Help:Dose Rate Constants++

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Air kerma

The dose rate from a point source of photons is given by

\dot{D} = A/r^2 \cdot \Gamma

where the dose rate constant \Gamma is defined as the dose rate from a point source with an activity A(Bq) at a distance r = 1m. The air kerma is given by (Appendix 4.2):

\dot{K}_{air} = \phi \cdot E_{\gamma} \cdot (1-g)^{-1} \cdot (\mu_{en}/{\rho})_{air}


 \phi is the photon fluence rate (photons per unit area per unit time)

 E_{\gamma} is the photon energy

 g is the fraction of energy of secondary electrons converted to bremsstrahlung

 (\mu_{en}/{\rho})_{air  is the mass energy absorption coefficient for air

The photon fluence rate from a point source with activity A is, assuming one photon per disintegration, given by

\phi = A/(4 \pi r^2)

In the more general case, photons with different energies  E_{i} are emitted, each with an emission probability of  p_{i} . The photon fluence rate for photons of energy  E_{i} is  \phi_{i} where

\phi_i = A/(4 \pi r^2) \cdot p_i

The air kerma rate for photons of energy  E_{i} is given by

\dot{K}_{air,i} = \phi_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot (\mu_{en}/{\rho})_{air,i}


\dot{K}_{air,i} = A/(4 \pi r^2)\cdot p_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot (\mu_{en}/{\rho})_{air,i}

Air kerma rate constant, ΓKair

From the previous section (see also Appendix 4.2)

\dot{K}_{air,i} = A/(4 \pi r^2)\cdot p_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot (\mu_{en}/{\rho})_{air,i}

The air kerma rate constant is given by

 \Gamma_{K_{air}} = \sum_{i} \Gamma_{K_{air},i}


\Gamma_{K_{air},i} = 1/(4 \pi)\cdot p_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot (\mu_{en}/{\rho})_{air,i}


\Gamma_{K_{air}} = 0.04590 \cdot \sum_{i} p_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot (\mu_{en}/{\rho})_{air,i}


 \Gamma_{K_{air}} has units µGy.m2.h-1.MBq-1

 E_{i} is the photon energy in keV

 (\mu_{en}/{\rho})_{air  is the mass energy absorption coefficient for air with units

Exposure rate constant, Γx

The conversion between the exposure rate and the air kerma rate for a given photon energy is given by

\dot{X} = \dot{K} \cdot (1-g) \cdot (W/e)^{-1}

where W/e = 33.97 ± 0.05 J/C is the ionisation constant for dry air. The exposure rate constant is then given by

\Gamma_{X} = 1.3510 \cdot 10^{-15} \cdot \sum_{i} p_i \cdot E_{i} \cdot (\mu_{en}/{\rho})_{air,i}


 \Gamma_{X} has units

 E_{i} is the photon energy in keV

 (\mu_{en}/{\rho})_{air  is the mass energy absorption coefficient for air with units

Ambient dose rate constant, ΓH*(10)

The ambient dose equivalent is related to the air kerma through

 H^*(10) = \sum_{i} K_{air,i} \cdot (H^*(10)/K_{air})_i

The ambient dose rate constant is then given by

\Gamma_{H^*(10)} = 0.04590 \cdot \sum_{i} p_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot (\mu_{en}/{\rho})_{air,i} \cdot (H^*(10)/K_{air})_i


\frac{H^*(10)}{K_{air}} = \frac{x_i}{(1.465 \cdot {x_i}^2 - 4.414 \cdot x_i + 4.789) } + 0.7006 \cdot arctan(0.6519 \cdot x_i)

and x_i = ln(E_i/E_0)


 \Gamma_{H^*(10)} has units µSv.m2.h-1.MBq-1.

 E_{i} is the photon energy in keV

 (\mu_{en}/{\rho})_{air  is the mass energy absorption coefficient for air with units

H^*(10)/K has units Sv/Gy

The ratio H*(10)/Kair as a function of the photon energy E(keV)

From the above dose rate quantities, it is straightforward to calculate the air kerma rate, the exposure rate, and the ambient dose rate by multiplying by A/r2 where A is the activity (MBq) and d(m) is the distance from the source.

