Answer to Question #1488 Submitted to "Ask the Experts"

The energy and intensity of the gamma or x rays of ²⁴¹Am can be found in tables in the NUDAT retrieval program made available on the Internet by Brookhaven National Laboratory.

A form will open asking various parameters. Fill in the mass number, nuclide, and radiation. (The radiation is G.) On the pull down for sort order, select the option beginning with intensity and energy of radiation. The sort will be ascending by intensity, so the most important contributions to exposure, most intense, will be at the bottom of the list. You will find these three photons:

All other photons are negligible relative to these three, so substituting the energy values in your equation will give the exposure from ²⁴¹Am. The formula you propose will give a larger number of photons at 30 feet the correct value because attenuation in air was not considered. Nevertheless, because incorrect energy absorption coefficients are used, the formula predicts less energy will be deposited per unit volume and the resulting exposure rate will be too low. The formula you propose should not be used for the case of ²⁴¹Am. The formula will give the answer of 0.093 mR/h. The actual result is higher, as will be shown below.

The distance of 30 feet (9.144 m) in air attenuates these photons significantly. The mass energy absorption coefficients (mE) to estimate the attenuation can be obtained from the National Institute of Science and Technology. It will be necessary to interpolate the coefficients for the ²⁴¹Am energies. The exposure rate by the simple formula should be reduced for each photon energy by the attenuation of 30 feet of air. The result must then be increased using the appropriate attenuation factor for the energy deposited in a unit volume.

Rather than attempting to correct a simple approximation formula, it is better to perform the full analysis. The analysis below is intended for use in a spreadsheet.

• The photons/s value for each energy is the intensity x 0.5 Ci x 3.7 x 10¹⁰ disintegrations/Ci.
   
• The photons/s-cm² at 914.4 cm for each energy is the previous column divided by 4 π 914.4².
   
• The attenuated photons/s-cm² is the previous column times exp (-μ_E 914.4).
   
• The photons absorbed/s-cm³ is the previous column times 1-exp (-μ_E1).
   
• The MeV absorbed/s-cm³ is the previous column times the first column.
   
• The total MeV absorbed/s-cm³ must be converted to MeV/h-kg by totaling the last column, multiplying by 3,600 s/h, dividing by 0.001205 g air/cm³, and then multiplying by 1,000 g/kg.
   
• MeV/h-kg is converted to eV by multiplying by 1 x 10⁶ to ion pairs/h-kg by dividing by 33.7 eV/ion pair.
   
• Coulomb/kg-h is obtained by dividing ion pairs/h-kg by 6.24 x 10¹⁸ ion pairs/coulomb, then R/h is finally obtained by dividing 2.58 x 10^-4 coulomb/R.
   
• The result is 0.254 mR/h.

A method to obtain mrad/h is to continue the above table after attenuated photons/s-cm² with the following:

• The kerma coefficient for each energy was interpolated from Table A-1 of ICRU Report 57, "Conversion Coefficients for use in Radiological Protection Against External Radiation," 1998.
   
• The kerma/s equals the product of the previous two columns.
   
• The mrad/h (in air) was converted from pGy/h (kerma/s times 3600 s/h).
   
• The total is 0.253 mrad/h.

Note: The result will be dependent upon the interpolated values. Also, buildup was not considered in these calculations and will likely increase the result about 30 percent.

Joe Alvarez