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

Q

Question: How much of a lead equivalent is 0.953 cm tempered glass? Is there a webpage that list common everyday materials and their lead equivalents?

A

The x-ray attenuating equivalence of tempered glass compared to lead glass depends on a number of considerations, including the specific compositions of the glasses and the energies of the x rays of interest. The National Institute of Standards and Technology (NIST) provides attenuation coefficients at this link for lead glass (click on Glass, Lead to see the values). The specific formulation and other properties assumed are given in an associated table on another page. The same pages also give data for a borosilicate glass. Tempered glass does not generally contain added boron, but the data provided for the borosilicate glass should provide a pretty good comparison for the unleaded and leaded glass cases. Most tempered glass differs from typical float glass only in the thermal treatment it undergoes to improve its safety characteristics.

From the data cited, you can see how the attenuation coefficients change with energy. Below are reproduced some of the mass attenuation coefficients for the two NIST glasses. The linear attenuation coefficients shown were obtained by multiplying the respective mass attenuation coefficients by the appropriate glass density (2.23 g cm-3 for borosilicate glass and 6.22 g cm-3 for the lead glass).

Energy, keV Mass att. Coeff., cm2 g-1 Linear Att. Coeff., cm-1
10 17.05 102.9 38.02 640.0
20 2.297 65.68 5.122 408.5
30 0.7987 23.05 1.781 143.4
50 0.3022 6.134 0.6739 38.15
100 0.1657 4.216 0.3695 26.22

The transmission of primary photons of a given energy through a linear thickness, x, of glass is given by T = e–µx. For a nominal glass thickness of 0.953 cm the transmissions at the above energy may then be estimated as below:

Energy, keV Transmisssion, T = e–µ(0.953)
10 1.84 × 10–16 1.30 × 10–265
20 7.59 × 10–3 8.49 × 10–170
30 1.83 × 10–1 4.46 × 10–60
50 5.23 × 10–1 1.62 × 10–16
100 7.03 × 10–1 1.41 × 10–11

It is clear and expected that the lead glass is much more effective at attenuating the noted x rays than is the unleaded glass. It is also clear that the effectiveness of the lead glass relative to the unleaded glass changes markedly with photon energy, with the effectiveness of the lead decreasing as the energy increases over the range of energies shown.

If you require the comparison for a specific glass composition you could calculate the attenuation coefficient for different energies by considering the mass fractions of the elements that make up the glass. The mass attenuation coefficient of the glass is determined by multiplying the mass fraction of each element in the glass by the mass attenuation coefficient for that element (this can be obtained from NIST at this page for the elements) and then summing such products for all the elements in the glass. As an example, see the previously asked question 8135 and its answer.

Since "lead equivalent" thickness of any material varies with the energy of the photons under consideration and with the specific compositions of the materials of interest, it is generally not meaningful to attempt to tabulate such equivalencies.

I hope this is helpful to you.

George Chabot, PhD

Ask the Experts is posting answers using only SI (the International System of Units) in accordance with international practice. To convert these to traditional units we have prepared a conversion table. You can also view a diagram to help put the radiation information presented in this question and answer in perspective. Explanations of radiation terms can be found here.
Answer posted on 1 October 2012. The information posted on this web page is intended as general reference information only. Specific facts and circumstances may affect the applicability of concepts, materials, and information described herein. The information provided is not a substitute for professional advice and should not be relied upon in the absence of such professional advice. To the best of our knowledge, answers are correct at the time they are posted. Be advised that over time, requirements could change, new data could be made available, and Internet links could change, affecting the correctness of the answers. Answers are the professional opinions of the expert responding to each question; they do not necessarily represent the position of the Health Physics Society.