Radiation Physics - Science topic

If you computational analysis for a similar system then here's a code that can do computational XPCS: https://gitlab.com/micronano_public/c-xpcs

Hello Thomas Cuff

In order to present the problem clearly, I have posted a picture of the chamber geometry with the thickness values of refractories. It is also noticeable that I have considered the radiation heat transfer inside the chamber.

High temperature on the Carbon steel layer is obtained !!!

https://www.xsplot.com can plot neutron cross sections for isotopes and I think it is the only one that can also plot materials as well. Materials cross sections are macroscopic cross section and require isotopes fractions and a material density which needs a bit more processing compared to simpler isotopes plots. Also if it doesn't do what you want it is open source so it can be modified and improved

معلومات قيمه

no, i am talking about polar angle that is used in General particle source (GPS) in Geant4.

in this we have to give theta maximum and theta minimum according to our object so most of radiation should incident on my object.

but i don't know how can i calculate this for my object.

Hello Recep Kurtulus ,

The statement by Gerhard Martens is not contradicted, and the statement by Nobuyuki Hamada is supported (with the caveat that CeO₂ is added to stabilize the color of the radiation shielding and radiation-resistant optical glasses) in the following book.

[1] Heinz G. Pfaender, (revised and expanded by) Hubert Schroeder; Schott Guide to Glass; Van Nostrand Reinhold Company; 1983; p. 146.

Regards,

Tom Cuff

:-Reyliegh Method or Thomson Method

:-Or in Quantum Electrodynamics by Feynman rules

:-Feynman -Schwinger theory

:-Julious Schwinger

Please check out the user manual especially the sections about tallies as well as Vid Merjlak mentioned a paper. Those all would be useful for you.

there are many and you can use the equations or some standard curves for every shielding material for software GEANT , Gorbit can be used

Francisco is correct in his comments. I will add a few as well.

Calculating MU/time and calculating dose are the inverse of one another. You need to be given MU to calculate dose, or you need prescribed dose to calculate MU. Typically one is tasked with calculating MU or time after the dose is prescribed by the Radiation Oncologist. In simple calculations the dose would be prescribed to a point in the body and the field apertures shaped to conform to the local anatomy, followed by normalizing to an isodose line that covers the desired area. All of this goes into the calculation of MU or time needed to get the prescribed dose. Modern techiniques involve the use of inverse planning utilizing arcs, dynamic MLC's, etc. It is always important to optimize the dose distribution, whether in linac based therapy or cobalt teletherapy. The ability to optimize the dose distribution is tied to the imaging available (2D, 3D, 4D, PET, MRI, etc.) the technology contained in the treatment planning system, and the skill of the planner.

This is simple calibration. Divide 6 mR/h by cpm. You are calibrated for that radionuclide at that particular geometry.

You can also read the attached useful paper.

I'm not sure, if someone can provide measured data as gamma (photon) spectroscopy for such X-ray tubes is hard to perform. Scintillator-based detectors mostly have a too rough energy resolution for this purpose. Semiconductor-based detectors are mostly too sensitive resulting in dead time problems. One can increase the distance to the X-ray tube or use collimation to decrease the photon flux photon, but then you will loose low energy photons which would otherwise contribute to your spectrum. In addition, low energy photons are also absorbed inside the cover of the detector (stainless steel cap for cooling).

But maybe you are lucky and someone solved these problems :-)

It is likely that FDM 1000 from Samyoung Unitech (Korea) is the energy compensated Guiger-Müller tube with integrated dose range from 0.01 µSv to 10 Sv. In your measurement, the dose you measured (31.73 µSv in 45 minutes) is in this range, but the source to detector/chamber distance (SDD/SCD) is 1 cm. So, simple conversion from the dose you measured to radioactivity of the source may not apply depending on the size and shape of the source.

It sounds like you have 0.11 R/hr, then you add 2 cm Pb shielding.  So you will calculate a new dose rate after the shielding.  Using 2 cm Pb, and Cs-137 gamma (0.66 MeV), I get a reduction of about 0.14: That is, the 0.11 rem/hr dose rate would be reduced to 0.14 of this, or about 0.016 mrem/hr.  In order to get a stay time, you need an allowable dose. If you assume 0.1 Rem (100 mrem or 1 mSv), then your stay time would be 0.1 rem / 0.016 rem/hr, or about 6 hours.

There are no stopping powers for neutrons, as neutrons are uncharged.  The stopping power is the energy loss per unit distance (cm or g/cm^2) due to the interaction of a charged particle with the electrons of the medium.

