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# Answer to Question #8135 Submitted to "Ask the Experts"

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

Q

How do I calculate half-value layer for brass?

A

Since brass is an alloy of two or more metals, its attenuation properties depend on the relative amounts of the constituents. The main components of brass are copper and zinc and, depending on the intended applications, the mass percentage of copper may range from about 55% to 95%, and zinc may range from about 5% to about 40%. Among some common brasses are red brass, with about 15% zinc, and yellow brass, with about 35% zinc. Other materials may also be present to alter physical properties of the brass. Such additives may include such elements as lead, tin, iron, arsenic, antimony, manganese, silicon, and phosphorus. The amounts of these generally do not exceed a few percent.

I assume the radiation of interest for you is ionizing photons. For a given energy photon the mass attenuation coefficient may be determined by taking into account the weight percentages of the elements that make up the brass. The generalized expression that is usually applied is

μ/ρ = Σ wi (μ/ρ)i ,

where μ/ρ is the mass attenuation coefficient for the brass, wi is the weight fraction of the ith element in the brass, and (μ/ρ)i is the mass attenuation coefficient of the ith element. The values of (μ/ρ)i may be found in various compilations, including tables available from the National Institute of Standards and Technology.

Once you have determined the value of the mass attenuation coefficient you can obtain the linear attenuation coefficient, μ, by multiplying by the density of the brass. The half-value layer, HVL, for the primary photon energy of interest may then be obtained using the simple relationship between linear attenuation coefficient and HVL—i.e.,

HVL = 0.693/μ.

As an example we can calculate the HVL for 500 keV photons in brass of mass density 8.4 g cm-3 containing 30% zinc and 70% copper. At 500 keV, the mass attenuation coefficient for zinc is 0.0845 cm2 g-1 and that for copper is 0.0836 cm2 g-1. We then obtain

μ/ρ = (0.0845 cm2 g-1)(0.30) + (0.0836 cm2 g-1)(0.70) = 0.0839 cm2 g-1, and

μ = (0.0839 cm2 g-1)(8.4 g cm-3) = 0.707 cm-1, whence

HVL = 0.693/0.707 cm-1 = 0.984 cm.

Because the zinc and copper differ in atomic number by only one unit, the mass attenuation coefficients of each at 500 keV are not very different, and the value of the mass attenuation coefficient calculated is not very different from either of the individual values. The differences between the two become greater at considerably lower photon energies. If other components are present in the brass you may have to include them in the calculation. Keep in mind that the above calculation of the HVL is for the primary energy photons. It does not take into account, for example, any secondary photons that might be present as a result of scatter in any shielding material.

I hope this answer is helpful to you.

George Chabot, PhD, CHP

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 3 April 2009. 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.