Inclusion of short-lived decay products in the evaluation of the dose rate constants

For the calculation of the dose rate constants, some nuclides are considered to be in equilibrium with daughter products. This is the case when a single radioactive decay chain in which radionuclides are present in their naturally occurring proportions, and in which no daughter nuclide has a half-life either longer than 10 days or longer than that of the parent nuclide, shall be considered as a single radionuclide. Such nuclides are denoted with an asterisk in the Nuclide Summary tab e.g. Cs-137*.

In the case of radioactive decay chains in which any daughter nuclide has a half-life either longer than 10 days or greater than that of the parent nuclide, the parent and such daughter nuclides shall be considered as mixtures of different nuclides.

Using dose rate constants for nuclides with short half-life daughters

(Qu.) How do I calculate the gamma dose rate for a nuclide which has short half-life daughters using the gamma dose rate constants?

Consider the decay of initially 1 MBq pure Cs-137. After a few minutes, secular equilibrium will be achieved. At this point the activities are as follows:

Cs-137: 1e6 Bq

Ba137m: 0.944 e6 Bq

Total: 1.94 e6 Bq.

The dose rate constant from Tschurlovits et al. for Cs-137 (in equilibrium with daughters) = 0.927e-1 mSv.m2/(h.GBq).

How do I calculate the gamma dose rate a 1m for 1 MBq Cs-137? Which activity do I multiply the dose rate constant by:

1. Cs-137

2. Ba137m

3. Total

(Ans.) The gamma dose rate only comes from Ba-137m. This gamma dose rate is included in the dose rate constant of Cs-137.

Therefore you have to multiply the Cs-137 activity with the Cs-137 dose rate constant i.e.

A = 1 MBq

H*(10) Doserate constant Cs-137 = 0.9271E-1 mSv.m2/(h*GBq)

Doserate H*(10) at 1 m = 0.9271E-4 mSv/h

To check this calculation one can consider the source (only for gamma) as a pure Ba-137m (as Cs-137 has no gamma). Then you can start with the activity of Ba-137m

A = 0.944 GBq

In this case you must calculate with the Ba-137m dose rate constant you'll also find in the Tschurlovits paper:

H*(10) Doserate constant Ba-137m = 0.9800E-1 mSv.m2/(h*GBq)

Doserate H*(10) in 1 m = 0,9251E-4 mSv/h

which is approx. the same value as calculated above for Cs-137.


- J. Kesten, private communication, 2016

- M. Tschurlovits et al., Dose rate constants for new dose quantities, Vol. 42 no. 2 pp.77-82 (1992).

- A. Steurer et al., Different values for dose rate constants in radiation protection literature – reasons and consequences in practice. Full paper: See also

- International Commission on Radiation Units and Measurements "Measurements of Dose Equivalent from External Photon and Electron Radiations", ICRU Report 47 (1992)

- ICRU Radiation Quantities and Units, Report 33 (Bethesda, MD: ICRU Publications) (1980)

- ICRU Determination of Dose Equivalents Resulting from External Radiation Sources. Report 39 (Bethesda, MD: ICRU Publications) (1985)

- DIN 6814-2 Terms in the field of radiological technique – part 3: Dosimetry (g-factors)

- S. R. Wagner et al., Unified Conversion Functions for the new ICRU Operational Radiation Protection Quantities, Radiation Protection Dosimetry, Vol. 12 No. 2 p.231-235 (1985).

- J. E. Turner, Atoms, Radiation, and Radiation Protection, 2nd Edition, Wiley 1995.

The Photon Dose Rate Constants application

The new Photon Dose Rate Constants++ application in Nucleonica allows the user to calculate the internationally accepted

  • Ambient Dose Equivalent Rate Constants,  \Gamma_{H*(10)}
  • Air Kerma Rate Constants,  \Gamma_{K_{air}}
  • Exposure Rate Constants  \Gamma_{X}}

for between 1300 and 1500 gamma and X-ray emitting nuclides (depending on which database is selected). These quantities are useful for estimating the dose rates from point sources where attenuation and scattering in the source and in air are negligible. From the dose rate constants, the ambient dose rates dH*(10)/dt, air kerma rates dKair/dt, and exposure rates dX/dt can then be obtained using the relations given in section 1.

