Calculation of the thickness of protection against photon radiation for various materials

The establishment of nuclear energy as the most promising source of electricity production, the use of power and research reactors, radionuclides and other sources of ionizing radiation in various areas of industry, science, technology and medicine are inextricably linked with the problem of radiation safety, with the tasks of designing and creating biological protection.

Most portable and laboratory metal and alloy analyzers use X-ray tubes, which are sources of ionizing radiation. Everyone knows the consequences of unsafe use of radioactive radiation for the health of the body. In order to ensure safe use of equipment for humans, it is necessary to know the thickness of a particular material used as a "radiation absorber".

No is the number of particles emitted from the X-ray tube.

N is the number of particles remaining after passing through a protection of thickness d.

To calculate, you need to know several concepts:

The linear attenuation coefficient μ characterizes the decrease in radiation intensity when passing through 1 cm of matter and is equal to the ratio of the fraction dN/N of photons of a given energy that have undergone interaction when passing an elementary path dl in the medium to the length of this path:

1 . dN

μ= N dl

The mass attenuation coefficient μ m is the ratio of the corresponding linear interaction coefficients to the density p of the medium through which the photon radiation passes:

μ

μm= p

The mean free path l in a medium is a value inversely proportional to the attenuation coefficient μ:

1

l = μ

N changes according to the law:

N of X-ray tubes is generally equal to 10 10 and let N=1, then

Since e≈ 2.7, then

μd=25

Next, using the tables of attenuation coefficients for various materials, knowing the radiation energy, we find the linear attenuation coefficient.

Example:

For an energy of 0.04 MeV, the attenuation coefficient for aluminum is , which means:

25
d = 1.33 = 18.8 cm

For lead µ= cm, which means:

25
d = 151 = 0.16 cm = 1.6 cm

That is, for safe use of the X-ray tube, it is necessary to install protection made of aluminum with a thickness of 18.8 cm and of lead with a thickness of 1.6 cm.