# Answer to Question #12659 Submitted to "Ask the Experts"

*Category: Radiation Basics*

The following question was answered by an expert in the appropriate field:

I have read that ^{137}Cs fallout from Fukushima has been estimated at 1.48 × 104 GBq km^{-2} in some areas. What kind of a dose rate in Sv hr^{-1} at one-meter height would this amount to? I tried to approximate this with a surface calculation:

Sv hr^{-1} = π G C ln[(r^{2} + h^{2})/ h^{2}],

where

G | = | gamma constant for ^{137}Cs through air (Sv m^{2} Bq^{-1} s^{-1}), |

C | = | concentration in soil (Bq m^{-2}), |

r | = | source radius (564 m), and |

h | = | detector height (1 m). |

The results seem rather odd as you vary the parameters. I am told that the above calculation is really a point-source type calculation and not amenable to the use for which I have employed it. Is there any truly planar calculation that can be used instead that would give real-world results?

The surface contamination level that you cite seems very high for most of the areas outside of highly controlled areas in Fukushima. I will not question this further and will attempt to address your concern regarding the calculation you are attempting.

If we assume a disc-shaped area as you appear to have done, and assume no photon attenuation in media between the source and the dose point, and further assume that the dose point lies directly above the source geometric center, we would find that the mathematical expression that you used is correct. It is based on a point kernel approach in which differential area source elements are used to define the differential exposure rates at the dose point from the respective elements. Integration over the entire source area results in the equation you cite.

You have not been specific as to how your results were "odd" in your judgment. One problem with application of the equation to large area sources is that the pathlength through air between differential source elements and the dose point continues to increase as the radial distance from the disc center increases. The result is that the attenuation of primary photons in the air will become significant enough that the outer portions of the disc source will not contribute significantly to the dose rate at the dose point. Even in the lack of attenuation, the outer portions of the disc contribute less to the total dose rate than do portions closer to the dose point because of the usual inverse square reduction of the radiation intensity with increased distance.

To account properly for photon attenuation using the point kernel approach in the differential equation we would include an exponential attenuation term (of the form e^{-µx}, where µ is the linear attenuation coefficient, and x is the pathlength through air) and also a buildup term that accounts for the fact that many photons will undergo scattering events with air molecules and some of these scattered photons will add to the dose rate at the dose point. I carried out such a calculation, varying the source diameter to show the change in dose rate with disc radius.

Results are shown below, based on your given contamination level of 1.48 × 104 GBq km^{-2} and a dose point location centered one meter above the disc. I used a gamma ray exposure rate constant of 8.8 × 10^{-5} Sv m^{2} h^{-1} Bq^{-1} for ^{137}Cs. Results are shown for the case with no attenuation (same equation you have) and results also accounting for attenuation and buildup. The linear attenuation coefficient I used for the photons of interest in air was 9.30 × 10^{-3} m^{-1}. I used an analytical form of the buildup factor attributable to Berger, B = 1 + aµx e^{bµx}, where a and b are fitted parameters dependent on photon energy and medium traversed, µ is the linear attenuation coefficient, and x is the variable pathlength through air. The values I used for a and b were 1.331 and 0.149, respectively. Results were obtained using numerical integration. The exposure rate results with both attenuation and buildup would be the most accurate; the results with attenuation but no buildup are shown so that you can see the significance of excess dose associated with scattered photons. Even though photon attenuation becomes more significant as distances traversed by the photons becomes large as the radius increases, at large radii the source is not contributing greatly to the dose rate because of the inverse square reduction, and the buildup adds progressively a somewhat larger fraction to the dose rate as distances increase. It is clear that once the radius exceeds about 200 meters the dose rate does not change appreciably.

Radius (m) | Exposure rate (C kg^{-1} hr^{-1}) with various attenuation characteristics |
||
---|---|---|---|

None | With Buildup | With No Buildup | |

10 | 4.64 × 10^{-07} |
4.64 × 10^{-07} |
4.46 × 10^{-07} |

100 | 9.29 × 10^{-07} |
9.29 × 10^{-07} |
7.74 × 10^{-07} |

200 | 1.06 × 10^{-06} |
1.06 × 10^{-06} |
8.15 × 10^{-07} |

400 | 1.21 × 10^{-06} |
1.14 × 10^{-06} |
8.26 × 10^{-07} |

564 | 1.29 × 10^{-06} |
1.14 × 10^{-06} |
8.26 × 10^{-07} |

I hope this is helpful to you. I am not sure what your background is so that I am not sure how much detail you desire or require. If you require more information, please contact me.

George Chabot, PhD