#### Tasks with hints and answers

##### Task #1

Estimate the total specific energy losses of electrons with an energy of 150 MeV in aluminum and lead.

Hint

^{๐๐ธ}⁄

_{๐๐ฅ})

_{๐ก๐๐ก}=(

^{๐๐ธ}⁄

_{๐๐ฅ})

_{๐๐๐}+(

^{๐๐ธ}⁄

_{๐๐ฅ})

_{๐๐๐}

Answer

^{๐๐ธ}⁄

_{๐๐ฅ})

_{๐๐๐}=22.6

^{MeV}⁄

_{cm}, in lead (

^{๐๐ธ}⁄

_{๐๐ฅ})

_{๐๐๐}=310

^{MeV}⁄

_{cm}

##### Task #2

Determine the specific ionization loss of muons in aluminum if their kinetic energy is: 1) 50 MeV, 2) 100 MeV, 3) 500 MeV.

Hint

โ^{๐๐ธ}โ_{๐๐ฅ}=3.1โ10^{5}(^{๐z2ฯ}โ_{A๐ฝ2})(11.2+๐๐(^{๐ฝ2}โ_{๐(1โ๐ฝ2)})โ๐ฝ^{2}) [^{๐๐}โ_{๐๐}]

Z – nuclear charge, z – particle charge, A – mass number, ฯ – substance density, ๐ฝ=^{V}โ_{c}.

Total particle energy: ๐ธ=๐ธ_{๐}+๐๐^{2}=^{๐๐2}โ_{โ1โ๐ฝ2}, ๐ฝ^{2}=^{๐ผ2+2๐ผ}โ_{๐ผ2+2๐ผ+1} (๐ผ=๐ธ_{๐}๐๐^{2})

Answer

1) ๐ฝ^{2}=0.539, ^{๐๐ธ}โ_{๐๐ฅ}=6.2 ^{MeV}โ_{cm}, 2) ๐ฝ^{2}=0.736, ^{๐๐ธ}โ_{๐๐ฅ}=5.1 ^{MeV}โ_{cm}, 3) ๐ฝ^{2}=0.97, ^{๐๐ธ}โ_{๐๐ฅ}=4.8 ^{MeV}โ_{cm})

##### Task #3

Calculate the threshold proton energy for the photoproduction of ฯ^{0} mesons in the interaction of a proton with a photon of CMB (cosmic microwave background) p+ฮณ โ> p+ฯ^{0}

Hint

_{ฯ0}= 134.98 MeV, m

_{p}= 938.27 MeV.

๐ธ_{๐กโ}=m_{ฯ0}๐^{2}(1+^{(mฯ0๐2)}⁄_{(2mp๐2)})

Answer

##### Task #4

What should be the thickness of the walls of the aluminum container so that they absorb no more than 1% of gamma rays with an energy of 10 keV?

Hint

^{cm2}โ

_{g}) for E

_{ฮณ}=10 KeV

ฯ(Al)[cm^{-1}] = ฯ(Al)[ ^{cm2}โ_{g}]รฯ[^{g}โ_{cm3}] = 24.3ร2.7 = 65.6 cm^{-1}

^{๐ผ}โ_{๐ผ0}=๐^{โฯ๐ฅ}=99%=0.99, ๐ฅโคโ^{1}โ_{ฯ}ln(^{๐ผ}โ_{๐ผ0})

Answer

##### Task #5

Determine specific radiative losses during the passage of electrons with an energy of 50 MeV by an aluminum target and to compare them to specific losses for ionization.

Hint

If ^{137}โ_{๐1โ3} < ^{๐ธ}โ_{๐๐2} (58<98), we use:

(^{๐๐ธ}โ_{๐๐ฅ})_{๐๐๐} = โ^{p๐๐ด}โ_{๐ด}๐ธ ^{๐2๐02}โ_{137} (4๐๐(^{183}โ_{๐1โ3})+^{2}โ_{9}) [^{๐๐}โ_{๐๐}]

If ๐ธ < ๐๐^{2}, we use:

(^{๐๐ธ}โ_{๐๐ฅ})_{๐๐๐} = โ^{16}โ_{3}(^{p๐๐ด}โ_{๐ด})๐ธ ^{๐๐02}โ_{137} [^{๐๐}โ_{๐๐}]

If ^{๐ธ}โ_{๐๐2} < ^{137}โ_{๐1โ3}, we use:

(^{๐๐ธ}โ_{๐๐ฅ})_{๐๐๐} = โ^{p๐๐ด}โ_{๐ด}๐ธ ^{๐2๐02}โ_{137} (4๐๐(^{2E}โ_{mec2})+^{4}โ_{3}) [^{๐๐}โ_{๐๐}]

^{๐๐ธ}โ

_{๐๐ฅ})

_{ion}= โ3.1โ10

^{5}โ

^{๐z2p}โ

_{๐ด๐ฝ2}(11.2+๐๐(

^{๐ฝ2}โ

_{๐(1โ๐ฝ2)})โ๐ฝ

^{2})

Answer

(^{๐๐ธ}โ_{๐๐ฅ})_{๐๐๐} = โ5.2 [^{M๐๐}โ_{๐๐}]

(^{๐๐ธ}โ_{๐๐ฅ})_{ion} = โ6 [^{M๐๐}โ_{๐๐}]

(^{๐๐ธ}โ_{๐๐ฅ})_{๐๐๐}/(^{๐๐ธ}โ_{๐๐ฅ})_{ion} ≈ 1.2

#### Tasks for independent solving

##### Task #6

Calculate the threshold electron energy for the photoproduction of an electron-positron pair: e^{–}+ฮณ โ> e^{–}+e^{–}+e^{+}

Answer

##### Task #7

Estimate the ratio of specific ionization losses in iron for protons and electrons with energies: 1) 10 MeV, 2) 100 MeV, 3) 1 GeV.

Answer

##### Task #8

Calculate the intensity of cosmic rays with kinetic energies > 1 GeV, based on the power-law form of the energy spectrum of cosmic rays with an index of 2.7 and their total energy density of ^{0.5 eV}โ_{cm3} (assume that the particles are relativistic and make the main contribution to the total energy density)

Answer

I = 2(^{Ek}/_{1GeV})^{-2.7} [^{particle}/_{(cm2*s*GeV)}]