In contrast to previously defined dose rates, the ambient dose rate H*(10) accounts for absorption and scattering in the human body by simulating the human body through a phantom (the ICRU sphere of 30 cm diameter made of tissue equivalent material).

Above the data grid (shown below), the selected nuclide is shown as an image and in the Nuclide selection box. Also shown are the

  • Low energy threshold (keV) below which energies are neglected in the calculations
  • Data set to be used in the calculations (currently JEFF3.1 and ENDF/B-VII.1 are available)

In the nearby check boxes. it is also possible to show details of the calculations and to include short-lived daughters. The points are discussed in more detail in the sections below.

At the bottom of the datagrid, the total number of nuclides with gamma and X-ray data is given. As can be seen, for the selected dataset (JEFF3.1), there are 1487 nuclides which have gamma/X-ray energies above the threshold energy (in this case 20 keV). Also shown are the total number of nuclides for which short-lived daughters (denoted by *) have been accounted for in the calculations. The number in ()following the * gives the number of daughter products included in the calculations. More details on the actual daughter products included are given in the tab Dose Rate Constants.

Dose Rate Constants: Nuclide summary

Nuclide Summary

In the main Nuclide Summary tab in the datagrid, the selected nuclide Ac 226 is shown highlighted. In the adjacent columns Z, and A together with the dose rate constants are given. In the nuclide list shown in the first column, an asterisk following the name of the nuclide indicates that short-lived daughters have been accounted for in the calculation. In the second column this asterisk is shown separately such that this column can be used to rearrange the table contents e.g. with the list of all nuclides with short-lived daughters shown at the top of the table.

Directly above the datagrid, there is a wiki link to "Short-lived decay products included" in the calculations.

To obtain the data for another nuclide, the nuclide name can be entered directly into the Nuclide drop-down menu or by scrolling through the datagrid and clicking on the nuclide of interest.

Dose Rate Constants

Once a nuclide has been selected, the Dose Rate Constants tab shows the short-lived daughters included in the calculations. In the example shown below for Ac 227, a total of 9 nuclides (1 parent + 8 daughters) are listed. The summed dose rate constants are shown at the bottom of the datagrid.

Dose Rate Constants tab showing the short-lived daughters of Ac 227 included in the calculations.

Using the Show details checkbox, further details of the calculations can be shown. As can be seen in the example below for Ac 227, in total 221 gamma energies and emission probabilities from 9 nuclides have be used in the calculations.

Dose Rate Constants tab showing the details of all the energies and emission probabilities used in the calculations.

Short-lived daughters

To see the daughters nuclides of Ac 226, the user should click on the decay tree icon adjacent to the nuclide box. Only daughter nuclides with a half-life less than 10 days or less than the half-life of the parent nuclide are accounted for in the calculations. Using the decay tree icon and the excerpt from the nulcide charge, the user can understand which nuclides are included in the calculations.

Decay tree for Ac 226 showing the daughters highlighted

Air absorption coefficients

Dose Rate Constants tab showing all energies and emission probabilities for Ac 226

Bremsstrahlung energy fraction, g

Dose Rate Constants tab showing all energies and emission probabilities for Ac 226


The results obtained by Nucleonica's Dose Rate Constants application have been compared with the results published by Tschurlovits et al; in 1992. The details are given in the link below.

In summary: 182 parent nuclides were considered. Of these 182 nuclides, there was agreement to better than 10% for 150 nuclides. For 29 nuclides, agreement was in the range 10-100%. For 3 nuclides, results disagreed by more than 100%.

Validation of the Dose Rate Constants results with the published values from Tschurovits et al, 1992

Shielding and buildup

If a shield material is present, the attenuation and buildup in the shield have to be accounted for. The photon fluence rate for each photon is then

\phi_i = A/(4 \pi r^2) \cdot p_i \cdot B_i \cdot exp(-\mu_{sh,i} d)

where B_i is the buildup factor for photons of energy E_i,

\mu_{sh,i} is the attenuation coefficient for photons of energy E_i,

and d is the shield thickness.