Physically, the absorbed dose is the same for mice and humans. Biologically, however, effects of the physically same absorbed dose are not necessarily the same. Moreover, the radiosensitivity varies among strains of mice, and its choice depends on your endpoint (e.g., which type of cancer).

The idea to mention the gamma radiation sensititvity in Rad/h is applied when neutron tubes are used in a combined neutron and gamma radiation field.

Neutron tubes, typically He3 proportional counter tubes, are somewhat sensitive to gamma radiation as well. If the gamma field is high, then the tube may falsely start to count the gamma's as neutron pulses. The Rad/h indication means that the tube is insensitive for gamma radiation field till that declared value.

This gamma radiation seen by a tube is typically less of energy than a neutron pulse that would be detected by the counter. Some solutions exist to make a tube less gamma sensitive. Suppliers can use special coating, specific less sensitive gazes or a tube base preamplifier with an LLD potentiometer to cut of that low energy part to obtain a neutron tube that is less sensitive to gamma radiation.

Looking at the relation of effective dose vs. equivalent dose, E/H*(10) is below 1 at least up to 10 MeV. So it is a good conservative estimate for photons < 10 MeV. For radiation > 10 MeV (as R. Schnell pointed out, such exposures are not very likely) this will more or less hold true because more and more of the radiation will go through the tissue without interaction. The half thickness  for 10 MeV photons  is already  around 30cm.

In Ge detectors, the depletion voltage is set as such (typical 500V in addition) that small voltage changes don't influence the charge collection, or energy calibration

In NaI, the HV is however set specifically to have a certain energy calibration value. Small changes in HV will influence the spectrum (channel to energy conversion), and most known phenomena is that indeed the charge collection is temperature depending (so you need to change the voltage or amplification to get back to "stabilize" the energy spectrum)

Ahmed is correct that you have to wait for HV to be stable, but most HVPS on NaI or Ge or anyhow using a ramp filter that is creating just that stability delay time before you can start an acquisition

Your question is unfortunately not fully complete. If the emission rate or fluence of a neutron source is known, we can calculate the same value at any distance using inverse square law. For the neutron sources mentioned by you, the fluence-to-dose equivalent conversion at a distance, d , can be computed using conversion factors for the particular neutron energy: Please refer to: NRC:10CFR 20.1004. Let me give you some values: For Am-Be, average energy of neutrons is 4.46 MeV and for Cf-252, average energy is 2.14 MeV and the corresponding fluence-to-dose equivalent factors are: Am-Be (QF=8.5): 1 rem = 24 x 10⁶ n/cm² and for Cf-252(QF=10); 1 rem = 28 x 10⁶ n/cm². The maximum permissible dose equivalent for a radiation worker is 20mSv/year or about 1mrem/hour (1 year = 52 weeks, 1 week = 40 working hours). From these values, you can work out whether, at the distance  away from the source you normally work, the dose is within the permissible limit. Otherwise, working backwards, calculate the distance at which you are safe. It may be of interest for you to know that you can design a shield for your neutron source by surrounding it by a paraffin cylinder on all sides to fully thermalize the fast neutrons which, in turn, is surrounded by a cadmium cylinder to absorb the thermal neutrons. The capture gamma rays produced by cadmium can be suppressed by surrounding this assembly by a further layer of lead on all sides. The high quality factor (QF) for neutrons implies neutrons are much more hazardous than gamma rays (QF=1). Personnel monitoring neutron badges are available from regulatory authorities. 

Hope the above analysis helps. For better accuracy, the energy spectrum of neutrons should be taken into account.

You can download software from website but you must pay for licence to activate it. It is not very expensive. 

I can try to attach Angle,but I am not sure that it is possible to attach .exe file even if it is "hide" as .rar. 

thanks pappu sir

Are interested in the radioactivity or the amount of chondroitin?

The radioactivity traces the chondroitin. The chondroitin was administered at 16 mg/kg. The radioactivity traces the fraction of the dose in various compartments. You will need mass of the feces and mass of the total administered (body mass x 16 mg/kg). If 25% went to the feces then concentration is 0.25 of total administered divided by fecal mass.

3H  has 28.8 Ci/mmol. 10^-15Ci = fCi. mmole = 90*10^-15/28.8

I just need some references showing the relationship between overdose ratio and product density/thickness. 

Thank you very much.

