7 -

Problem 1 (20 points) : Explain the following terms and their corresponding units in detail :

Solution

a) absorbed dose (6 points)

① The energy absorbed per unit mass of absorber medium due to radiation exposure.

② Hence, it can be used for any materials including the human body.

③ Units

- conventional unit : rad ≡ "energy of 100 ergs absorbed by a gram of medium"

≡ 1 rad = 100 ergs/g

- SI-unit : Gy(gray) ≡ "energy of 1 Joule absorbed by a kilogram of medium"

≡ 1 Gy = 1 J/□ = 100 rads

b) dose equivalent (7 points)

① The absorbed dose modified by the biological effectiveness to human due to radiation

type and other modifying factors ; [eqn: H`=`SUM from H ``D_R ``w_R ``N_R]

② Units

- conventional unit : rem(roentgen equivalent in man)

□ in case DR(absorbed dose) is given by rad

- SI-unit : Sv(sievert)

□ in case DR is given by Gy ≡ 1 Sv = 100 rems

c)  exposure (7 points)

① The number of ion pairs produced in air by □- or X- radiation

② Units

- conventional unit : roentgen(R, r) ≡ exposure quantity of □- and x-radiation which produces

ion pairs which are equivalent to 1 esu /□ of STP air

- SI-unit : Xunit(= C/□ of air) ≡ 1 Xunit = 3880 r

Problem 2 (20 points) : Show that the exposure rate is given by

[eqn: bold{dot X `=`left(1 over 34 right)left(1.6times10^-13 right)`Sum_i` left({mu_e (E_i )} over {rho}right)_air `E_i `phi_i ~~~[Xunit/sec]}]

Solution

The absorption rate of radiation is given by : [eqn: F `=`int_E `Sigma_a (E) `phi(E) `dE ~~~[absorptions/sec cdot cm^3 ]]

where, [eqn: Sigma_a (E)] = absorption cross-section,  [eqn: [cm^-1 ]]

[eqn: phi(E)] = differential beam flux of radiation, [eqn: ["#" /sec cdot cm^2 cdot MeV ]], and

[eqn: E] = energy of the beam, [eqn: [MeV ]]

The absorption rate is : [eqn: F `=`int_E `E`Sigma_a (E) `phi(E) `dE ~~~[MeV/sec cdot cm^3 ]]

For discrete photons, the energy absorption rate in the absorber medium m is

[eqn: F_E^m `=`Sum_i `E_i ``mu_e^m (E_i )`phi_i ~~~[MeV/sec cdot cm^3 ]]

where, [eqn: E_i] = energy of photon i (discrete),  [eqn: [MeV ]]

[eqn: mu_e^m (E_i )] = linear energy absorption coefficient of medium m for photon energy [eqn: E_i], [eqn: [cm^-1 ]], and

[eqn: phi_i] = beam flux of photon i, [eqn: ["#" /cm^2 cdot sec ]]

Hence, the absorbed dose rate by the medium is

[eqn: dot D `=`{F_E^m} over rho_m `=` Sum_i` left({mu_e (E_i )} over {rho}right)_m `E_i `phi_i ~~~[MeV/kg cdot sec]]

or in Joules, [eqn: dot D `=`left(1.6times10^-13 right)`Sum_i` left({mu_e (E_i )} over {rho}right)_m `E_i `phi_i ~~~[J/kg cdot sec ~(=`Gy/sec)] ~~cdotscdots~①]

where, [eqn: left(1.6times10^-13 right) `=`Joules/MeV]

[eqn: left({mu_e (E_i )} over {rho}right)_m] = Mass energy absorption coefficient of medium

for photon energy [eqn: E_i , ``[cm^2 /kg]], and

[eqn: E_i ``phi_i] = Energy flux of photon i, [eqn: [MeV /sec cdot cm^2 ]]

The exposure rate is given by : [eqn: dot X ~[Xunit/sec] `=`{1 over 34} dot D ~[Gy/sec-air]]

Therefore, inserting Eq. ① for air, the exposure rate is given by

[eqn: bold{dot X `=`left(1 over 34 right)left(1.6times10^-13 right)`Sum_i` left({mu_e (E_i )} over {rho}right)_air `E_i `phi_i ~~~[Xunit/sec]}]

