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
Answer
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
โ๐๐ธโ๐๐ฅ=3.1โ105(๐z2ฯโA๐ฝ2)(11.2+๐๐(๐ฝ2โ๐(1โ๐ฝ2))โ๐ฝ2) [๐๐โ๐๐]
Z – nuclear charge, z – particle charge, A – mass number, ฯ – substance density, ๐ฝ=Vโc.
Total particle energy: ๐ธ=๐ธ๐+๐๐2=๐๐2โโ1โ๐ฝ2, ๐ฝ2=๐ผ2+2๐ผโ๐ผ2+2๐ผ+1 (๐ผ=๐ธ๐๐๐2)
Answer
1) ๐ฝ2=0.539, ๐๐ธโ๐๐ฅ=6.2 MeVโcm, 2) ๐ฝ2=0.736, ๐๐ธโ๐๐ฅ=5.1 MeVโcm, 3) ๐ฝ2=0.97, ๐๐ธโ๐๐ฅ=4.8 MeVโcm)
Task #3
Calculate the threshold proton energy for the photoproduction of ฯ0 mesons in the interaction of a proton with a photon of CMB (cosmic microwave background) p+ฮณ โ> p+ฯ0
Hint
๐ธ๐กโ=mฯ0๐2(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
ฯ(Al)[cm-1] = ฯ(Al)[ cm2โg]รฯ[gโcm3] = 24.3ร2.7 = 65.6 cm-1
๐ผโ๐ผ0=๐โฯ๐ฅ=99%=0.99, ๐ฅโคโ1โฯln(๐ผโ๐ผ0)
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 137โ๐1โ3 < ๐ธโ๐๐2 (58<98), we use:
(๐๐ธโ๐๐ฅ)๐๐๐ = โp๐๐ดโ๐ด๐ธ ๐2๐02โ137 (4๐๐(183โ๐1โ3)+2โ9) [๐๐โ๐๐]
If ๐ธ < ๐๐2, we use:
(๐๐ธโ๐๐ฅ)๐๐๐ = โ16โ3(p๐๐ดโ๐ด)๐ธ ๐๐02โ137 [๐๐โ๐๐]
If ๐ธโ๐๐2 < 137โ๐1โ3, we use:
(๐๐ธโ๐๐ฅ)๐๐๐ = โp๐๐ดโ๐ด๐ธ ๐2๐02โ137 (4๐๐(2Eโmec2)+4โ3) [๐๐โ๐๐]
(๐๐ธโ๐๐ฅ)ion = โ3.1โ105โ๐z2pโ๐ด๐ฝ2(11.2+๐๐(๐ฝ2โ๐(1โ๐ฝ2))โ๐ฝ2)
Answer
(๐๐ธโ๐๐ฅ)๐๐๐ = โ5.2 [M๐๐โ๐๐]
(๐๐ธโ๐๐ฅ)ion = โ6 [M๐๐โ๐๐]
(๐๐ธโ๐๐ฅ)๐๐๐/(๐๐ธโ๐๐ฅ)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 eVโcm3 (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)]