Tasks

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

(𝑑𝐸⁄𝑑π‘₯)π‘‘π‘œπ‘‘=(𝑑𝐸⁄𝑑π‘₯)π‘–π‘œπ‘›+(𝑑𝐸⁄𝑑π‘₯)π‘Ÿπ‘Žπ‘‘

Answer

In aluminum (𝑑𝐸⁄𝑑π‘₯)π‘–π‘œπ‘›=22.6 MeV⁄cm, in lead (𝑑𝐸⁄𝑑π‘₯)π‘–π‘œπ‘›=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

βˆ’π‘‘πΈβ„π‘‘π‘₯=3.1βˆ—105(𝑍z2ρ⁄A𝛽2)(11.2+𝑙𝑛(𝛽2⁄𝑍(1βˆ’π›½2))βˆ’π›½2) [π‘’π‘‰β„π‘π‘š]

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

Total particle energy: 𝐸=πΈπ‘˜+π‘šπ‘2=π‘šπ‘2β„βˆš1βˆ’π›½2, 𝛽2=𝛼2+2𝛼⁄𝛼2+2𝛼+1 (𝛼=πΈπ‘˜π‘šπ‘2)

Answer

1) 𝛽2=0.539, 𝑑𝐸⁄𝑑π‘₯=6.2 MeV⁄cm, 2) 𝛽2=0.736, 𝑑𝐸⁄𝑑π‘₯=5.1 MeV⁄cm, 3) 𝛽2=0.97, 𝑑𝐸⁄𝑑π‘₯=4.8 MeV⁄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 –> p+Ο€0

Hint

mΟ€0 = 134.98 MeV, mp = 938.27 MeV.

πΈπ‘‘β„Ž=mΟ€0𝑐2(1+(mΟ€0𝑐2)⁄(2mp𝑐2))

Answer

144.7 MeV

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

Linear absorption coefficient is Ο„(Al)=24 (cm2⁄g) for EΞ³=10 KeV

Ο„(Al)[cm-1] = Ο„(Al)[ cm2⁄g]×ρ[g⁄cm3] = 24.3Γ—2.7 = 65.6 cm-1

𝐼⁄𝐼0=π‘’βˆ’Ο„π‘₯=99%=0.99, π‘₯β‰€βˆ’1⁄τln(𝐼⁄𝐼0)

Answer

≀ 0.0015 mm

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⁄𝑍1⁄3 < πΈβ„π‘šπ‘2 (58<98), we use:

(𝑑𝐸⁄𝑑π‘₯)π‘Ÿπ‘Žπ‘‘ = βˆ’p𝑁𝐴⁄𝐴𝐸 𝑍2π‘Ÿ02⁄137 (4𝑙𝑛(183⁄𝑍1⁄3)+2⁄9) [π‘’π‘‰β„π‘π‘š]

If 𝐸 < π‘šπ‘2, we use:

(𝑑𝐸⁄𝑑π‘₯)π‘Ÿπ‘Žπ‘‘ = βˆ’16⁄3(p𝑁𝐴⁄𝐴)𝐸 π‘π‘Ÿ02⁄137 [π‘’π‘‰β„π‘π‘š]

If πΈβ„π‘šπ‘2 < 137⁄𝑍1⁄3, we use:

(𝑑𝐸⁄𝑑π‘₯)π‘Ÿπ‘Žπ‘‘ = βˆ’p𝑁𝐴⁄𝐴𝐸 𝑍2π‘Ÿ02⁄137 (4𝑙𝑛(2E⁄mec2)+4⁄3) [π‘’π‘‰β„π‘π‘š]

(𝑑𝐸⁄𝑑π‘₯)ion = βˆ’3.1βˆ—105βˆ—π‘z2p⁄𝐴𝛽2(11.2+𝑙𝑛(𝛽2⁄𝑍(1βˆ’π›½2))βˆ’π›½2)

Answer

(𝑑𝐸⁄𝑑π‘₯)π‘Ÿπ‘Žπ‘‘ = βˆ’5.2 [Mπ‘’π‘‰β„π‘π‘š]

(𝑑𝐸⁄𝑑π‘₯)ion = βˆ’6 [Mπ‘’π‘‰β„π‘π‘š]

(𝑑𝐸⁄𝑑π‘₯)π‘Ÿπ‘Žπ‘‘/(𝑑𝐸⁄𝑑π‘₯)ion ≈ 1.2

Tasks for independent solving

Task #6

Calculate the threshold electron energy for the photoproduction of an electron-positron pair: e–+Ξ³ –> e–+e–+e+

Answer

8066 MeV

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

1) 17.5, 2) 2.4, 3) 0.6

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⁄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)]

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