#### 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

^{ππΈ}⁄

_{ππ₯})

_{π‘ππ‘}=(

^{ππΈ}⁄

_{ππ₯})

_{πππ}+(

^{ππΈ}⁄

_{ππ₯})

_{πππ}

Answer

^{ππΈ}⁄

_{ππ₯})

_{πππ}=22.6

^{MeV}⁄

_{cm}, in lead (

^{ππΈ}⁄

_{ππ₯})

_{πππ}=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

β^{ππΈ}β_{ππ₯}=3.1β10^{5}(^{πz2Ο}β_{Aπ½2})(11.2+ππ(^{π½2}β_{π(1βπ½2)})βπ½^{2}) [^{ππ}β_{ππ}]

Z – nuclear charge, z – particle charge, A – mass number, Ο – substance density, π½=^{V}β_{c}.

Total particle energy: πΈ=πΈ_{π}+ππ^{2}=^{ππ2}β_{β1βπ½2}, π½^{2}=^{πΌ2+2πΌ}β_{πΌ2+2πΌ+1} (πΌ=πΈ_{π}ππ^{2})

Answer

1) π½^{2}=0.539, ^{ππΈ}β_{ππ₯}=6.2 ^{MeV}β_{cm}, 2) π½^{2}=0.736, ^{ππΈ}β_{ππ₯}=5.1 ^{MeV}β_{cm}, 3) π½^{2}=0.97, ^{ππΈ}β_{ππ₯}=4.8 ^{MeV}β_{cm})

##### Task #3

Calculate the threshold proton energy for the photoproduction of Ο^{0} mesons in the interaction of a proton with a photon of CMB (cosmic microwave background) p+Ξ³ β> p+Ο^{0}

Hint

_{Ο0}= 134.98 MeV, m

_{p}= 938.27 MeV.

πΈ_{π‘β}=m_{Ο0}π^{2}(1+^{(mΟ0π2)}⁄_{(2mpπ2)})

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}β

_{g}) for E

_{Ξ³}=10 KeV

Ο(Al)[cm^{-1}] = Ο(Al)[ ^{cm2}β_{g}]ΓΟ[^{g}β_{cm3}] = 24.3Γ2.7 = 65.6 cm^{-1}

^{πΌ}β_{πΌ0}=π^{βΟπ₯}=99%=0.99, π₯β€β^{1}β_{Ο}ln(^{πΌ}β_{πΌ0})

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}β_{π1β3} < ^{πΈ}β_{ππ2} (58<98), we use:

(^{ππΈ}β_{ππ₯})_{πππ} = β^{pππ΄}β_{π΄}πΈ ^{π2π02}β_{137} (4ππ(^{183}β_{π1β3})+^{2}β_{9}) [^{ππ}β_{ππ}]

If πΈ < ππ^{2}, we use:

(^{ππΈ}β_{ππ₯})_{πππ} = β^{16}β_{3}(^{pππ΄}β_{π΄})πΈ ^{ππ02}β_{137} [^{ππ}β_{ππ}]

If ^{πΈ}β_{ππ2} < ^{137}β_{π1β3}, we use:

(^{ππΈ}β_{ππ₯})_{πππ} = β^{pππ΄}β_{π΄}πΈ ^{π2π02}β_{137} (4ππ(^{2E}β_{mec2})+^{4}β_{3}) [^{ππ}β_{ππ}]

^{ππΈ}β

_{ππ₯})

_{ion}= β3.1β10

^{5}β

^{πz2p}β

_{π΄π½2}(11.2+ππ(

^{π½2}β

_{π(1βπ½2)})βπ½

^{2})

Answer

(^{ππΈ}β_{ππ₯})_{πππ} = β5.2 [^{Mππ}β_{ππ}]

(^{ππΈ}β_{ππ₯})_{ion} = β6 [^{Mππ}β_{ππ}]

(^{ππΈ}β_{ππ₯})_{πππ}/(^{ππΈ}β_{ππ₯})_{ion} ≈ 1.2

#### Tasks for independent solving

##### Task #6

Calculate the threshold electron energy for the photoproduction of an electron-positron pair: e^{–}+Ξ³ β> 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}β_{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)}]