SCI 270: On Nuclear Technology Practice Problems |
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**At 273 K and 1 atm pressure, 1.0 mol of N**_{2}occupies 22.4 L. Calculate the number of N_{2}molecules in 1.0 L. At 277 K, 1.0 L of water weighs 1.0 kg. Calculate the number of moles of water in 1.0 L. Calculate the number of H_{2}O molecules in 1.0 L at 277 K. The density of lead (Z, 82; at. wt. 207.2) is 11.29 g/mL, calculate the number of lead atoms in 1.0 L. (1.0 mole of N_{2}has 6.02x10^{23}N_{2}molecules).The purpose is to compare the particle density of solid, liquid, and gaseous states of some material so that you can picture or estimate the number of collisions an ionizing particle has over its path.

*Hint...*

N_{2}gas: (6.023E23 N_{2}molecules)/22.4 L = 2.69E22 N_{2}molecules/L

Water: (1000 g / 18g/mol) 6.023E23/ 1.0 L = 3.35E25 H_{2}O molecules/L

Lead: (11290 g/L) (6.023E23 ) / (207.2 g/mole) = 3.28E25 Pb atoms/L

**Discussion...**

The number of lead atoms and number of water molecules are approximately the same, but molecular densities of solid and liquid are 100 times more than those of gasses at STP.**Explain the process by which beta particles lose energy when they pass by a nucleus. What radiation is produced in this process?***Hint...*

There are three processes by which beta particles lose energy:

Bremsstrahlung radiation-

Annihilation with positrons-

Ionization-

**Discussion...**

For details, consult the lecture notes.**Describe the mechanisms by which gamma-rays interact with matter. Explain each mechanism separately.***Hint...*

The mechanisms by which gamma rays interact with matter are:

Compton effec-

Photoelectric effect-

Pair production-

**Discussion...**

For details, consult the lecture notes.**What is the quality factor or rbe? A person weighing 100 kg received 1 Gy of fast neutron radiation dose. How much radiation in unit Sv is exposed to?***Hint...*

See page 316 for a discussion of the quality factor or rbe.

Dose = Q'D = 10'1 Gy = 10 Sv.**Calculate the dose (in rem and Sv) received on the first day by a patient weighing 50 kg who has been injected with 1.0 microcurie of**^{131}I, half-life 8 days. Consider the energy released per disintegration being 2 MeV, all absorbed by the tissues of the patient.*Hint...*

A sample of 1 microcurie of 132I has an initial radioactivity of 3.7e4 dps, A_{0}= 3.E4.

Since the half-life is 8 days, the decay constant is,

l = ln 2 / (8*24*3600) = 1.0E-6 /s

Total disintegration I,

I = ò*A dt*

= ò A_{0}e-l*t dt*

= A_{0}ò e-l*t dt*

= (A_{0}/ -l) e-l*t*|_{0}^{8640}(note integration over 1 day = 8648 s)

= (3.7e4 / 1e-6) (1 - e-8640*(1e-6)) (note two signs changed here)

= (3.7e10) (1 - 0.917)

= (3.06e9) disintegrations or decays

Total energy released = 2 MeV*3.06E9

= 6.12E9 MeV(1.6E-13 J/1 MeV)

= 9.79E-4 J (1 kg/1 J)(1/50 kg) (Note 1 Gy = 1 J/1 kg)

= 1.96e-5 Gy

Dosage = 1.96e-5 Sv (since Q = 1)

or 2.E-5 Sv if integration is not used

**Discussion...**

The integration is simple, but if you do not know how or do not understand the integration, you may use disintegration per day.

3.7e4 * 86400 = 3.2e9 integrations (instead of 3.06e9 disintegrations)

Dose = 1.E-5 Sv if integration is not used, and 9.8e-6 if using integration.

The result is slightly higher, because at the end of the day, the rate is lower. That is why the integration method is more reasonable.You may also calculate a decay constant with unit per day from a half life of 8 days. Then you can calculate the number of nuclei, and use

*I = N*l to calculate total number of decays.We have made the assumption that all the energy is dissipated in the patient. In reality, some energy is released into the air.