This relation can also be expressed as

\phi_i = A/(4 \pi r^2) \cdot p_i \cdot B_i \cdot exp[-(\mu_{sh,i}/\rho)_{sh} \cdot (\rho d)_{sh}]

However in the following sections, the simplified version exp(-\mu_{sh,i} d) is used.

Air kerma rate constant and air kerma rate

It follows from

\Gamma_{K_{air}} = 1/(4 \pi)\cdot \sum_{i} p_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot B_i \cdot exp(-\mu_{sh,i} d) \cdot (\mu_{en}/{\rho})_{air,i}


\Gamma_{K_{air}} = 0.04590 \cdot \sum_{i} p_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot B_i \cdot exp(-\mu_{sh,i} d) \cdot (\mu_{en}/{\rho})_{air,i}

the air kerma rate is given by

\dot{K}_{air} = A/r^2\cdot \Gamma_{Kair}

with units

E_i (keV),

\mu / \rho (m2/kg),


\Gamma_{K_{air}} (µSv.m2.h-1.MBq-1))

Exposure rate constant and exposure rate

Similarly, it follows

\Gamma_{X} = 1.3510 \cdot 10^{-15} \cdot \sum_{i} p_i \cdot E_{i} \cdot B_i \cdot exp(-\mu_{sh,i} d) \cdot (\mu_{en}/{\rho})_{air,i}

The exposure rate is given by

\dot{X} = A/r^2\cdot \Gamma_{X}

with units

E_i (keV),

\mu / \rho (m2/kg),


\Gamma_{X} (µGy.m2.h-1.MBq-1))

Ambient dose rate constant and ambient dose rate

From a previous section

\Gamma_{H^*(10)} = 0.04590 \cdot \sum_{i} p_i \cdot E_{i} \cdot (1-g_i)^{-1} \cdot B_i \cdot exp(-\mu_{sh,i} d) \cdot (\mu_{en}/{\rho})_{air,i} \cdot (H^*(10)/K_{air})_i

The ambient dose rate is given by

\dot{H}^*(10) = A/r^2\cdot \Gamma_{H^*(10)}

with units

E_i (keV),

\mu / \rho (m2/kg),


H^*(10) (µGy.m2.h-1.MBq-1))

Photon dose equivalent rate, HX

The photon dose equivalent Hx (measured in Sv) is a quantity introduced in Germany in 1980. It became the legal quantity in Germany on 1 January 1986. Hx was an interim solution because at that time no international agreement on dose equivalent quantities had been achieved. In Germany Hx was replaced by SI quantities such as H*(10) on 1 August 2001. Hx was not accepted internationally. Nevertheless, it is interesting to describe the background behind Hx.

Most instruments available up to the early 1980s were designed for Exposure (rate) and calibrated in R(/h). The question arose whether these instruments could still use them to measure dose equivalent. The answer was yes, because Exposure is a good estimate for the dose equivalent of photons in tissue. Therefore Hx was defined as

H_X(Sv) = 0.01(Sv/R) \cdot X(R).

Using the relation, 1 R = 2.58 x 10-4 C/kg of air,

H_X(Sv) = 0.01Sv/(2.58 \cdot 10^{-4} C \cdot kg^{-1}) \cdot X(C \cdot kg^{-1}),


H_X(Sv) = 38.76 \cdot X(C \cdot kg^{-1})

Since this conversion does not depend on photon energy, Hx and Exposure X are strongly related quantities; they just differ by the factor 100. This is why some instruments allow the user to select either R or Sv as the unit. Basically Hx was not really a new quantity. It was more the old quantity Exposure in a new wrapping.

Using the relation X(C/kg) = K_{air}(Gy) \cdot(1-g_a) \cdot (W/e)^{-1} gives

H_X(Sv) = 38.76 \cdot K_{air}(Gy)·(1-g_a) \cdot (W/e)^{-1}


H_X(Sv) = 1.141 (Sv/Gy) \cdot K_{air}(Gy)

Nucleonica (previous) photon dose equivalent rate, Htis

The dose rate in tissue calculated in Nucleonica is

\dot{H}_{tis} = \phi \cdot (\mu_{en}/{\rho})_{tis}


\phi is the energy fluence (energy per unit area per unit time),

(\mu_{en}/{\rho})_{tis} is the mass energy absorption coefficient in tissue.