Problem 3 (20 points) : Consider a point isotropic source of q (Bq), and a detector located

at r meter away from the source.

a) Show that the exposure rate from a point □-source at r meter away is

[eqn: bold{dot X `=`{1.043times10^-12}{q over r^2}`Sum_i f_i `E_i `left{mu_e^air right}_i ~~~[Xunit/hr]}]

b) Source strength, □ is defined as the gamma-radiation exposure from 1 MBq point source at 1 meter away. Show that the source strength becomes :

[eqn: bold{Gamma `=`{1.043times10^-6}`Sum_i f_i `E_i `left{mu_e^air right}_i ~~~[Xunit/hr]}]

Solution

(a) (12 points)

If the attenuation of gamma-rays in the air is neglected, the photon flux of energy [eqn: E_i]

at r meter away from a point source q(Bq) is : [eqn: phi_i `(r) `=`{q over {4`pi`r^2}}f_i ~~[photons/sec cdot m^2 ]]

where, [eqn: q] = activity of the point source, [eqn: [Bq]]

[eqn: f_i] = yield of photon i per transmission, and

[eqn: r] = distance between the source and detector point, [eqn: [m]]

Using the result of problem #2, the exposure rate from a point □-source at r meter away is

[eqn: dot X `=& left(1 over 34 right)left(1.6times10^-13 right)`Sum_i` left({mu_e (E_i )} over {rho}right)_air `E_i `phi_i `(r) ~~~[Xunit/sec] ## =& {{(1.6times10^-13 )(3600)} over {34}}{q over {4`pi`r^2}}`Sum_i` f_i `E_i `left({mu_e (E_i )} over {rho}right)_air ~~~[Xunit/hr] ## =& 1.35times10^-12 {q over {r^2}}`Sum_i` f_i `E_i `left({mu_e (E_i )} over {rho}right)_air ~~~[Xunit/hr]]

where, [eqn: left({mu_e (E_i )} over {rho}right)_air] = Mass energy absorption coefficient of air for photon energy [eqn: E_i , ``[m^2 /kg]]

[eqn: therefore ~dot X `=` {{1.35times10^-12}over 1.293} {q over {r^2}}`Sum_i` f_i `E_i `left{mu_e^air right}_i bold{={1.043times10^-12}{q over r^2}`Sum_i f_i `E_i `left{mu_e^air right}_i ~~[Xunit/hr]}}]

where, [eqn: left{mu_e^air right}_i] = linear energy absorption coefficient of air for [eqn: E_i , ``[m^-1 ]]

[eqn: rho_air] = [eqn: 1.293``kg/m^3]

(b) (8 points)

"Specific gamma-emission (source strength : □)" is defined as the gamma-radiation exposure from 1 MBq point source at 1 meter away, i.e., [eqn: Gamma `=`dot X `(r=1m`,`q=1MBq)]

Using the result of problem #3-(a), the source strength becomes :

[eqn: dot X `=` {1.043times10^-12}{{10^6 `Bq} over (1m)^2}`Sum_i f_i `E_i `left{mu_e^air right}_i ~~[Xunit/hr]}]

or, [eqn: bold{dot X `=` {1.043times10^-6} `Sum_i f_i `E_i `left{mu_e^air right}_i ~~[Xunit/hr]}}]

Problem 4 (20 points) :

a) Calculate the exposure rate (R/hr) at 1 meter away from a Na-24 point source of 1 Ci.

b) What is the absorbed dose (mrad) that a man received if he stayed there for 10 minutes?

Data □  Na-24 :[eqn: bold{T_{1/2} `=`14.96``hour}]

□[eqn: bold{E_gamma}] = 1.369 MeV (100%) & 2.754 MeV (100%)

□ [eqn: bold{left({mu_e (1.369MeV)} over rho right)_air `=`0.027`cm^2 / g}], [eqn: bold{left({mu_e (2.754MeV)} over rho right)_air `=`0.021`cm^2 / g}]

□ [eqn: bold{left({mu_e (1.369MeV)} over rho right)_tissue `=`0.030`cm^2 / g}], [eqn: bold{left({mu_e (2.754MeV)} over rho right)_air `=`0.025`cm^2 / g}]

