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

Hint

(𝑑𝐸𝑑𝑥)𝑡𝑜𝑡=(𝑑𝐸𝑑𝑥)𝑖𝑜𝑛+(𝑑𝐸𝑑𝑥)𝑟𝑎𝑑

In aluminum (𝑑𝐸𝑑𝑥)𝑖𝑜𝑛=22.6 MeVcm, in lead (𝑑𝐸𝑑𝑥)𝑖𝑜𝑛=310 MeVcm

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

Hint

𝑑𝐸𝑑𝑥=3.1∗105(𝑍z2ρA𝛽2)(11.2+𝑙𝑛(𝛽2𝑍(1−𝛽2))−𝛽2) [𝑒𝑉𝑐𝑚]

Z – nuclear charge, z – particle charge, A – mass number, ρ – substance density, 𝛽=Vc.

Total particle energy: 𝐸=𝐸𝑘+𝑚𝑐2=𝑚𝑐2√1−𝛽2, 𝛽2=𝛼2+2𝛼𝛼2+2𝛼+1 (𝛼=𝐸𝑘𝑚𝑐2)

1) 𝛽2=0.539, 𝑑𝐸𝑑𝑥=6.2 MeVcm, 2) 𝛽2=0.736, 𝑑𝐸𝑑𝑥=5.1 MeVcm, 3) 𝛽2=0.97, 𝑑𝐸𝑑𝑥=4.8 MeVcm)

Calculate the threshold proton energy for the photoproduction of π0 mesons in the interaction of a proton with a photon of CMB (cosmic microwave background) p+γ –> p+π0

Hint

mπ0 = 134.98 MeV, mp = 938.27 MeV.

𝐸𝑡ℎ=mπ0𝑐2(1+(mπ0𝑐2)(2mp𝑐2))

144.7 MeV

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

Linear absorption coefficient is τ(Al)=24 (cm2g) for Eγ=10 KeV

τ(Al)[cm-1] = τ(Al)[ cm2g]×ρ[gcm3] = 24.3×2.7 = 65.6 cm-1

𝐼𝐼0=𝑒−τ𝑥=99%=0.99, 𝑥≤−1τln(𝐼𝐼0)

≤ 0.0015 mm

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 137𝑍1⁄3 < 𝐸𝑚𝑐2 (58<98), we use:

(𝑑𝐸𝑑𝑥)𝑟𝑎𝑑 = −p𝑁𝐴𝐴𝐸 𝑍2𝑟02137 (4𝑙𝑛(183𝑍1⁄3)+29) [𝑒𝑉𝑐𝑚]

If 𝐸 < 𝑚𝑐2, we use:

(𝑑𝐸𝑑𝑥)𝑟𝑎𝑑 = −163(p𝑁𝐴𝐴)𝐸 𝑍𝑟02137 [𝑒𝑉𝑐𝑚]

If 𝐸𝑚𝑐2 < 137𝑍1⁄3, we use:

(𝑑𝐸𝑑𝑥)𝑟𝑎𝑑 = −p𝑁𝐴𝐴𝐸 𝑍2𝑟02137 (4𝑙𝑛(2Emec2)+43) [𝑒𝑉𝑐𝑚]

(𝑑𝐸𝑑𝑥)ion = −3.1∗105𝑍z2p𝐴𝛽2(11.2+𝑙𝑛(𝛽2𝑍(1−𝛽2))−𝛽2)

(𝑑𝐸𝑑𝑥)𝑟𝑎𝑑 = −5.2 [M𝑒𝑉𝑐𝑚]

(𝑑𝐸𝑑𝑥)ion = −6 [M𝑒𝑉𝑐𝑚]

(𝑑𝐸𝑑𝑥)𝑟𝑎𝑑/(𝑑𝐸𝑑𝑥)ion ≈ 1.2

Calculate the threshold electron energy for the photoproduction of an electron-positron pair: e+γ –> e+e+e+

8066 MeV

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