What is the thinnest material that will normally stop a 0.5 MeV beta particle? A

Question

What is the thinnest material that will normally stop a 0.5 MeV beta particle? A sheet of notebook paper A 1 cm thick aluminum plate A 1 m thick concrete wall What is the thinnest material that will significantly attenuate a 0.5 MeV gamma ray? A sheet of notebook paper A 1 cm thick aluminum plate A 1 m thick concrete wall What is the thickness of a copper plate of density 8.9 g/cm^3 needed to stop 0.4 MeV beta-rays? What is the thickness of an aluminum plate of density 2.8 g/cm^3 needed to stop beta-rays of the same energy? If a beam of 0.4 MeV gamma-rays is attenuated by 4% by that slab of aluminum, what is the mean free path of these y-rays in aluminum?

Explanation / Answer

Part a)

The beta radiation are high speed electrons emitted by radioactive nuclei decay time

The penetrating power of this radiation is medium and can be absorbed by a sheet of aluminum 1 cm thick

Part b)

Gamma radiation is electromagnetic waves of very high energy, it is a very penetrating radiation and need a high mass material thickness for absorption, and therefore need a concrete thickness 1m

Part c)

The analysis with beta rays is more delicate because the energy of these is highly variable assume that the given value is the maximum 0.4 MeV, in this case the exhaustion of copper is 0.04

I = Io e-m x

I/Io= e-1 = 0.368

0.368 = e-m x

ln 0.368 = -m x

x = - ln 0.368 / m

x = 1 / 0.04 8.9

x = 2.225 cm

Part d)

Part e)

The expression for the absorption of gamma rays by matter is

I = Io e-x = Io e-mx

Where

It is linear attenuation coefficient

m It is the mass attenuation coefficient

is the density

the beam is attenuated 4%

I/Io = 0.04

= 2.7 gr/cm3 aluminium

m = 0.09 to radiation energy in the order of 0.4 MeV

0.04 = e -m x

0.04 = e -0.09 2.7 x

ln 0.04 = -0.09 2.7 x

x= -ln( 0.04) / ( 0.09 2.7)

x = 3.219/0.243

x = 13.25 cm

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