Part II P 5 points for each question Show your work/calculations in the space pr

Question

Part II P

5 points for each question

Show your work/calculations in the space provided

Box your answer whenever possible

        Part 1. Iron-56 (56 over 26 Fe) has a binding energy per nucleon of 8.79 MeV. (1MeV is 1.60 ´ 10–13 J).

                   Determine the difference in mass between one mole of iron-56 nuclei and the component nucleons of which it is made.

Part 2 Breeder reactors are used to convert the nonfissionable nuclide 238 over 92 U to a fissionable product. Neutron capture of the 238 over 92 U is followed by two successive beta decays. What is the final fissionable product?

              Write a balanced equation for each of the following reactions:

Part 3 Cesium metal with Cl2(g).

Explanation / Answer

Part 1:

Atomic mass = Mass of protons + mass of neutons

Atomic number = Mass of protons.

Number of neutrons in Fe-56 = 30

Number of protons = 26

Difference in mass = MN + MP - MO

MN = mass of neutron = 1.008664916 u, MP = 1.007276467 and MO atomic mass of Fe = 55.9349375 u

For Fe, Difference in mass = (30 x MN + 26 x MP) - MO

                                           = (30 x 1.008664916 + 26 x 1.007276467 ) - 55.9349375

                                           = 56.46339748 - 55.9349375   = 0.52845998 u

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