Estimate the radiation rate at a new distance, or the distance for a target rate, from one reference reading using the inverse square law.
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The Inverse Square Law
Radiation from a point source spreads equally in all directions across an expanding sphere. Because the surface area of a sphere equals 4ฯrยฒ, the same total energy is distributed over a larger area as distance increases. Intensity therefore drops as the inverse square of distance rather than linearly. This geometric model applies when the source and detector are small relative to their separation, with comparable measurement response and negligible intervening attenuation or scatter. Extended sources, background and shielding can change the relationship.
Key limitation: source and detector size must be small compared with both measurement distances. A single source-size multiplier does not guarantee accuracy or safety for every geometry. This calculator does not model source shape, attenuation, shielding, scatter or background; verify conditions and measurements with qualified radiation-protection personnel.
Radiation Distance Formula
The following formula calculates the ideal-model distance corresponding to a supplied target rate from a known measurement point. The supplied target is not a regulatory or safety determination.
- Drad = estimated model distance corresponding to the target rate (same unit as D1)
- D1 = original measurement distance
- R1 = radiation rate measured at D1, using the same quantity and unit as R2
- R2 = supplied target rate, using the same quantity and unit as R1
Distance Multiplier and Dose Reduction Reference
The table below shows how dose rate scales relative to a baseline measurement at distance Dโ. These are geometric ratios under the point-source assumptions. They do not account for attenuation, scatter, background, detector response or shielding, and they do not establish an absolute safe rate.
| Distance from Source | Dose Rate Factor | % of Original | Practical Context |
|---|---|---|---|
| Dโ (baseline) | 1/1 | 100% | Reference survey point |
| 2 x Dโ | 1/4 | 25% | One doubling; dose drops 75% |
| 3 x Dโ | 1/9 | 11.1% | Geometric example; no safety boundary implied |
| 4 x Dโ | 1/16 | 6.25% | One sixteenth of the reference rate |
| 5 x Dโ | 1/25 | 4.0% | One twenty-fifth of the reference rate |
| 7 x Dโ | 1/49 | 2.0% | Approximately 2.04% of the reference rate |
| 10 x Dโ | 1/100 | 1.0% | 1% of the reference rate; absolute hazard still depends on the source and conditions |
Radiation Protection Context
Time, distance and shielding are components of radiation protection. This calculator models distance alone and cannot establish compliance, an exclusion boundary or individual health risk. Applicable limits depend on the exposure category and jurisdiction; use current requirements and qualified radiation-protection guidance. Exposure in R, absorbed dose in Gy, equivalent dose in Sv and instrument counts are distinct quantities and cannot be universally interchanged.
For a hypothetical unchanged point source measuring 60 mR/hr at 1 ft, the ideal model gives 60 / 3ยฒ โ 6.67 mR/hr at 3 ft. A total of 200 periods of 6 minutes is 20 hours, giving integrated exposures of 1,200 mR and about 133.3 mR at those respective rates; these are not automatically equivalent dose in mrem. Similarly, a hypothetical 2,000 mR/hr measurement at 1 m corresponds to 5 mR/hr at 20 m under the model. That 5 mR/hr value is an illustrative target, not a universal controlled-area limit; an extended patient source need not obey point-source scaling nearby.
How to Calculate Radiation Distance
The following examples demonstrate how to use the radiation distance formula.
Example Problem #1
- Determine the original distance: 50 ft
- Determine the original dose rate: 6 rem/hr
- Determine the required dose rate: 2 rem/hr
- Apply the formula: Drad = SQRT(50ยฒ x 6/2) = SQRT(7,500) = 86.60 ft
The model gives approximately 86.60 ft for the supplied 2 rem/hr target, a distance multiplier of SQRT(6/2) = SQRT(3) โ 1.732. This arithmetic example does not define a safe location or authorize exposure.
Example Problem #2: Illustrative Rate Target
Suppose a compatible radiation survey measurement is 120 mrem/hr at 2 ft from a point-like source. For an illustrative target of 5 mrem/hr, what distance does the inverse square model predict? This target is not a universal controlled-area limit.
Variables: D1 = 2 ft, R1 = 120 mrem/hr, R2 = 5 mrem/hr
Drad = SQRT(2ยฒ x 120/5) = SQRT(4 x 24) = SQRT(96) = 9.80 ft
The estimated model distance is approximately 9.80 ft, a multiplier of SQRT(24) โ 4.899 and a rate reduction factor of 24. Rounded numbers explain why 4.9ยฒ = 24.01 differs slightly from 24. Qualified surveys and applicable protection procedures must determine any actual boundary.