Hello;

I'am not really expert, but I'am a user of MCNP code and I dealt few years ago with F Tally card and I think that for symmetry reasons your sphere center should correspond with a middle of the target, in aim to obtain normalized number of particles / surfaces, elsewhere you should make some geometrical calculations to plot a accurate particle spectra.

for plot graphic and geometry, there are two ways:

- direct ay ith Vised editor: an graphical interface of MCNP5 (under windows and I think there an available release under other OS like linux...)

- if you are working with line command terminal, you have to use line command as:

> mcnp ip inputfile.name 

many options are available on MCNP user guide, it just depends which release are you using (MCNP5 or MCNPX)

Probably there will be more complete answers than mine, hope I could help a litte;

good luck;

Saladin

First of all, the pure (external) exposure of wood to nuclear radiation causes no problem as no radioactivity is generated in the wood by this radiation. However, wood from high contaminated areas will have a certain uptake of radionuclides, especially  Cs-137, which will still be present today.

There are estimations in IAEA Tecdoc1376 about different scenarios using wood from the most contaminated areas in the Chernobyl region (see link). Some of these scenarios can cause exposures above radiological permissible limits (table XVII). However, these scenarios  comprise mosty the use of wood ash for ferilisation and/or an extensive use of wood for building etc.

For all practical reasons, a single piece of wood from the Chernobyl region won't cause harmful doses, even at the highest radioactivity levels for wood given in table XVI. A piece of wood, let's say, one kg with an activity of 10000 Bq/kg  at a distance of 10 cm will roughly cause a dose of 0.1 mSv per year.  However, depending on its exact activity, you might be obliged to handle it within a controlled area. 

OK, if the reconstruction using FBP method looks good, you are probably right that the mesh grid comes from the algorithm. Maybe it has something to do with the starting point of the iteration process, if the mesh grid is less visible when increasing step lenght. So, you could have a look at interim reconstruction results. In addition, you could also try to change/increase your sinogram data/resolution (more angular steps) to check influence of your input data. But I don't know this reconstruction algorithm, so I'm only guessing...

Good luck!

Why not an air flush for evacuating radon and avoiding its accumulation close to the detector? 

Fixing radon Inside the sample is possible for powder samples, by mixing 10% activated charcoal.

A large component of the radiation dose to the human population arises from cosmic radiation entering the Earths atmosphere (Grasty and La Marre, 2004), The Intensity of the radiation due to cosmic rays depends mainly on the thickness of the atmosphere above the measurements location, and therefore mainly on altitude above sea level. Changes in barometric pressure and temperature and associated differences in atmospheric attenuation also cause small fluctuations of short term nature.

The annual effective dose rate due to the outdoor cosmic effect was calculated from a digital terrain model averaging topographic height within a cell size of 0.829 Km x 0.890 for the studied region using the following

Cosmic (mSv/a)= 37*exp (0.38*Altitude (km))* 0.877)/100

Dear Nelson. The content of the attachment could help you.

Dear Moradi,

I am providing below the HVL calculation for all energies for typical absorber materials used for gamma ray attenuation

Refer to Handbook of Chemisty and Physics (Google search), latest Edition, 2016. Here you find table of mass attenuation coefficients of important elements for energies from 10 keV to 1 GeV. Taking lead (Z=82) as the attenuating element, we find:

µ/ρ for 1MeV = 7.1 x 10^-2 cm2/gm;

µ/ρ for 1GeV = 11.5 x 10^-2 cm2/gm;

Now HVL = 0.693/µ, where µ is the linear attenuation coefficient and µ/ρ is the mass attenuation coefficient.

Hence HVL in lead (ρ, density of lead = 11.34) for 1MeV is (0.693)/( 7.1 x 10^-2 x 11.34) = 0.861 cm of Pb or 8.61 mm;

And  HVL in lead (ρ, density of lead = 11.34) for 1GeV is (0.693)/( 11.5 x 10^-2 x 11.34) = 0.531 cm of Pb or 5.31mm.

The TVL (tenth value layer) values can be calculated using the relation, TVL = 3.32 HVL

As for your other question: how many HVL values are required, it depends on the radiation dose equivalent rate level at a particular point of the beam and the transmitted level desired after attenuation by the absorber (Remember, the radiation attenuation is exponential and no amount of absorber thickness will stop the beam completely). One HVL will reduce the dose rate level by one-half; one TVL will reduce the level by one-tenth.

