Final answer:
The radius of a deuterium nucleus is approximately 2.591 fm, the volume is approximately 24.541 fm^3, the density is approximately 0.082 amu/fm^3, the density of a deuterium atom in amu per cubic meter is approximately 8.2 x 10^42 amu/m^3, the total bond energy of a deuterium nucleus is approximately 1.36 MeV, and the bond energy per nucleon is approximately 0.68 MeV.
Step-by-step explanation:
The radius of a deuterium nucleus can be calculated using the formula:
radius = 1.2 x 10 ^ -15 * (A ^ 1/3)
where A is the mass number of the nucleus. For deuterium, A = 2. Taking A = 2, we can calculate the radius to be approximately 2.591 fm.
The volume of the deuterium nucleus can be calculated using the formula:
volume = (4/3) * pi * radius^3
Using the calculated radius of 2.591 fm, we can find the volume to be approximately 24.541 fm^3.
The density of the deuterium nucleus can be calculated using the formula:
density = mass/volume
Substituting the given mass of the nucleus (2.013547 amu) and the calculated volume (24.541 fm^3), we can find the density to be approximately 0.082 amu/fm^3.
The density of a deuterium atom in units of amu per cubic meter (m^3) can be calculated by converting the density from amu/fm^3 to amu/m^3. Since 1 fm^3 is equal to (10^-15)^3 = 10^-45 m^3, we can calculate the density to be approximately 8.2 x 10^42 amu/m^3.
The total bond energy of a deuterium nucleus can be calculated using the formula:
bond energy = mass defect x c^2
where c is the speed of light (3 x 10^8 m/s). For deuterium, the mass defect is given as 0.002388 amu. Substituting the values, we can find the bond energy to be approximately 2.181 x 10^-13 J or 1.36 MeV.
The bond energy per nucleon in the deuterium nucleus can be calculated by dividing the total bond energy by the number of nucleons in the nucleus, which is 2 for deuterium. Dividing the bond energy by 2, we can find the bond energy per nucleon to be approximately 0.68 MeV.