The above formula is value for a thin layer of tissue in which there is no attenuation and no backscattering from other layers. It follows

\dot{H}_{tis} = \phi \cdot (\mu_{en}/{\rho})_{air} \cdot [(\mu_{en}/{\rho})_{tis}/(\mu_{en}/{\rho})_{air}] = K_{air} \cdot (1-g_a) \cdot [(\mu_{en}/{\rho})_{tis}/(\mu_{en}/{\rho})_{air}]

Since the ratio of the mass absorption coefficients in air to tissue is approx. 1.10, then H_{tis} \cong  1.10 \cdot K_{air}


Relationship between Exposure (X) and Absorbed Dose (D) in air

The definition of Exposure (X) is given by

X = \frac{dQ}{dm}

where dQ is the charge produced in a mass dm of air. Consider the creation of 1 Coulomb of charge in 1 kg air i.e. X = 1C/kg air. This corresponds to the input of a certain amount of energy into the mass dm. The average energy dissipated in the production of a single ion pair is 34 eV. The charge on an ion is 1.6x10-19 C.

Number of ion pairs formed = 1C/(1.6x10-19 C).

For each ion pair 34 eV energy is required. So the total amount of energy required to produce 1 C of charge is

Total energy required to produce 1 C = 34eV / 1.6x10-19 = 34 . 1.6x10-19 CV / 1.6x10-19 = 34 J

It follows that the absorbed dose in 1 kg air is 34 J / 1 kg air or 34 Gy (in air).

In summary:

If 1 C of charge is produced in air

Exposure X is given by

X = \frac{dQ}{dm} = 1^{-1}

Absorbed dose D is given by

D = \frac{dE}{dm} = 34^{-1} = 34 Gy

General relationship:

D = 34 \cdot X

where X is in units of C/kg and D is in Gy. Note also, if the dose rate in air is known, then the exposure follows from:

X = D/34 = 2.941\cdot 10^{-2} \cdot D

Relationship between Kerma (K), Absorbed Dose (D), and Exposure (X)

The basic relations for the kerma and the absorbed dose are given by:

K = I \cdot (\frac{\mu_{tr}}{\rho})

D = I \cdot (\frac{\mu_{en}}{\rho})

where \mu_{tr} is the linear energy transfer coefficient, \mu_{tr} is the linear energy absorption coefficient, ρ is the density of the medium, D the absorbed dose and K the kerma, and I the energy intensity of the radiation (Jm-2.s-1) The above relations are valid for any media. K is the kinetic energy released or transferred per unit mass to the medium. K refers to the energy removed from the beam and D refers to the energy absorbed locally in the medium. K can be considered as consisting of two parts, i.e.

K= K_{coll}+K_{rad}= K_{coll}+gK

where g is the fraction of energy lost by radiative process (e.g. bremsstrahlung) photons. The quantity (1-g) is the fraction of energy lost by collisional process. It follows that

K(1-g)= K_{coll}

or g = 1-\frac{\mu_{en}}{\mu_{tr}}

Now Kcoll corresponds exactly to the energy absorbed locally in the medium, hence

K(1-g)= D

Again this relation is valid in any media. When the medium is air we have:

K_{air}(1-g)= D_{air} = (W/e)\cdot X

or alternatively

X = K_{air}\cdot (W/e)^{-1}(1-g)

This is the basic relation between the exposure in air and air kerma. (W/e) is just the energy required to produce an ion pair in air (i.e. the 34 eV mentioned previously). In this relation, X is given in C/kg, (W/e) in eV, and Kair in Gy.

Mass attenuation µ/ρ, Energy transfer µtr/ρ, and Energy absorption µen/ρ coefficients

The linear attenuation coefficients for photons of a given energy in a given material comprises the individual contributions from the photoelectric effect (τ), Compton scattering (σ) and pair production (κ), i.e.

