Self assessment radiochem

NSE 516 HW 3 Radiation and 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 λ=304 nm, increased from 0.01 (initial) to 0.203 (irradiated). The molar absorp- tivity of ferric ions for this wavelength is ε304 = 2187 Lmol-1cm-1, 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 J-1. From the lecture,

DF = ΔAϵ*GFe3+*ρ*L With our plugged in values,

DF = 0.203 -0.012187  Lmol-cm *1.6E-6 molJ *1.024  gml *1 [cm] → 0.203 mol

L -0.01 mol

L 

2187  Lmol-cm *1.6E-6 molJ *1.024  kgL *1 [cm] = 53.86  Jkg  For the dose rate then,

D  = ΔDΔt = 53.86 [Gy]3 [hr] = 17.95 Gyhr 

2. Calculate the dose rate and the absorbed dose for the ceric sulfate dosimetric solu- tion 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 λ=320 nm, decreased from 1.01 (initial) to 0.21 (irradiated). The molar absorp- tivity of ceric ions for this wavelength is ε320 = 561 Lmol-1cm-1, density of the dosimet- ric 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.

Similar setup,

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D = ΔAϵ*GCe3+*ρ*L → 0.21 -1.01561  Lmol-cm *0.25E-6 molJ *1.025  gml *1 [cm] = 5564.97[Gy]

Dose rate is then,

D  = ΔDΔt = 5564.97 [Gy]3 [hr] = 1854.99 Gyhr 

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 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+]

First finding the initial activity of 239Pu,

A = Nλ = 0.03 mol L  *1 [L] *6.022 E23  atommol  * ln27.604 E11s-1 = 1.65 E10 diss 

Dose can then be measured as the integral over a week of the dose rate:

D = ∫0604 800 A0 e-λt*Em ⅆt = 1.65 E10  diss  * 8.395E - 13  Jdis   e-9.11E-13 s-1*604 800 [s]-e-9.11E-13s-1*0[s]-9.11E-13 s-1  0.03 [mol] =

8377.54 [J] /0.03 [mol] = 279 251.3  Jmol  Now to find the amount of hydrogen radicals possible from this:

0.2E-6 mol J  * 8377.54  J

L  = 0.0017 mol

L 

2 radicals per reduced Pu means that there are 0.00017 mol L /2= 8E-4 mol

L *1 [L] = 8E-4 [mol] of Pu(VI)

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 calcu-

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AgNO3. Identify late their initial activity (DE = 40%). Recall that precipitating of both the unknown radionuclides with silver nitrate indicates that both unknown elements are halogens.

t (h) 0 0.5 1 1.5 2 3 4 5 6 7 8 10 12 c /m 6000 3043 1698 1062 741 455 321 237 177 132 99 56 31

We know that at t=0,

AT = 6000 * .4 = 2400 = A0,1 +A0,2 Similarly,

AT {t} = A0,1 e-λ1 t +A0,2 e-λ2 t Assuming one radionuclide has decayed entirely by the last point,

12.4[cpm] = 22.4[cpm] *e-λ1*120[min] → 0.005min-1 Or a half life of about 2.31 hrs. Knowing this we can solve for A0,1,

12.4[cpm] = A0,1 *e-0.005*720[min] → 453.81[cpm] Which allows us to solve for A0,2

2400 = 453.81+A0,2 → A0,2 = 1946.19 [cpm]

Solving for λ2 then, 1217.2 = 453.81[cpm] *e-0.005min-1*30[min] +1946.19[cpm] e-λ2*30[min] →λ2 = 0.029 min-1

Or a half life of about 0.398 hrs.

Thus the first radionuclide is likely 83Br and the second is likely 128I

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1000

Activity of 2 unknown β emitters

100 Total

Isotope #1

Isotope #2

10

y2 = 43.975e-0.295t

y1 = 227.3e-1.834t

10 12 14

Time (hrs)

A (b

q)