomic scale, the atomic mass unit (amu) is a standard unit of mass that quantifies the mass of an atom or a molecule. It's defined as one-twelfth of the mass of a carbon-12 atom, which is approximately \(1.6605 \times 10^{-24}\) grams. When dealing with nuclear densities, you will often need to convert the mass of a nucleus from amu to grams to use standard units in your density calculations.
For instance, a nucleus of Magnesium-24 \(^{24}_{12} Mg\) has an atomic mass of 24 amu. Multiplying the atomic mass by the amu value in grams gives the mass in grams:\[ 24 \times 1.6605 \times 10^{-24} \mathrm{g} \]. It's essential to be comfortable with handling scientific notation and understand that even though these numbers seem extremely small, they are quite significant in the realm of physics and chemistry.
For instance, a nucleus of Magnesium-24 \(^{24}_{12} Mg\) has an atomic mass of 24 amu. Multiplying the atomic mass by the amu value in grams gives the mass in grams:\[ 24 \times 1.6605 \times 10^{-24} \mathrm{g} \]. It's essential to be comfortable with handling scientific notation and understand that even though these numbers seem extremely small, they are quite significant in the realm of physics and chemistry.
Density Formula
Density is a measure of how much mass is contained within a particular volume. The general density formula used for any substance is given by \( \rho = \frac{m}{V} \), where \( \rho \) represents the density, \( m \) is the mass, and \( V \) is the volume. This fundamental concept allows us to compare how much matter is packed into different substances or objects, regardless of their size or shape.
In application to our nuclear density calculation, we utilize the mass of the Magnesium-24 nucleus (determined in grams from the amu) and the volume of the sphere (calculated from the radius of the nucleus). By positioning these numbers into the density formula, we obtain the density in grams per milliliter (g/mL), which is the standard unit for density in laboratory measurements. Understanding how to apply this formula is not only key to solving physics problems but also in understanding the relative matrices of different materials in various scientific fields.
In application to our nuclear density calculation, we utilize the mass of the Magnesium-24 nucleus (determined in grams from the amu) and the volume of the sphere (calculated from the radius of the nucleus). By positioning these numbers into the density formula, we obtain the density in grams per milliliter (g/mL), which is the standard unit for density in laboratory measurements. Understanding how to apply this formula is not only key to solving physics problems but also in understanding the relative matrices of different materials in various scientific fields.
Other exercises in this chapter
Problem 3
For the following isotopes that have missing information, fill in the missing information to complete the notation (a) \(^{34}_{14} \mathrm{X}\) (b) \(^{36}_{X}
View solution Problem 5
Write the nuclide notation, including charge if applicable, for atoms with the following characteristics: (a) 25 protons, 20 neutrons, 24 electrons (b) 45 proto
View solution Problem 7
What are the two principal differences between nuclear reactions and ordinary chemical changes?
View solution Problem 11
Write a brief description or definition of each of the following: (a) nucleon (b) \(\alpha\) particle (c) \(\beta\) particle (d) positron (e) Y ray (f) nuclide
View solution