5 -

Problem 1 (20 points) : Explain the following terms.

Solution

a) dose commitment (4 points)

The absorbed dose accumulated by the organ until the introduced activity is completely eliminated from the organ, i.e., the internal dose accumulated t = 0 to t = ∞.

[eqn: D_inf `=`int_{0}^{inf} dot D (t) ``dt ~~[Gy]]

b) committed dose equivalent (4 points)

The □dose equivalent to target tissue T accumulated over 50 years following the intake of nuclide j in S. :

[eqn: H_50 `(T←S)_j `≡`int_{t_o}^{t_0 +50y} dot H (T←S)_j ``dt ~~[Sv]]

c) specific effective energy (4 points)

The □effective energy of radiation, biologically adjusted, absorbed by unit mass of T, which is emitted by a nuclear transformation of nuclide j in S.

[eqn: therefore`` SEE(T ``←``S)_j `=`{Sum_R `f_R `E_R `AF(T←S)_R `w_R `N_R} over {M_T} ~~[MeV/g cdot t]]

d) effective half-life (4 points)

Once the radioactive isotope is taken into the body and goes to an organ, it stays in the organ and affects the organ until they are completely cleared out of the organ. Hence, the "effective elimination constant (□E )" is defined in order to incorporate radiological decay as well as biological elimination of the radioactive isotope from the organ.

The effective half-life becomes : [eqn: T_E `=`ln2 over lambda_E `=`{T_R times T_B}over{T_R + T_B}]

(where[eqn: bold{T_R}] = radiological half-life,[eqn: bold{T_B}] = biological half-life)

e) absorbed fractions of various radiation types (4 points)

① AF(T←T)R = 1  (for all types of radiation except □ and n)

② AF(T←S)R = 0  (□, □, □, □,…)

③ Tables are given for AF of □-radiation.

Problem 2 (60 points) : Following the intake of 1 MBq of I-131 through inhalation,

a) Calculate the effective half-life of I-131 in the thyroid.

b) Find the time-dependent activity of I-131 in kBq in the thyroid.

c) Find the absorbed dose rate of the thyroid in mrad/hour at t=0.

d) Find the dose commitment to the thyroid in rad.

e) Find the specific effective energy of I-131 to the thyroid : SEE(Thyroid←Thyroid)I-131 .

f) Find the committed dose equivalent (mSv) to the thyroid.

Data :

□ thyroid weight = 20 grams □ Effective radius of the thyroid = 3 cm □ biological half-life of iodine in the thyroid = 138 days □ radiological half-life = 8.05 days □ fraction of iodine reaching the thyroid after inhalation = 0.2 □ iodine radiation energies (in MeV/t)
	Beta radiations		Gamma radiations
	fR	[eqn: bold{bar E _beta^R}]	fR	[eqn: bold{E _gamma^R}]	AF(T←T)R
	0.016	0.0701	0.026	0.080	0.111
	0.069	0.0955	0.054	0.284	0.120
	0.005	0.1428	0.820	0.364	0.122
	0.904	0.1917	0.068	0.637	0.124
	0.006	0.2850	0.016	0.723	0.120

Solution

(a) (5 points) [eqn: T_E `=`{T_R times T_B} over {T_R + T_B} `=` {8.05times138} over {8.05+138} `=`bold{7.61``days}]

(b) (15 points)

The rate of change in the number of I-131 nuclides in the thyroid is give by :

[eqn: {dN_{thyroid`;`I-131}(t)} over dt `=`-lambda_E (thyroid`;`I-131)``N_{thyroid`;`I-131}(t)]

Solving the equation, the time-dependent I-131 activity in the thyroid becomes :

[eqn: therefore~ {A_{thyroid`;`I-131}(t)} `=`A_{thyroid`;`I-131}(0) exp left[ -lambda_E (thyroid`;`I-131)`t right]]

The effective elimination constant of I-131 from the thyroid is :

[eqn: lambda_E (thyroid`;`I-131) `=`ln2 over 7.61days `=`0.091 day^-1 `=`1.05times10^-6 `sec^-1]

Hence, the time-dependent I-131 activity in the thyroid following the instantaneous uptake of 1 MBq at t=0 is :

[eqn: bold{therefore~{A_{thyroid`;`I-131}(t)} `=`(10^3 times0.2 ) `e^{-0.091`t} `=`200`e^{-0.091`t} ~~[kBq] }]

(where t is given in days)