We will take an illustrative example. A 1 MeV source (close to a Co-60 source, with an average energy of 1.25 MeV) is emitting gamma radiation whose dose equivalent rate is 1Sv/hr (roughly an air kerma rate of 100 rads/hr) at a specified distance from the source. A barrier of lead is required so that the dose equivalent rate on the other side of the barrier is safe for an occupational worker (20 mSv/yr or 10 µSv/hr, assuming the total working hours/year is 50 weeks/yr and 40 hrs/week). How many TVLs and HVLs are required?

1 TVL reduces the dose equivalent rate by a factor of 10

Therefore, 5 TVLs reduce the dose equivalent rate by a factor of 100000, i.e., the dose rate reduces to 10 µSv/hr

So the thickness of lead barrier required = 5 x 8.61 x 3.32 = 14.3 cm (about 17 HVLs)

The above computation assumes narrow beam geometry, for which the mass attenuation coefficients are given in the Handbook mentioned above.  In practice, there will be a lot of scatter from the absorber,etc. for a broad beam for which build-up corrections are required.

Note: I did not elaborate on the mathematics of the above computations because they are available in any book on radiation protection.

I hope the above will be of help. Please give your feedback.

It is not clear what characteristic of GM counter Mr Shahid refers. If it is the graph between the charge collected by the anode wire versus the applied voltage for an ionizing event, it is a steadily rising curve. In the GM region the gas amplification is very large: 10⁸  to 10¹⁰for a single ionizing particle of any kind (alpha, beta or gamma). That is why the GM counter, unlike an ion chamber or proportional counter, cannot be used to identify the ionizing particle. What really happens here is the primary avalanche initiates further avalanches due to ionization and excitation of the atoms of the gas, producing UV photons which are also ionizing. The +ve ions being heavy are initially localised at one point of the anode wire.  Ultimately due to secondary electron production due to photoionization by UV, the whole anode will be covered by a sheath of +ve ions which results in the reduction of electric field, thus terminating the discharge. The +ve ions then drift toward the cathode releasing electrons from the cathode wall,etc and the discharge will become self-perpetuating. To arrest this recycling, quenching agents are used. Dr Pekko has elaborated on these aspects well.

The other characteristic plotted for the GM counter is the relation between count rate and applied voltage for a given radioactive source placed below the counter, Here, there is a slow rise, followed by a plateau where the count rate remains fairly constant over a range of voltages. Further increase of voltage results in the production of multiple counts. 

I can discuss quenching agents and other concepts like dead time, recovery time, etc but these are available from standard texts (see e.g., Nuclear Radiation Detection by Wlliam J Price or Radiation Detection and Measurement by Glenn F.Knoll).

There are various formulae for equivalent square.  The simplest is E=4A/P where A is the area of the field and P is the perimeter.  This works OK for slightly irregular fields, especially ones defined with blocks.  For MLC-defined fields this tends to underestimate the equivalent square, since the perimeter of MLC-defined fields is larger than a similar block-defined one.

Clarkson integration works by dividing the field up into a large number of segments (in circular geometry, centered on the point of interest), and for each segment looking up some suitable scatter factor (SAR can be used, or the PSF from BJR25) for a circular field of the same radius as the segment.  Averaging round the circle gives an average value of the scatter factor.  Looking back into the original data to determine what field size would give the same scatter factor gives you the final answer. (usually as an equivalent diameter, which then needs to be converted back to equivalent square using the methods in BJR25).  Various computational tricks are needed for complex shapes if the segment at a agiven angle has gaps in it.

A simpler method for MLC-defined fields is to average the MLC-defined width (for open leaf pairs in the jaw defined field only) to give W, and to add up the number of such leaf pairs times the MLC spacing to give L, and use the formula E=2LW/(L+W).  This gives very similar answers to the Clarkson method, for considerably less effort.

Dr. Sankaran Ananthanarayanan,  1. YOU ARE POOR IN BASIC RADIATION PHYSICS. 2. YOUR ANSWER LACKS CLARITY AND VERY VAGUE.  I am very sure about my experimental discovery of UV dominant optical emission following Cu or Ag X-rays from within the same excited metal atoms of Variable Energy X-ray Source, AMC 2084, U.K. (Braz. J. Phy., 40, no 1, 38-46,2010)

http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-97332010000100007).  I have already  explained with unprecedented detail how the optical emission takes place in radioisotopes and XRF sources in the same paper.   

