Consider the gold isotope 197 Au.
a. How many electrons, protons, and neutrons are in a neutral 197 Au atom?
b. The gold nucleus has a diameter of 14.0 fm. What is the den-sity of matter in a gold nucleus?
c. The density of gold is 19,300 kg/m³. How many times the density of gold is your answer to part b?
The Correct Answer and Explanation is :
Part a. How many electrons, protons, and neutrons are in a neutral (^{197}\text{Au}) atom?
A neutral gold atom has the same number of electrons and protons.
- Electrons: Since gold has an atomic number of 79, there are 79 electrons in a neutral atom.
- Protons: The number of protons is also equal to the atomic number of gold, which is 79.
- Neutrons: The number of neutrons can be determined by subtracting the atomic number from the mass number. For gold-197 ((^{197}\text{Au})), the mass number is 197, so the number of neutrons is: [
\text{Neutrons} = 197 – 79 = 118
]
So, in a neutral (^{197}\text{Au}) atom, there are:
- 79 electrons
- 79 protons
- 118 neutrons
Part b. The gold nucleus has a diameter of 14.0 fm. What is the density of matter in a gold nucleus?
The density of matter in the nucleus can be calculated using the formula for the density of a sphere:
[
\text{Density} = \frac{\text{Mass}}{\text{Volume}}
]
The mass of the nucleus is approximately the mass of the gold atom, which is 197 atomic mass units (u), or 197 g/mol. To convert this to kilograms, we use the conversion factor 1 u = (1.6605 \times 10^{-27}) kg:
[
\text{Mass of gold nucleus} = 197 \times 1.6605 \times 10^{-27} \, \text{kg} = 3.27 \times 10^{-25} \, \text{kg}
]
The volume of the nucleus is given by the formula for the volume of a sphere:
[
V = \frac{4}{3} \pi r^3
]
where (r) is the radius of the nucleus, which is half the diameter. Given that the diameter of the nucleus is 14.0 fm, the radius is:
[
r = \frac{14.0 \, \text{fm}}{2} = 7.0 \, \text{fm} = 7.0 \times 10^{-15} \, \text{m}
]
Now, we calculate the volume:
[
V = \frac{4}{3} \pi (7.0 \times 10^{-15} \, \text{m})^3 = 1.436 \times 10^{-43} \, \text{m}^3
]
Now, we can calculate the density:
[
\text{Density} = \frac{3.27 \times 10^{-25} \, \text{kg}}{1.436 \times 10^{-43} \, \text{m}^3} = 2.28 \times 10^{18} \, \text{kg/m}^3
]
Thus, the density of the gold nucleus is (2.28 \times 10^{18} \, \text{kg/m}^3).
Part c. How many times the density of gold is your answer to part b?
The density of gold is given as 19,300 kg/m³. To find how many times the density of gold is the density of the gold nucleus, we divide the density of the gold nucleus by the density of gold:
[
\frac{2.28 \times 10^{18} \, \text{kg/m}^3}{19,300 \, \text{kg/m}^3} = 1.18 \times 10^{14}
]
Thus, the density of the gold nucleus is approximately (1.18 \times 10^{14}) times the density of gold.
Explanation
The nucleus of an atom is incredibly dense because the mass of the atom is concentrated in a very small volume. In this case, the nucleus of a gold atom has a diameter of only 14.0 femtometers (fm), which is extremely small compared to the overall size of the atom. Despite this small size, the mass of the nucleus is nearly the same as the mass of the whole atom, which results in an extremely high density for the nucleus.
In contrast, the density of gold, as a solid metal, is much lower because the atoms are spaced much farther apart in the solid state. Therefore, the nucleus of a gold atom is much denser than the gold metal itself by a factor of around (1.18 \times 10^{14}).
Now, I will generate an image to visualize the density of gold compared to the density of its nucleus.
Here is an image comparing the density of the gold nucleus to that of solid gold metal. It illustrates the significant difference in scale and density between the tiny, dense nucleus and the atoms spaced farther apart in the solid metal form. This visual should help clarify the concept. Let me know if you’d like any further explanation!
