**ABSTRACT**

This article explains the use of the “6CE” equation for estimating exposure rates from point sources of known quantities of various specific gamma emitting radioisotopes. A table is provided listing principle particle and photon emissions from many commonly used radioisotopes.

**THE ****“****6CE” EQUATION **

It Is sometimes necessary to estimate the exposure rate which can be expected from a known quantity of a single gamma-emitting isotope. Such a situation might occur when ordering a significant quantity of radioactive material prior to an experiment, when purchasing a new calibration source, or in radiography work. An equation which addresses this need very nicely is

R = 6CE (f) / r^{2}

Where:

R • exposure rate at distance r (in Roentgens/hour)

6 = a unit conversion constant

C = activity (in Curies)

E = total photon energy (in MeV)

f = decimal fraction of photon yield

r = distance from the point source (in feet).

**SOME ****EXAMPLES **

*Example 1. *

This example deals with an Isotope which has only one photon energy listed in Table 1. Major Particle and Photon Radiations from Various Isotopes.

Your facility has ordered 5 Curies or Cesium-137 to be used in a density gauge. What would the expected unshielded exposure rate be at a distance of one foot? At three feet?

R = 6CE(f) / r^{2}

Where:

C = 5 Curies

E = 0.662 MeV (from Table I)

F = 0.889 (from Table I)

R = 1 foot and 3 feet

R = 6 (5) (0.662) (0.889) / 1^{2
}= 17.66 / 1

= 17.66R / hr

The exposure rate one foot away from a 5-Curie point source of Cs·137 will be approximately 17.66 R/hour.

At a distance of 3 feet the equation becomes

R = 6 (5)(0.662)(0.889) / 3^{2
}= 17.66 / 9

= 1.96 R/hr

The exposure rate three feet away from a 5-Curie point source of Cs-137 will be approximately 1.96 R/hour.

*Example 2:*

This example addresses a situation where an isotope has a more complex where an isotope has a more complex decay scheme, with several different¬ energy photons being emitted with various frequencies. In this case, the energy of each photon Is multiplied by its yield, then these products are added together to determine the value for “E(f).”

You have just received a shipment which contains 3 Curies of i-131. This small container must be transferred out of the shipping container. Estimate the exposure rate from this unshielded source at distances of 1 foot and 3 feet.

R = 6 C E (f) / r^{2}

Where:

C = 3 curies

E_{1} = 0.365 MeV (Table 1)

F_{1} = 0.812 MeV (from Table 1)

E_{2} = 0.637 MeV (from Table 1)

F_{2} = 0.073 MeV (from Table 1)

E_{3} = 0.284 MeV (from Table 1)

F_{3} = 0.061 MeV (from Table 1)

r = 1 foot and 3 feet.

As shown in the example below, the exposure rate one foot away from a 3-Curie point source of 1-131 will be approximately 6.46 R/hour.

R = 6 (3) [(0.365) (0.B12) + (0.637) + (0.073) + (0.284) + (0.061)] / 1^{2
}= 6 (3) [(0.296) + (0.046) + (0.0171] / 1

= 6 (3) (0359) / 1

= 6.46R/hr

At a distance of 3 feet, the equation becomes

R = 6 (3) (0.359) / 3^{2
}= 6.46 / 9

= 7.18 R/hr

The exposure rate three feet away from a 3-Curie source of 1·131 will be approximately 0.718 R/hour or 718 mR/hour.

**D****ISTAN****C****E IN METERS**

A variation of this equation can be used if the distance from the source is measured in meters rather than in feet.

If this is the case. the equation is stated

R = 0.5 C E (f) / r^{2}

R = exposure rate at distance r (in Roentgens/hour)

0.5 = a unit conversion constant

C = activity (in Curies)

E = total photon energy (in MeV)

F = decimal fraction of photon yield

R = distance from the point source (in meters)’

**LO****W ENERGY X·RAYS**

Due to the disruption of the electron cloud of some isotopes during decay (Auger electron and conversion electron emission), some x rays will also be emitted. A number of isotopes listed in Table 1 indicate the emission of x rays.

usually from the k- or I- shell. For use in this equation, include *only *x rays with energy greater than 70 keV (0.070 MeV). The reason for this has to do with now Roentgen is defined and measured. Also, the attenuation of photons in air becomes significant at these low energies.

