Question

If you have a deuterium atom that has a spherical nucleus and the mass of that nucleus is 2.013547 amu, then if you know that the mass of the proton is equal to 1.007590 amu and the mass of the neutron is equal to 1.008976 amu, find:

The radius of a deuterium nucleus equals:
A) 1.512 fm
B) 2.195 m
C) 2.591 fm
The volume of the deuterium nucleus in fermi cubic units (fm³) equals:
A) 8.664
B) 14.469
C) 24.541
The density of the deuterium nucleus in units (amu) per (fm³) is equal to:
A) 0.751
B) 0.425
C) 0.139
The density of a deuterium atom in units (amu) per cubic meter (m³) is equal to:
The total bond energy of a deuterium nucleus in MeV units:
A) 4.561
B) 3.782
C) 2.810
The bond energy per nucleon in the deuterium nucleus in units of MeV:
A) 1.405
B) 4.104
C) 2.280

Answer 1

The mass of a deuterium nucleus is 2.013547 amu. The radius of a deuterium nucleus is 2.591 fm and its volume is 24.541 fermi cubic units. The density of a deuterium nucleus is 0.139 amu per fm³.

Step-by-step explanation:

The mass of a deuterium nucleus is equal to the sum of the masses of a proton and a neutron, which is 2.013547 amu. The mass of the proton is 1.007590 amu, and the mass of the neutron is 1.008976 amu. Therefore, the mass defect of the deuterium nucleus is 2.013547 amu - 2.016566 amu = -0.003019 amu. The radius of the deuterium nucleus can be calculated using the formula r = r0 * A1/3, where r0 is the radius of a single nucleon (1.2 fm) and A is the mass number. For deuterium, A = 2, so the radius of the deuterium nucleus is r = 1.2 fm * 21/3 = 2.591 fm.

1. The radius of a deuterium nucleus equals 2.591 fm.
2. The volume of the deuterium nucleus in fermi cubic units (fm³) equals 24.541.
3. The density of the deuterium nucleus in units (amu) per (fm³) is equal to 0.139.