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Problem 1 (20 points) : a) Find the uncollided and b) observed fluxes through a 1 cm Pb shield at 10 cm away from the source. The photon source is a point isotropic with 1 MeV energy. The source strength is 1010 photons per second. Consider the linear attenuation factor of Pb for 1 MeV photon is approximately (E = 1 MeV) = 1 cm-1 and the corresponding dose buildup factor Br(□r = 1,E = 1 MeV) for a point source is 1.37. Neglect any attenuation by the air. (including the unit of results)

Solution

a) The uncollided flux (7 points)

[eqn: PHI_u ` (t=1□`;`r=10□) ~=& {S_0 exp(-mu `t) } over {4pi`r^2} ## =&{(10^10 photons/sec) exp[-(1□^-1 ) (1□)]} over {4 pi`(10□)^2} ## =&bold{2.93times10^6 ``[photons/seccdot□]}]

b) The observed flux (7 points)

[eqn: PHI_o ` (t=1□`;`r=10□) ` =& B_r (mu t=1 `;`E=1Mev) PHI_u (t=1□`;`r=10□) ## =& (1.37) times (2.93times10^6 ) ## =& bold{4.01 times 10^6 ``[photons/seccdot□]}]

Problem 2 (20 points) : Briefly explain the following terms.

a) LET (3 points)

Solution

- The acronym of Linear Energy Transfer

- The energy loss per unit length due to collisions, which give rise to excitation and ionization

- Special interest in connection with the biological effects of radiation

b) stopping power (4 points)

Solution

The total energy lost per path length by a charged particle

( = The total rate of decrease in the energy of the particle along its path )

c) specific ionization (3 points)

Solution

- [eqn: "S.I." `=` {dE over dx} over w ~~["ion pairs"/cm]]     where, w = mean energy expenditure per ion pair.

- The number of ion pairs produced per unit path traveled by the particle

d) Bremsstrahlung (4 points)

Solution

The radiative loss of energy which appears as the continuous electromagnetic radiation produced by inelastic scattering of beta particle with nuclei

e) photo-electric effect (3 points)

Solution

The total absorption of photon by absorber atoms and subsequent ejection of orbital electron

f) Compton effect (3 points)

Solution

The elastic scattering of photon with atomic orbital electron

Problem 3 (20 points) : Briefly explain the interactions of beta, alpha, and

electromagnetic radiations with matter.

Solution

(1) □-ray interaction with matter (7 points)

① Collision with atomic electrons : ionization, excitation (dominates in few Mev region)

② Collision with nucleus : bremsstrahlung

- the radiative loss of energy which appears as the continuous electromagnetic radiation

produced by inelastic scattering of beta particle with nuclei

- becomes significant with high energy beta particles in high Z materials

- In particular, we should be very careful about using most common Pb shield since

shielding of high energy beta using lead creates an accompanying problem of secondary

radiation of electromagnetic radiation. Extra shielding should be considered for photons

due to bremsstrahlung

③ Annihilation : positron + negatron

(2) □-ray interaction with matter (6 points)

① Inelastic collision with atomic electrons : ionization, excitation

② Inelastic collision with nucleus : bremsstrahlung

③ Very high energy interaction : nuclear reaction(400∼500Mev), star reaction(Bev)

(3)  Electromagnetic radiation interaction with matter (7 points)

① Photo-electric effect

- total absorption of photon by absorber atoms and subsequent ejection of orbital electron

- the ejected photon carries the kinetic energy of incident photon subtracted by the

binding energy of the ejected orbital electron.

② Compton effect : elastic scattering of photon with atomic orbital electron

③ Pair production : the creation of (e+ + e-) pair due to photon interaction with the electric

field formed around atomic nucleus

④ Neutron production (Photo-neutron) : threshold reaction

Problem 4 (20 points) : Find the thickness of Al foil to stop 210Po-□ (E = 5.3 MeV). The range of the alpha particle in the STP air is 3.95 meters. Densities of aluminium and air are 2.7 and 1.293×10-3 g/□, respectively. Mass numbers of aluminium and air are 27 and 14.5, respectively. (including the unit of results)

Solution

By Bragg-Kleeman Rule, the ranges of alpha in different absorber media are

[eqn: R_1 over R_2 `=`rho_2 over rho_1 sqrt{A_1} over sqrt{A_2}]	where, □ = density of absorber medium, g/□, and A = atomic number

[eqn: rho_Al `=`2.7`g/cm^3]	[eqn: rho_Air `=`1.293times10^-3 `g/cm^3]
[eqn: A_Al `=`27]	[eqn: A_Air `=`14.5]

Therefore, the range of □ in the aluminium becomes,

[eqn: R_Al `=& rho_Air over rho_Al sqrt{A_Al} over sqrt{A_Air} R_air ## =& 1.293times10^-3 over 2.7 sqrt{27} over sqrt{14.5} (3.95times10^2 □) ## ≒&0.258□ `=`bold{2.58times10^-3 `m}]

Problem 5 (20 points) : Assuming the rate of linear energy loss of 0.1 MeV beta particle through STP air and through graphite are 5×10-3 MeV/cm and 8.67 MeV/cm, respectively.

a) Find the S.I. (specific ionization : number of ion-pairs per cm). (8 points)

b) Find the relative mass stopping power of graphite for 0.1 MeV beta particle. The density of graphite is :	[eqn: bold{rho_graphite}] = 2.25 g/□  (12 points)

Solution

a) [eqn: "S.I." `=`{5000`eV/cm} over {34`eV/i.p} `=`bold{147`"ion pairs"`/cm}]

b) [eqn: S_graphite `=`{8.67`MeV/cm} over {2.25`g/cm^3} `=`3.85`MeV/(g/cm^2 )]

[eqn: S_air `=`{5times10^-3 `MeV/cm} over {1.293times10^-3 `g/cm^3} `=`3.87`MeV/(g/cm^2 )]

[eqn: therefore ~ rho_m `=`S_graphite over S_air `=`3.85 over 3.87 `=`bold{0.995}]