Lead can be used to shield against harmful radiation. The amount of lead required for shielding depends on the type of radiation as well as its energy. In general, lead is used to shield living things from radioactive sources in laboratory, medical, and industrial environments. For example, X-rays are routinely used for diagnostics in medical centres, and the patients are required to cover their body parts that are not to be scanned with lead shields. Being efficient as well as cost effective, lead is the most widely used radiation shielding material.

- Skill level:
- Easy

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## Instructions

- 1
Determine the type and energy of the radioactive source you want to shield against. If the radiation is coming out of an X-ray machine, it will have a whole energy spectrum consisting of two types of X-rays: bremsstrahlung and characteristic X-rays. Bremsstrahlung spectrum is continuous while the characteristic X-rays have narrow peaks. Since attenuation of X-rays in a material is not linear, it is best to use different energy points to calculate the shielding. However, for most practical purposes one uses the highest energy X-rays for calculating the shielding since, in general, if the shielding is good for high energy X-rays it will be good for lower energy X-rays too. The maximum energy X-rays are equal to the maximum voltage applied to the X-ray tube. For example, a 100kV X-ray tube will produce X-rays of maximum energy of 100keV. Here k stands for 1000 and eV (electron volts) is a unit for energy.

- 2
Find the values of attenuation coefficient for the radiation at the energies of interest for lead using a data tables, for example the ones published by National Institute of Standards and Technology (NIST). Note that the NIST attenuation coefficient data is usually in dimensions of cm^2/g and must be multiplied by the density of lead before being used in calculations. For example, at the X-rays used in radiography have energies in the range of 12-120keV. Let us find the attenuation coefficient for 100keV X-rays in lead. Looking at NIST's online database we find that the value is 130.6cm^2/g. Multiplying this with lead's density of 11.3g/cm^3, we get the attenuation coefficient u=130.6*11.3=1475.78 per cm. You can determine the values at other energies of interest using this method.

- 3
Calculate thickness of the lead required for shielding at a certain energy using the formula: d = 9.21/u, where u is the attenuation coefficient you just calculated. This formula is for thickness needed to stop 99.99% of the X-ray photons. For 100keV X-rays, this thickness comes out to be d=9.21/1475.78 = 0.006cm. Therefore only a thin sheet of lead of about 0.6mm would be enough to stop 99.99 per cent of the photons.

#### Tips and warnings

- These calculations are based on the photon attenuation formula: I = I_0 exp(-u*d), where I is the intensity of photons (number of photons per unit time) coming out of the material, I_0 is the intensity of the incident photons, u is the photon attenuation coefficient, and d is the thickness of the material. It is not specific to lead and can be used for any material for as long as the correct attenuation coefficient is used.
- One should be careful with the above calculations since they are based on statistical nature of interaction of photons with matter. It is a good practice to add some thickness to the value calculated. For example, instead of the thickness of 0.6mm calculated here, one may use a 1.5mm thick sheet of lead to ensure that no X-ray photons reach the other side.