If you know that the radioactive source is, for example, Cesium-137, is it possible to extrapolate a relationship between the count rate and radiation intensity? If it is not possible, what is the minimum information you need to determine this, e.g. activity?

Furthermore, are the units for radiation intensity still $\text{W m}^{-2}$?

  • 1
    $\begingroup$ Intensity and count rate are directly proportional. $\endgroup$
    – DanielSank
    Aug 27, 2014 at 0:08
  • $\begingroup$ Daniel got it right, the count rate should be proportional to the intensity of the source. The problem is, that the proportionality constant is very hard to estimate from geometry alone. One usually needs a calibrated source to calibrate the absolute sensitivity of a counter (and it is usually good to avoid an absolute measurement to begin with). And near the saturation limit of the counter, the measured count rate is lower than the theoretical count rate, of course. $\endgroup$
    – CuriousOne
    Aug 27, 2014 at 0:46

2 Answers 2


The activity of a radioactive source is measured (SI units) in Bq - Becquerels. One Bq = 1 disintegration per second. Frequently you will see the Curie (Ci) which is $37 \cdot 10^9 Bq$.

The energy of radiation depends on the decay scheme. For example, for Cs-137 you find (source: http://upload.wikimedia.org/wikipedia/commons/6/66/Caesium-137_Decay_Scheme-de.svg)

enter image description here

Here you see that there are two different ways for the Cs-137 to decay: one gives rise to Ba-137 with the emission of a $\beta -$ (electron) with energy up to 1.17 MeV, while the other goes through an intermediate (metastable) 137-Ba which subsequently decays to stable 137-Ba with the emission of a gamma ray with 662 keV of energy.

If you want to include the beta energy in your "intensity" calculation you will find that extremely hard since there is a LOT of self-absorption (betas don't travel very far in matter). If you are only interested in the gamma radiation, then you can find the total energy per second (emitted into $4\pi$ steradians) as

$$energy = activity (Bq) * 0.95 * 662 * 10^3 * 1.6 * 10^{-19} J$$

At a given distance $r$, the total area is of course $A = 4\pi r^2$ so you can divide energy by area and get intensity. It would be unusual to express this in $W/m^2$.

When you look at radiation damage, you actually do use a measure of energy. The Gray is the SI unit, expressed as $J/kg$ - in other words, the amount of energy deposited per kg of material absorbing. That depends not only on the radiation emitted, but also on the material receiving. Materials with higher Z will typically stop more energy per unit mass and thus have higher values for radiation dose. Note that the Gray is used for non-living materials. For biological materials, the Sievert (Sv) is preferred since it represents "damage" and not just "absorbed energy". For a more complete explanation see for example http://en.wikipedia.org/wiki/Gray_(unit)

  • $\begingroup$ I interpreted the question to be about counting and answered it that way as well as adding the appropriate tags, but the clearly draw distinction between energy and activity here is important as the OP's text is ambiguous. $\endgroup$ Aug 27, 2014 at 1:50
  • $\begingroup$ @dmckee - "count rate" is a terribly loose concept; after seeing your answer I realize that one could go into completely different directions upon reading the question. Let's see how OP responds to our respective answers -- this will clarify what the question really was... $\endgroup$
    – Floris
    Aug 27, 2014 at 1:52

As noted in the comments the count rate is (to first order) proportional to the activity: $$ \text{rate} = k \cdot \text{activity} \,.$$

The constant of proportionality is

$$ k = \text{acceptance} \times \text{efficiency} \times \text{fractional live time} \,,$$

where acceptance is a geometric factor, quantum efficiency is a property of the detector and live time is related to both the detector and the actual rate of count arrivals (i.e. $\text{activity} \times \text{acceptance} \times \text{efficiency}$).

That means that live time can actually introduce a degree of non-linearity, but as long as you keep live time near 1.0 (fully live) the level of non-linearity is generally negligible.

All of these terms can be estimated to one degree or another, but for high precision you generally have to measure them in situ. No one said experimental science was easy.

First order estimation:

  • The acceptance is approximated from the active area $a$ presented to the source and the distance $r$ from the source to the detector by $$ A \approx \frac{a}{4\pi r^2} \,. $$ The approximation is very good when $r$ is much larger than both the linear size of the source and the linear size of the active area on the detector.

  • Efficiency. You generally have to trust the manufacturer on this, and you may have to multiply together several efficiencies of sub-components or do some calculating of the expected number of photons arriving at the PMT face or the like. A bit of an art.

  • Live time. Define the fractional live time as $$\ell = \frac{\text{time the detector was available to record an event}}{\text{length of the whole data collection period}} \,.$$ As long as the fractional dead time $d = 1 - \ell$ is much less than one you can estimate the actual dead time as $D = N \tau$ where $N$ is the total number of counts recorded and $\tau$ is the time the detector spends in active as a result of a single count. Getting the live time correction for more complicated situation can be very difficult indeed.

  • $\begingroup$ Completely valid, and a totally different direction than I took my answer... $\endgroup$
    – Floris
    Aug 27, 2014 at 1:52

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