FIRST AND BEST REVIEW OF THE RESEARCH PAPER

Margaret West,*a Andrew T. Ellis,b Philip J. Potts,d Christina Streli,c Christine Vanhoof,e Dariusz Wegrzynekf and Peter Wobrauschekc, Atomic spectrometry update-X-ray fluorescence spectrometry, J. Anal. At. Spectrom., 2011, 26, 1919.

DOI: 10.1039/c1ja90038b http://pubs.rsc.org | doi:10.1039/C1JA90038B

Refer under the title: 2.3 Spectrum analysis, matrix correction and calibration procedures

Words of citation: The phenomenon of optical emission predominantly in the UV, which accompanies the emission of X-rays, gamma rays, and beta radiation from radioisotope sources and X-ray tubes was investigated by Rao. It was the first work in which the emission of UV radiation was confirmed experimentally and a possible explanation for the mechanism of the UV emission was given by the author.  https://www.researchgate.net/publication/273124068_24_FIRST_AND_BEST_REVIEW_OF_THE_RESEARCH_PAPER_publis

The reviewers already cited my plausible explanation for UV emission following X-rays. PLEASE NOTE IN MY QUESTION, I DID NOT ASK FOR ALTERNATE EXPLANATION, YET YOU TRIED TO PROVIDE ONE. 

I never came across any reference on what you said, 'photo and Compton electrons exciting valence electron to emit UV'.  And you have not cited any reference, when you said, "(photo and Compton) electrons which, in turn, in favourable cases, can excite the valence electrons to emit UV or visible light.In a particular metal, these are decided by transitions of energy levels allowed  by quantum mechanical rules.  

When I mentioned my work with Cu or Ag X-rays from AMC2084,UK, you provided your own explanation  on gamma rays. Your comment is totally irrelevant and very vague. 

You said, "The line emission spectrum of various elements are available from reference data published by National Standard Laboratories".  What you quoted is optical spectrum of metals at high temperatures.   What I have reported is UV dominant  optical emission from Cu, Ag, and Mo XRF sources and 57-Co  notably at room temperature. Please note radioisotopes and XRF sources emit a new class of atomic spectra of  solids (radioisotopes and  XRF sources) unprecedented at room temperature. If I understand correctly, you are trying to downplay the work done and confuse the readers. In this process, your ignorance of the subject  is exposed.

 In addition to Hanno's answer, the mass attenuation coefficient is energy dependent. The detector's response to different energies will be different. The response of the detector will depend upon its size and shape. Further, the electronic configuration of the detector will alter the response.

2-3 x 10 ^{- 6} mb

In a simple harmonic model, the vibrational frequency is square root (bond strength/effective mass). A reduction can be interpreted as a reduction of the effective mass of the bond constituents, which could be consistent with a chain scission.

Maybe I can add a more intuitive reply. It depends on weather the FOV you refer to is a scanning or reconstruction FOV. The latter does not affect the dose since it is executed through reconstruction after the scan is finished. The former, however, involves adjusting the diameter of the radiation field during the scan and this will definitely affect the effective dose to the patient.

Xrays and gamma rays are nothing but photons of different energies. X rays are emitted by atoms when electrons jump from higher to lower energy states. Gamma rays, on the other hand are emitted by nuclei. Using the equation E=hν we see that higher energy photons have higher frequencies and hence smaller wavelengths. Roughly speaking  X rays have 10^-8 > λ > 10^-12  meters and gamma rays have 10^-10 > λ > 10^-14 meters. As you have noticed, these ranges do have an overlap. There is no deep physical reason for the fact that the wavelength ranges overlap. It is a matter of nomenclature.

Half-order Bragg peaks mean a double-fold periodicity in real space. In perovskites a common source for that is indeed octahedral rotations, that can double the unit cell (i.e. each opposing-ly two rotated octahedra can be seen as one unit cell). For this case it's common to use synchrotron radiation due to the low relative intensity of the half order peaks.

Other sources for half order peaks could be periodic oxygen vacancies, see the Brown-Millerite structure for example.

For further reading on octahedral rotations (someone should come up with a convenient acronym for that already), I recomment the well-written work by May, Rondinelli, Spaldin and co-workers

arXiv version: http://arxiv.org/pdf/1002.1317.pdf

Attach papers for your answer

Hi,

It is not only messy, it seems that you have some points missing. You might have a nonideal buffer subtraction for the messy file, leading to an over subtraction. You can have a look at these two papers: Protein Sci. 2010 Apr;19(4):642-57. doi: 10.1002/pro.351 and Methods Mol Biol. 2014 ; 1091: 245–258. doi:10.1007/978-1-62703-691-7_18.

