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Nuclear radii, binding energy, isotope, uranium, fission, fusion

1. Compare the nuclear radii of the following nuclides: (2 1H), (60 27Co), (197 79Au), (129 94Pu).

2. Calculate the average binding energy per nucleon of (a)(23 11Na) Sodium, (b) (60 27Co) cobalt, and (c) (107 47Ag) Silver

3. The peak of the stability curve occurs at 56Fe, which is why iron is prominent in the spectrum of the Sun and stars. Show that 56Fe has a higher binding energy per nucleon than its neighbors 55Mn and 59C0.

4. The half-life of an isotope of phosphorus is 14 days. If a sample contains 3.0 x 10^16 such nuclei, determine its activity. Express your answer in curies.

5. Burning 1 metric ton(1 000 kg) of coal can yield an energy of 3.30 x 10^10 J. Fission of one nucleus of uranium-235 yields an average energy of approximately 208 MeV. What mass of uranium produces the same energy as 1 metric ton of coal?

6. Identify the missing nuclides (X) in the following decays: (a) X->(65 28Ni)+ y(b)(215 84Po)->X + (c)X->(55 26Fe) + e+ v.

7. A person whose mass is 75.0 kg is exposed to a whole-body dose of 25.0 rad. How many joules of energy are deposited in the person's body?

8. An x-ray technician works 5 days per week, 50 weeks per year. Assume that the technician takes an average of eight x-rays per day and receives a dose of 5.0 rem/yr as a result. (a) Estimate the dose in rem per x-ray taken. (b) How does this result compare with the amount of low-level background radiation the technician is exposed to?

9. Find the energy releaeed in the fission reaction (1 0n) + (235 92U) -> (88 38Sr) + (136 54Xe) + (121 0n).

Solution Preview

1.

r = (1.2 x10-5 m) A1/3

H : r = (1.2 x10-5 m) (2)1/3 = 1.5 x10-5 m
Co : r = (1.2 x10-5 m) (60)1/3 = 5.87 x 10-5 m
Au : r = (1.2 x10-5 m) (197)1/3 = 8.73 x 10-5 m
Pu : r = (1.2 x10-5 m) (129)1/3 = 6.0 x 10-5 m

2.

Na : total mass = 11(1.007825) + 12(1.008665) = 23.190055 u
Mass from appendix = 22.9899 u
Difference in mass = 0.200155 u
Binding energy = 0.200155 u (931.5 MeV / u) = 186.44 MeV
Binding energy per nucleon = 186.44 MeV / 23 = 8.106 MeV

Co : total mass = 27(1.007825) + 33(1.008665) = 60.4972 u
Mass from appendix = 55.9338 u
Difference in mass = 0.56342 u
Binding energy = 0.56342 u (931.5 MeV / u) = 524.83 MeV
Binding energy per nucleon = 524.83 MeV / 60 = 8.747 MeV

Ag : total mass = 47(1.007825) + 60(1.008665) = 107.887 u
Mass from appendix = 106.9051 ...

Solution Summary

Nuclear radii, binding energy, isotope, uranium, fission, and fusion is examined.

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