\mu = \tau + \sigma + \kappa

In terms of the mass attenuation coefficients, the above relation becomes:

\mu = \frac{\tau}{\rho} + \frac{\sigma}{\rho} + \frac{\kappa}{\rho}

Small contributions from photonuclear reactions and Rayleigh scattering are neglected.

Note that not all the energy of the incoming photon that interact with a small region are necessarily absorbed there. This leads to the concepts of energy transfer coefficients (µtr/ρ) and energy absorption coefficients (µen/ρ).

Photoelectric Effect:

Here the absorption of a photon of energy hv produces a secondary electron with initial kinetic energy T = hν – B where B is the binding energy of the ejected electron. Following ejection of the photoelectron, the inner shell vacancy is filled immediately by an electron from an upper lever. This and subsequent electronic transitions are accompanied by the simultaneous emission of photons or Auger electrons. The fraction of incident energy transferred to the electrons is (1-δ/hν) where δ is the average energy emitted as fluorescent radiation. The mass energy transfer coefficient is therefore:

\frac{\tau_{tr}}{\rho} = \frac{\tau}{\rho} \cdot (1-\frac{\delta}{h\nu})

where the quantity τtr/ρ gives the fraction of the intensity that is transferred to electrons. To the extent that the photoelectron and Auger electrons subsequently emit photons (as bremsstrahlung), the energy transfer coefficient does not adequately describe energy absorption in the region.

Compton scattering:

For Compton scattering of mono-energetic photons, the mass energy transfer coefficient is given by

\frac{\sigma_{tr}}{\rho} = \frac{\sigma}{\rho} \cdot \frac{T_{avg}}{h\nu}

where the factor T/hν gives the average fraction of the incident photons energy that is converted into initial kinetic energy of the Compton electrons. As with the photoelectric effect, the energy transfer coefficient takes no account of subsequent bremsstrahlung by the Compton electrons.

Pair production:

A photon of energy hν produces an electron-positron pair with a total initial kinetic energy of hν – 2mc2, where 2mc2 is the rest energy of the pair. Thus the mass energy transfer coefficient for pair production is given by

\frac{\kappa_{tr}}{\rho} = \frac{\kappa}{\rho} \cdot (1-\frac{2mc^2}{h\nu})

The total mass energy transfer coefficient µtr/ρ is then given by

\frac{\mu_{tr}}{\rho} = \frac{\tau_{tr}}{\rho} + \frac{\sigma_{tr}}{\rho} + \frac{\kappa_{tr}}{\rho}


\frac{\mu_{tr}}{\rho} = \frac{\tau}{\rho}(1-\frac{\delta}{h\nu}) + \frac{\sigma}{\rho} \cdot \frac{T_{avg}}{h\nu} + \frac{\kappa}{\rho}(1-\frac{2mc^2}{h\nu})

This coefficient determines the total initial kinetic energy of all electrons produced by the photons, both directly (as in photoelectric absorption, Compton scattering, and pair production) and indirectly (as Auger electrons).

Except for the subsequent bremsstrahlung that the electrons might emit, the energy absorbed in the immediate vicinity of the interaction site would be the same as the energy transferred there.

Letting g represent the average fraction of the initial KE transferred to the electrons that is subsequently emitted as bremsstrahlung, the mass energy absorption coefficient is defined as

\frac{\mu_{en}}{\rho} = \frac{\mu_{tr}}{\rho} \cdot (1-g)


\mu = \frac{\tau}{\rho} + \frac{\sigma}{\rho} + \frac{\kappa}{\rho}

\frac{\mu_{tr}}{\rho} = \frac{\tau_{tr}}{\rho} + \frac{\sigma_{tr}}{\rho} + \frac{\kappa_{tr}}{\rho}


\frac{\mu_{tr}}{\rho} = \frac{\tau}{\rho}(1-\frac{\delta}{h\nu}) + \frac{\sigma}{\rho} \cdot \frac{T_{avg}}{h\nu} + \frac{\kappa}{\rho}(1-\frac{2mc^2}{h\nu})

\frac{\mu_{en}}{\rho} = \frac{\mu_{tr}}{\rho} \cdot (1-g)

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