Self assessment radiochem

SELF ASSESSMENT-

Estimated Solved Correct Problems: problems (%)

Estimated Time to Re-solve: minutes

CHEMICAL DOSE:

1. Calculate the dose rate and the absorbed dose for the Fricke dosimetric solution that was irradiated with Co-60 source during 3 hrs. The absorbance of the irradiated dosimetric solution (Fe3+), measured with UV/Vis-spectrophotometer in 1 cm cuvettes at the wavelength =304 nm, increased from 0.01 (initial) to 0.203 (irradiated). The molar absorptivity of ferric ions for this wavelength is ε304 = 2187 L mol1cm1, the density of the dosimetric solution is ρ=1.024 g/mL, and the tabulated radiation chemical yield for gamma source 60Co is 1.6 µmol J1.

T = 3hrs

d =1cm

ʎ =320 nm

ε320 = 561 L mol1cm1

ρ =1.025 g/Ml

G =2.3 molecules/100 eV

A = εcl

A= 0.6821

2. Calculate the dose rate and the absorbed dose for the ceric sulfate dosimetric solution that was irradiated with Co-60 source during 3 hrs. The absorbance of the irradiated dosimetric solution (Ce4+), measured with UV/Vis-spectrophotometer in 1 cm cuvettes at the wavelength =320 nm, decreased from 1.01 (initial) to 0.21 (irradiated). The molar absorptivity of ceric ions for this wavelength is ε320 = 561 L mol1cm1, density of the dosimetric solution is ρ=1.025 g/mL and the G value for the cerous-ion yield is 2.3 molecules/100 eV for the gamma source 60Co.

T = 3hrs

d =1cm

ʎ =320 nm

ε320 = 561 L mol1cm1

ρ =1.025 g/Ml

G =2.3 molecules/100 eV

A = εcl

A= 0.6821

3. Estimate the effect of alpha-autoradiolysis in the acidic aqueous solution of 239Pu with concentration of 0.03 mol/L of 239Pu:

a) Considering the half-life T1/2=24111 y (=7.60364x1011 sec) and the alpha energy of 5.24 MeV, what is the weekly dose absorbed in 1L solution? E(alpha/Pu) = 5.24 MeV x 1.602.10-19 J/eV = 8.395x10-13 J

Solution

ʎ (week-1) = ln2/ T1/2 = ln2/(7.60364x1011/604800)

= 5.5134 x 10-7 /week

N(atoms) x ʎ (week-1) = 0.03 x 5.5134 x 10-7 x 6.022 x1023

=9.9604 x 1015 Bq

· k = 5.76 x 10-7

· S = 9.9604 x 1015 Bq

· E = 5.24 MeV

· μt/ρ =  0.0326 cm2/g (values are available at NIST)

· μ =  1.289 cm-1 (values are available at NIST)

· D = 1

· r = 1

1489862 x107

b) The plutonium is originally in its hexavalent state, but is reduced to its tetravalent state (then precipitated) by the reaction with hydrogen radicals. If the radiation yield of hydrogen radicals is 0.2 umol/J, how much Pu(VI) is reduced in one week? Hint:

[Pu6+ + 2H* = Pu4+ +2H+]

Using the reaction ratio

1 : 2 = 1 : 2

If 2 = 0.2 umol/J

1=?

=½ x 0.2 umol/J

=0.1 umol/J of Pu6+ per reaction

4. The count rates listed in the table below were measured for a mixture of two genetically independent radionuclides. Both radionuclides are beta- emitters, and they both precipitate in the solution of AgNO3. Identify the unknown radionuclides in the measured mixture, and calculate their initial activity (DE = 40%). Recall that precipitating of both the unknown radionuclides with silver nitrate indicates that both unknown elements are halogens.

Fluorine F 18 decays by positron, (β+) emission and has a half-life of 109.7 minutes. This half-life is close to obtained in the experiment; therefore one mixture must be containing chlorine. The other element might be Tennessine because it half-life less than 30 minutes.

HINT:

The concentrations of colored solution are usually determined spectrophotometrically. Method is based on the Lambert-Beer law, which binds the absorbance A (decrease of the light beam intensity) with the molar concentration of the analyte C, its molar absorptivity ε (constant at given wavelength) and the length of the light path (thickness of the cuvette with measured liquid):

A = ε.c.l