Problem 93
Question
The nucleus of \({ }^{6} \mathrm{Li}\) is a powerful absorber of neutrons. It exists in the naturally occurring metal to the extent of \(7.5 \%\). In the era of nuclear deterrence, large quantities of lithium were processed to remove \({ }^{6} \mathrm{Li}\) for use in hydrogen bomb production. The lithium metal remaining after removal of \({ }^{6} \mathrm{Li}\) was sold on the market. (a) What are the compositions of the nuclei of \({ }^{6} \mathrm{Li}\) and \({ }^{7} \mathrm{Li}\) ? (b) The atomic masses of \({ }^{6} \mathrm{Li}\) and \({ }^{7} \mathrm{Li}\) are \(6.015122\) and 7.016004 amu, respectively. A sample of lithium depleted in the lighter isotope was found on analysis to contain \(1.442 \%^{6} \mathrm{Li}\). What is the average atomic weight of this sample of the metal?
Step-by-Step Solution
VerifiedKey Concepts
Atomic Mass
The atomic mass is usually expressed in atomic mass units (amu), where 1 amu is defined as one twelfth of the mass of a carbon-12 atom. This allows chemists to compare the masses of different atoms on a relative scale.
For isotopes, the atomic mass reflects the number of protons and neutrons in the nucleus. For example, the atomic mass of the isotope extsuperscript{6}Li is 6.015122 amu. This reflects that the nucleus of this isotope contains six nucleons in total—three protons and three neutrons.
Nuclear Composition
- extsuperscript{6}Li: This isotope has an atomic number (Z) of 3, meaning it has 3 protons. To determine the number of neutrons, subtract the number of protons from the isotope's mass number (A): extsuperscript{6}Li has 3 neutrons ( A = 6 - Z = 6 - 3 = 3).
- extsuperscript{7}Li: Similarly, extsuperscript{7}Li has 3 protons and 4 neutrons ( A = 7 - Z = 7 - 3 = 4).
Average Atomic Weight
In the case of a sample containing less extsuperscript{6}Li, as described in the exercise, the average atomic weight needs to be recalculated. It must account for the different proportions of extsuperscript{6}Li and extsuperscript{7}Li.
To find the average atomic weight, you can use the following formula:
- Determine the percentage of each isotope in the sample. For example, extsuperscript{6}Li at 1.442% and extsuperscript{7}Li at 98.558%.
- Use a weighted average: \(\text{Average Atomic Weight} = (\text{Percentage of } \text{ extsuperscript{6}Li})\times(\text{Atomic Mass of } \text{ extsuperscript{6}Li}) + (\text{Percentage of } \text{ extsuperscript{7}Li})\times(\text{Atomic Mass of } \text{ extsuperscript{7}Li})\)
Lithium Isotopes
- extsuperscript{6}Li: Comprising about 7.5% of natural lithium, this isotope has an atomic mass of 6.015122 amu. It contains 3 protons and 3 neutrons. extsuperscript{6}Li is well-known for its neutron absorption capabilities, making it critical for certain nuclear applications including hydrogen bomb production.
- extsuperscript{7}Li: This isotope is more abundant, making up around 92.5% of natural lithium. With an atomic mass of 7.016004 amu, it has 3 protons and 4 neutrons. This isotope is more stable and less reactive compared to extsuperscript{6}Li.