Solution

(a) (10 points) The energy absorption rate by the medium is given by :

[eqn: dot D `=`left(1.6times10^-6 right)`Sum_i` left({mu_e (E_i )} over {rho}right)_m `E_i `phi_i ~~~[ergs/g cdot sec]]

where, [eqn: left({mu_e (E_i )} over {rho}right)_m] = Mass energy absorption coefficient of medium for photon energy [eqn: E_i , ``[cm^2 /g]]

[eqn: E_i ``phi_i] = Energy flux of photon i, [eqn: [MeV /sec cdot cm^2 ]], and

[eqn: left(1.6times10^-6 right) `=`ergs/MeV]

At r cm away from a point source of q (Bq), since the photon flux is given by : [eqn: phi_i `=`{f_i ``q} over {4 `pi`r^2}] ,

the energy absorption rate in the medium becomes :

[eqn: dot D `=`left(1.6times10^-6 right){q over {4`pi`r^2}}`Sum_i` left({mu_e (E_i )} over {rho}right)_m `E_i `f_i ~~~[ergs/g cdot sec]]

Hence, the absorbed dose rate of air becomes :

[eqn: dot D `=& left(1.6times10^-6 right){q over {4`pi`r^2}}`Sum_i` left({mu_e (E_i )} over {rho}right)_air `E_i `f_i ## =& left(1.6times10^-6 right){{(1)(3.7times10^10 )} over {4`pi`(100)^2}}`[(0.027)(1.369)(1)+(0.021)(2.754)(1) ] ## =& 4.47times10^-2 ~ergs/g cdot sec `=`4.47times10^-4 ~rads/sec `=`1.61`rad/hr]

Hence, the exposure rate becomes :

[eqn: bold{dot X`(R/hr) `=`{1 over 0.877}{dot D `(rad/hr)} `=`{1 over 0.877}(1.61 `rad/hr) `=`1.84`R/hr}]

(b) (10 points)

If the soft tissue is located in this radiation field, the absorbed dose of tissue becomes :

[eqn: dot D `=& left(1.6times10^-6 right){q over {4`pi`r^2}}`Sum_i` left({mu_e (E_i )} over {rho}right)_tissue `E_i `f_i ## =& left(1.6times10^-6 right){{(1)(3.7times10^10 )} over {4`pi`(100)^2}}`[(0.030)(1.369)(1)+(0.025)(2.754)(1) ] ## =& 5.18times10^-2 ~ergs/g cdot sec `=`5.18times10^-4 ~rads/sec `=`1.86`rad/hr]

Hence, if one stays in the radiation field for 10 mins, the absorbed dose by the soft tissue becomes :

[eqn: bold{D `=`(1.86 `rad/hr)left({1 over 6}hr right) `=`0.31`rad `=`310`mrad}]

Problem 5 (20 points) : Explain the differences between :

Solution

a) external and internal doses (5 points)

① External Dose : Portion of the dose equivalent received from radiation sources outside the body

② Internal Dose : Portion of the dose equivalent received from radioactive material taken into the body

b) individual and collective doses (5 points)

① Individual Dose : The dose equivalent measured by a personal dosimeter

② Collective Dose : The sum of the individual doses received in a given period of time

by a specified population from exposure to a specified source of radiation

c) whole body and organ dose (5 points)

① Whole Body Dose : The dose resulting from what is regarded as uniform exposure of the whole body

② Organ Dose : Average absorbed dose in a specified tissue or organ of the human body

d) occupational and public doses (5 points)

① Occupational dose means the dose received by an individual in the course of employment in which the individual's assigned duties involve exposure to radiation or to radioactive material from licensed and unlicensed sources of radiation, whether in the possession of the licensee or other person. Occupational dose does not include doses received from background radiation, from any medical administration the individual has received, from exposure to individuals administered radioactive material and released, from voluntary participation in medical research programs, or as a member of the public.

② Public dose means the dose received by a member of the public from exposure to radiation or to radioactive material released by a licensee, or to any other source of radiation under the control of a licensee. Public dose does not include occupational dose or doses received from background radiation, from any medical administration the individual has received, from exposure to individuals administered radioactive material and released, or from voluntary participation in medical research programs.