Idem for the second file, your background subtraction might lead to small differences. Radiation damage is usually seen in the low q region. You might need to apply some correction factors (J. Phys.: Condens. Matter 25 (2013) 383201 (24pp) doi:10.1088/0953-8984/25/38/383201 ).

Hop it helps.

By using f4 you can calculate number of electrons in the specific volume

Stopping power is generally used for charged particle interactions, whereas the mass energy absorption coefficient is specific for photons.  If you look into cavity theory, you will see a factor 'd' in the calculations that controls for the relative weighting of the stopping power as opposed to the mass attenuation coefficients.  The Attix textbook should be a good reference.

The iron

In addition to Dr Newhauser's comments, here are some other considerations:

Use high density coarse aggregates, for eg, granite in place of limestone

High workability of concrete and sufficiently spaced reinforcement

Monitoring and controlling heat of hydration during casting and subsequently. Cement replacement materials like fly ash and silica fume are good choices as they not only decrease the rate of hydration, but have filler properties resulting in fewer pores in the matrix.

Sufficient shrinkage reinforcement

Substantial and sustained curing

Well, we are using this film for sample holder purpose while studing Proton induced X-ray emission or EDXRF of liquid sample. As mentioned in your paper related to soft X-ray microprobe, the special type proportional counter have 0.5 micrometer film.  Kepton foil (8 micron) is a good choice for our proton accelerator experiment. We are using this film, Kepton foil is quite strong and radiation stability is high, you can search for the low thickness for your purpose

Hanno

There is no of binding of electrons to the alpha during emission. Electrons are ejected from the cloud on a time scale similar to your estimate of the time of passage.

My description of ionization in vacuo was very simplified. True isolation would mean the alpha and the parent would move apart indefinitely requiring the electrons to seek the ground state by sorting the Coulombic attraction between the two. (Another simplification.)

Alpha decay is usually described by considering a bare parent nucleus. You correctly stated that electrons must be added to calculate alpha recoil. The energy of the alpha exiting the bare nucleus is different from the energy of the alpha exiting the electron cloud. The alpha particle is moving slower than the tightly bound electrons. The electrons are perturbed by the passing alpha and are excited or ejected. The parent is undergoing nuclear transformation with two excess electrons, some excited and all undergoing changes in their wave functions.

The claim for up to 10 comprises the two excess electrons and those excited enough for ionization. The time scale for ionization to persist is the time scale of the alpha passage. Nevertheless, experimental results show the 3E-3 occurrence of +1 ionization much longer than the alpha passage. (See R. Roy and M.L. Goes, 

Comptes Rendus de l'Académie des Sciences, Paris 237, 1515 (1953).)

Dear Erkki,

you are Always welcome,

my formulas of the LT begins simply with the + instead of the minus, just a matter of defininig the positive direction of the x and so v.

I could have put the - in the forward trasformation and the + in the reverse nothing would have changed basically.

The thing is that time does not reverses in the Doppler loop, how can it? The Doppler loop is a sequence of transformations so there no way time reverses.

You can reverse all the doppler loop like in a movie just with L^-2 but that would not mean much..only bring the system to its initial conditions at the emission.

If I get the identity I, the Doppler loop does not have anymore that form,

that ratio of the frequency f₁/fo=(1+beta)/(1-beta) is the one predicted by the Doppler Radar, moving mirrors, Gravity Probe A experiment and so on. So with the identity it would not fit the experimental evidence anymore..

Maybe  my paper was obscure then...

The complex transformation Matrix is interesting anyway..

Best

Stefano

Good point.  Go stick a stick in the ocean and see what happens!  It's probably easier in a pond or a wave table, but you get the idea.

Yes. I have done this and measured also the PDDs in a stationary water phantom, from 0,5 cm depth to 5 cm depth.

At the end of the day you need to work according to a standard, so te TRS398 what you have at hand. But, if you do research and you are not restrained to a standard, you can also do it differently: use a virtual water slab phanton, and you can work your way from the surface down to any depth you like (the slabs have widths between 1 mm and 1 cm, so you can measure the PDDs much more acurrately than in real water. You can take a look at the RW3 slab phantom from PTW, from instance.

Reptran has much more molecules than the (very) old LOWTRAN. I strongly encourage you to use the reptran option.

