Problem 11
Question
(II) Approximately how many nucleons are there in a \(1.0-\mathrm{kg}\) object? Does it matter what the object is made of? Why or why not?
Step-by-Step Solution
Verified Answer
There are approximately 6.022 x 10^{26} nucleons in a 1 kg object. The composition of the object does not significantly change this estimate because most materials consist of elements with similar atomic masses.
1Step 1: Understand the Concept of a Nucleon
A nucleon is a term that encompasses both protons and neutrons, found in the nucleus of an atom. The mass number of an atom is essentially the sum of protons and neutrons, and this mass is approximately equal to the mass of a nucleon.
2Step 2: Understand Atomic Mass
The atomic mass of an element is the average mass of atoms of an element, measured in atomic mass units (u). It approximately equals the mass number, with one nucleon having a mass of approximately 1 u or 1.67 x 10^{-27} kg.
3Step 3: Calculate Number of Nucleons per Kilogram
To estimate the number of nucleons in 1 kg of material, use the average atomic mass of common elements. A gram of material with an average atomic mass of about 1 u contains Avogadro's number of atoms (6.022 x 10^{23}). Therefore, 1 kg (1000 g) contains approximately 6.022 x 10^{26} nucleons.
4Step 4: Determine the Role of Composition
Most common materials (e.g., carbon, silicon, etc.) have similar nucleon counts per unit mass because their atomic masses are close to one another. Therefore, the exact composition does not significantly change the estimate of nucleon count; the material's typical atomic mass provides a good approximation.
Key Concepts
Atomic MassMass NumberAvogadro's Number
Atomic Mass
Atomic mass is an essential concept in chemistry and physics. It represents the average mass of atoms in an element, measured in atomic mass units (u). The atomic mass reflects isotopic distribution in a naturally occurring element.
Understanding atomic mass can be more intuitive with the following points:
Understanding atomic mass can be more intuitive with the following points:
- Atomic mass is typically not a whole number because it's an average of isotopes with different masses.
- The value is based on the mass of carbon-12, which is exactly 12 u.
- It's close to the mass number but varies slightly due to the presence of isotopes.
Mass Number
Mass number is another key aspect of understanding atomic structure. This value indicates the sum of protons and neutrons in the nucleus of an atom. Each proton or neutron is considered one nucleon.
A few important considerations about mass number:
A few important considerations about mass number:
- The mass number is always an integer since it counts whole particles (protons and neutrons). For example, the mass number of Helium is 4, indicating it has 2 protons and 2 neutrons.
- Mass number can vary among isotopes of an element. For example, Carbon-12 and Carbon-14 have different neutron counts.
- Mass number provides a basis for defining isotopes, which have the same atomic number but different mass numbers due to neutron variation.
Avogadro's Number
Avogadro's Number stands as a fundamental constant in chemistry, essential for mole calculations. Avogadro's Number is defined as the number of constituent particles, usually atoms or molecules, contained in one mole of a substance.
Key attributes of Avogadro's Number include:
Key attributes of Avogadro's Number include:
- The value of Avogadro's Number is approximately \(6.022 \times 10^{23}\), indicating a huge quantity that's difficult to visualize.
- This constant allows chemists and physicists to shift from atomic scale to macroscopic quantities when calculating masses or concentrations.
- While the number appears large, it's essential for converting between the number of atoms and macroscopic measurements like grams and kilograms.
Other exercises in this chapter
Problem 4
(II) (a) What is the approximate radius of a \(148 \mathrm{~d}\) nucleus? (b) Approximately what is the value of \(A\) for a nucleus whose radius is \(3.7 \time
View solution Problem 5
(II) What is the mass of a bare \(\alpha\) particle (without electrons) in MeV/\(c^{2} ?\)
View solution Problem 17
(II) Compare the average binding energy of a nucleon in \({ }_{11}^{23}\) Na to that in \({ }_{11}^{24} \mathrm{Na}\).
View solution Problem 22
(I) Show that the decay \({ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{10} \mathrm{~B}+\mathrm{p}\) is not possible because energy would not be conserved.
View solution