Answer to Question #10370 Submitted to "Ask the Experts"
Category: Radiation Basics
The following question was answered by an expert in the appropriate field:
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?
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|
|Borosilicate||Lead Glass||Borosilicate||Lead Glass|
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(0.3)|
|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