Hi Fayzan, Could you send your email to discuss the the curve fitting for EBT2/ EBT3 films. Prof Tomas Kron forward your request to me.

Derrick

As far as i understand, geometry at this level is not important. This assumes, that your spheres are "small" in the sense that fluctuations in beam direction randomly another sphere.

So just create a new material and add the elements H, O, and Au (assuming you have gold spheres) in the correct weight fractions.

Dear Joseph,

I first read our old discussion again, in order not to repeat to much.

I think you had a look at my chapter about doses and dose units. The central point there is, you can calibrate a dose meter using exaggerating geometries like enlarged and adjusted fields in special phantoms. That means that your calibration factors contain  attenuation and scattering from the phantom and maximal enlarged fields. If you now add the radiation quality factors Q you indeed can factor in the kind of radiation (light or dense ionizing).

I have strange problems to use the unit Sievert (Sv) for "physical" operational doses used for measurements and body doses like organ dose and effective dose which are used to estimate risks.

For the information of other participants in this question I add again my textbook chapter, where you can find the calibration geometries, phantoms and dose definitions etc.

Dear Christoph,

take care for Co-59 contaminations of your structures. Using high energy photon beams you will produce neutrons by nuclear photo reactions. These neutrons could be captured by metalls containing natural Cobalt. The result would be 60- Co with the known problems.

Thanks Lung-Kwang Pan for your views, but the facilities which really use only GM for practical applications, it may not be appropriate to use data of other monitors/detectors.

My first though is that you are trying to make things too complicated. An alpha particle is best thought of as a cannon ball loosing its kinetic energy by impact. Electron holes are a lattice phenomena; the alpha particle has two positive charges not electron holes. There is no evidence that an alpha particle experiences drag in free space.

The actual cost of a passive radon measurement campaign depends on many factors but basically on your relation with the dealers so I have no suggestion about this;

However, E-perm system is available also in a version usable for 3-7 days instead of a minimum of 30 days of trace detectors such as CR-39 (e.g. Radosys Ltd).

The long term version of E-perm  claims to be usable at higher doses (integration time*concentration ) than trace detectors but I have no direct proof about this.

The e-perm system is theoretically subject to discharge caused by strong electric fields and other environmental factors. Moreover once you read an e-perm the dose value is lost while a trace detector can be stored (in principle) indefinitely for future recording.

Another non negligible issue is that trace detectors are lighter and smaller than e-perm which is of some relevance if you have to deploy  a large number of detectors in some environments.

I started annual radon monitoring in an important road tunnel in year 2000  with e-perm and swithched to trace detectors in 2003

You should also read the seminal monograph published by Martin Berger in 1963. Its title is  "Monte Carlo Calculation of the Penetration and Diffusion of Fast Charged Particles."

It dependent to your protocol.  For estro is 5 cm lateral and 10 cm depth of phantom. Brass used for 10 mv and higher. In iaea protocol use of build up phantom. You can use of fc65 instead of ebt2.

Rather than answering your question in frequency, it would be more realistic to answer in energy, simply the frequency is so high that gamma ray behave more like a particle (photon) than a wave. The CGRO and other satellite experiment can detect gamma ray only up to several 100s GeV or < 1TeV = 1.E12 eV. Higher than that, the flux is so low that satellite instruments loss detection power or their discrimination power to separate gamma from much higher flux of cosmic rays. The Ground gamma ray telescope can detect gamma photons interaction with atmosphere via indirect measurement. The highest energy of gamma ray  of those experiments can reach approximately 1.E14 eV, in terms of frequency ~ 2.4E28 Hertz.

Dear Patrick, thanks, it´s worth to read these historical textbooks. I did also some looking around and found again

Rutherford first edition 1904

Rutherford second edition 1905

I append the first edition.

This is called secular equilibrium or for Rn-222: Rn=k*[1-exp(-t*ln2/half-life(Rn))], where Rn is radon activity, k is proportional to the radium-226 activity, and t is time, and the half-life(Rn) (of radon) is 3.82 days, or more simply, Rn=k*[1-1/2**(t/3.82 days)]. In other words, independent of the chamber size, in 3.82 days Rn activity is at 1/2 of its terminal value, at twice the time; 7.64 days, Rn activity is at 3/4 of its terminal activity, at 3*3.82 days or 11.46 days, Rn is at 1-1/2**3=1-1/8=7/8ths of its terminal activity and so forth. In other words, Rn activity is at 99% of its terminal activity in 609 hours (=3.82*24*ln100/ln2 hrs).