*Example 3.*

Estimate the exposure rate 2 feet away from a 2·Curie point source of T1-201.

Table 1 shows that 11·20 1 emits 1 y per decay plus 2 x-rays with energies greater than 70 keV.

As shown in the example below, the exposure rate two feet away from a 2, Curie point source of T1-201will be approximately 201 mR/hour.

R = 6CE(f) / r^{2
}= 6 (2) [(0.167)(0.1) + (0.07)(0.46) + (0.08)(0.2)] / 2^{2
}= 6 (2)[(0.017) +(0.03) + (0.02)] / 4

= 6(2)(0.067) / 4

= 0.201R / hr

= 201mR / hr

**HI****G****H·****E****N****E****R****GY ****GAMMA RAY****S**

Some Isotopes emit relatively high energy gamma photons during decay. The equation only holds for gamma

energies up to 3.0 MeV. This is because the Roentgen unit is not used for photon energies greater than this value.

**POS****I****T****R****O****N EMITT****ERS**

For each positron (ß+) emitted during radioactive decay, eventually *two *gamma rays are produced. This occurs when the ß+ expends all of its kinetic energy and combines with a free electron. At this Instant, the masses of both of these particles is annihilated and converted into two photons. Since the rest mass of a positron (and of an electron) is approximately 5.486 E-4 atomic mass units (AMU), and since the energy equivalent of 1 AMU = 931.5 MeV. Then

(5.486 E – 4 AMU) (931.5 MeV / AMU)

= 0.511 MeV

The complete mass-lo-energy conversion of these two annihilating particles yields *two *gamma rays. Each with 0.511 MeV (511 keV) of energy. Since two photons are produced per ß+ emission, the yield listed in Table 1 usually approaches 2 per disintegration.

**S****UMMA****R****Y OF LIMITATI****O****N****S FO****R A****“CE” ****E****Q****UATI****O****N**

This equation provides a fairly accurate estimate of the exposure rate from a point source of a known quantity of radioactive material, for x- and gamma-ray energies (E) in the range

0.07 MeV < E < 3 MeV.

I In order for the answer to work out in units of Roentgens per hour, the activity *must *be given in *Curies, *and the photon energy in MeV. If *either *the activity is given in milliCuries, or the energy is in keV, the answer will be in units of milliRoentgens per hour. If *both *milliCuries *and *keV are used, the answer will be in microRoentgens per hour (µR/hour). As always, *watch **your **units.*

**C****O****NC****LUSION****S**

The “6CE” equation is a very useful tool which the Radiation Safety Officer can use to estimate exposure potentials and quantities of radioactive material present in various sources of gamma radiation.

**REFRENCES**

Table 1: Major Particle and Photon Radiations from Various Isotopes

*Cember, Herman. Introduction to HeaM Physics. New York: Pergamon Press, 1969, p. 151.**Kocher, David C. Radioactive Decay Data Tables – A Handbook of Decay Data for Application 10 Radiation Dosimetry and Radiological Assessments, DOE/TIC- JI026. Springfield, VA: National Technical Information Service. 1981 .*

**THE AUTHOR**

*K. Paul Steinmeyer is President of Radiation Safety Associates, Inc ., a firm specializing in radiological consulting and decontamination services. He is Editor of RSO Magazine. and Radiation Protection Management – The Journal of Applied Health Physics. He started his career in 1962 as part of the Navy nuclear submarine program. Originally published in RSO Magazine, August 1996. Radiation Safety Associates, Inc., 19 Pendleton Drive, P.O. BOX 107. Hebron, CT 06248*

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