HEALTH PHYSICS AND RADIATION PROTECTION 93 



layers of absorber, (c) the reduction of intensity as a function of distance 

 (inverse-square law), and (d) the half-layer of absorber as a function of 

 energy. 



The following equation gives a relationship between milli roentgens and 

 millicuries which holds fairly well for gamma energies from 0.3 to 3 Mev: 



Rf = QCE (3-2) 



where /?/ = dosage rate, mr/hr at 1 ft 



C = number of millicuries of activity 

 E = average quantum energy per disintegration, Mev 

 Values for R/ per millicurie will be listed for the individual isotopes in 

 Chap. 6. It should be noted that, if more than one gamma ray is emitted 

 per disintegration, the calculation is more involved. If the radiations are 

 emitted in cascade, the two energy values are added to give the value of 

 E; for Co^", E = l.l -{- 1.3 = 2.4. If there are several modes of disin- 

 tegration, then the weighted fractions of the energy values are used. 

 Attention is called to the fact that uncertainties in the decay scheme may 

 introduce errors in the calculated Rf values. 



In order to calculate the thickness of a given absorber required to 

 reduce an intensity by a certain factor, it is convenient to use the half- 

 value layer according to the following equation: 



N = 3.32 log X (3-3) 



where A^ = number of half-value layers reciuired to reduce the intensity 

 by a factor of X. 



One other relationship must be employed : the application of the inverse- 

 square law, which states that the radiation intensity varies inversely as 

 the square of the distance. Thus a dosage rate of 100 mr/hr at 1 ft would 

 be reduced to 100 X 1-/3' = 11 mr/hr at 3 ft. 



Table 3-5 presents some half-value layers for various materials as a 

 function of the gamma-ray energy. 



These data and relationships may be very easily used to estimate the 

 shielding required for any particular circumstance. The following is an 

 illustration: 



A sample containing 50 mc P^' is received and is to be shielded in steel 

 so that the radiation does not exceed 2 mr/hr at 6 in. From Chap. 6 it is 

 seen that P^^ has an i?/-per-millicurie value of about 2.5, which by Eq. 

 (3-2) gives a dosage of 125 mr/hr at 1 ft for the 50 mc. From the inverse- 

 square law this would be 500 mr/hr at 6 in. Therefore, sufficient shield- 

 ing is required to reduce the intensity by a factor oi ^^% = 250. From 

 Eq. (3-3) it is calculated that the number of half-layers required would be 