I am sorry, but no one can share you beaminputifles. For constructing a real model, a "non-disclosure agreement" is signed with the company's.

But you can find phsp files at the IAEA database.

kind regards

Back in 1953 when I first went to work for the old Atomic Energy Commission, we used film badges for personnel monitoring.  Calibration, development and density reading with a densitometer were all pretty obvious, and judging from the discussion, irrelevant to the question being asked.  I can say that temperature control in the darkroom "soup" was critical and that the calibration film and the personnel film returned to us were developed at the same time in the same "soup."  But that is also rather obvious.  Calibration was against a radium source traceable to the National Bureau of Standards (now NIST).

A. For water: Internal consumption leads to higher stomach dose. External irradiation has several by effects. My team has published on bot issues:

Internal

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Rad. Prot. Dosim. 2003 106(3): 219-226

Rad Prot.Dosim. 2004 112(2): 251-258

Radiat. Meas. 2008 43(7): 1305-1314. (better approach)

External

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J.Environ. Radioactiv. 2004 76(3):283-294

Sci. Tot. Environ. 2007 373: 82-93

Sci. Tot. Environ. 2008 405 (1-3): 36-44.

Sci. Tot. Environ. 2010 408: 495-504 (better approach)

Environ. Sci.: Processes Impacts, 2013,15, 1216-1227

From Soil:

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To my knowledge there has not been a concern on that. Indeed the contribution from soil radon is (a) radon diluted in atmosphere (2) radon diluted in water. For the first we employ several approaches that depend on (I) F-factor (II) EEPC (III) OF (IV) aerosol size distribution and so on.

A recent review of me and my colleague on the subject is

Frontiers in Public Health 10/2014; 2:207.

Therefore: I would suggest (i) measure radon in water content (ii) measure radon and progeny concentrations outdoors and indoors (iii) calculate dose to stomach region & effective dose due to water consumption (iv) calculated dose due to radon inhalation.

Some authors use emanation and exhalation interchangeably. The correct use is shown in your diagram. Radon in the pores is exhaled by diffusion if there is no movement (convection) of the gas in the pores. Radon is removed by convection if there is movement in the pores, gaps, and cracks in the material. Convection can be caused by temperature differences in the material, water intrusion, or pressure difference caused by movement of air across the surface of the substance or atmospheric pressure changes. Pressure differential in a structure can be caused by air movement (advection) or temperature differentials.

My point was that we do not understand the mechanism of emanation. Those measurements assume the mechanism is diffusion. Different situation is when measuring permeability - then it is forced (Darcy) flow. When forced flow is present there is always some diffusion (sometimes called dispersion).If we don't understand diffusion, then no meaningful measurement can be made. 

Thank you 

for your additions. 

Regards 

For measuring correctly you need a good calibration before you conduct the measurements.  You must be careful about the leakage of gas from the can or container in which you measure the concentration. 

It is also depends on the procedure of sample collection. 

There is a ratio between indoor radon and Radon releasing from water depending on the radium content in the water.

First: What is a primary standard: See "International vocabulary of

metrology — Basic and general concepts and associated terms (VIM)" at BIPM website.

5.4 (6.4)

primary measurement standard

primary standard measurement standard established using a primary reference measurement procedure, or created as an artifact, chosen by convention

In case of Rn-222 you need a radon gas standard done by an absolute method. This would be for example 

J.L. Picolo, Nucl. Instr. and Meth. A 369 (1996) 452

R. Dersch, Appl. Radiat. Isot. 60 (2004) 387

See also for further information: Dersch, R., 1998.  Appl. Radiat. Isotopes, 49, pp.1171-1174.

PTB and METAS have implemented the absolute method.

You are probably interested in measuring radon exhalation (or emanation) rates from rocks or soil.  We are using both CR-39 and LR-115 in cups with crushed rock to measure Alpha tracks from 222Rn. These are slightly different SSNTD, and the availability and etching is different for each.  As is widely done in this technique, we use a plastic barrier to impede diffusion, so we effectively remove decay from 220Rn, which has a very short half life.  There is considerable literature on this technique.  As Parviz notes, the exposure time is important, and we are exposing SSNTD above crushed granites and black shales for at least 3 months to get enough tracks.  The measurements of uranium, thorium, radium need to be approached from different methods, but is likely that radon measurements are directly related to radium (but not necessarily uranium, nor thorium).  Measurements can also be made using scintillation counting.