[eqn: bold{therefore~ {A_{thyroid`;`I-131}(t)} `=`(10^3 times0.2 ) `e^{-1.05times10^-6`t} `=`200`e^{-1.05times10^-6`t} ~~[kBq]}]

(where t is given in seconds)

(c) (15 points)

The absorbed dose rate of the thyroid consists of following two parts ;

1) For □-radiation □[eqn: dot D_beta`` (Gy/ sec) `=`(1.6times10^-13 ) ``q `bar E _beta over M_T]

where, [eqn: 1.6times10^-13 ~J/MeV] , [eqn: q] = activity in the organ, Bq

[eqn: bar E _beta] = energy, MeV/t , [eqn: M_T] = mass of the organ, □

[eqn: therefore ```dot D_beta `=&(1.6times10^-13 ) ``(2times10^5 ) `{{Sum_R`` f_R ``barE_beta^R } over {(2times10^-2 )}} ## =&(1.6times10^-13 ) ``(2times10^5 ) `{{(1.83times10^-1 )} over {(2times10^-2 )}} `=`bold{2.93times10^-7 ``Gy`/`sec}]

[eqn: left({Sum_R`` f_R ``barE_beta^R } ~=~ 0.016times0.0701 + 0.069times0.0955 + 0.005times0.1428 + 0.904times0.1917 + 0.006times0.285 `=` 0.183right)]

2) For □-radiation □

[eqn: dot D_gamma`` (Gy/ sec) `=&(1.6times10^-13 ) ``q `SEE(T←T)_{I-131}^gamma ## =&(1.6times10^-13 ) ``(2times10^5 ) `(2.25) ## =&bold{7.2times10^-8 ``Gy`/`sec}]

[eqn: left("cf. solution of problem-2-e" ~:~SEE(T←T)_{I-131}^gamma `=`{0.045`Mev/t} over {0.02`kg} `=`2.25`Mev / □ cdot t `right)]

By the results of 1) & 2)

[eqn: therefore ~dot D `=& dot D_beta `+`dot D_gamma `=`(29.3+7.2)times10^-8 `=`3.65times10^-7 ~Gy/sec ## =&3.65times10^-2 ~mrad`/sec `=`bold{131.4`mrad`/hour}]

(d) (5 points)

The dose commitment to the thyroid becomes ;

[eqn: D_inf `=` {dot D_0} over {lambda_E} `=`{3.65times10^-2 `mrad`/sec} over {1.05times10^-6 ``sec^-1} `=`3.48times10^4 `mrad `=`bold{34.8 `rad}]

(e) (10 points)

[eqn: SEE(Thyroid←Thyroid`)_{I-131} `=`SEE(T←T)_I-131^beta `+`SEE(T←T)_I-131^gamma]

[eqn: =`{{Sum_R ``[`f_R^beta `E_R^beta `w_R^beta `N_R^beta ]}+{Sum_R ``[`f_R^gamma `E_R^gamma ` AF(T←T)_R^gamma `w_R^gamma `N_R^gamma ]}} over M_Thyroid]

The value of the numerator is  ;[eqn: {{Sum_R ``[`f_R^beta `E_R^beta `w_R^beta `N_R^beta ]}+{Sum_R ``[`f_R^gamma `E_R^gamma ` AF(T←T)_R^gamma `w_R^gamma `N_R^gamma ]}}]

= {0.016×0.0701×1×1 + 0.069×0.0955×1×1 + 0.005×0.1428×1×1 + 0.904×0.1917×1×1 +0.006×0.285×1×1}

+ {0.026×0.080×0.111×1×1 + 0.054×0.284×0.120×1×1 +0.82×0.364×0.122×1×1 + 0.068×0.637×0.124×1×1 + 0.016×0.723×0.120×1×1}

= 0.183 + 0.045 = 0.228 MeV / t

[eqn: therefore ~SEE(Thyroid←Thyroid`)_{I-131} `=`{0.228 ``MeV`/`t} over {0.02`kg} `=`bold{11.4 Mev / □ cdot t}]

(f) (10 points)

Let [eqn: U_{Thyroid ; I-131} ~≡` int_{0}^{50y} A_{thyroid`;`I-131} (t) `dt ~~["# of transformations"]]

The committed dose equivalent to the thyroid due to I-131 uptake is ;

[eqn: H_50 (Thyroid←Thyroid)_{I-131} `=`(1.6times10^-10 )times(U_{Thyroid ; I-131 })timesSEE(Thyroid←Thyroid)_{I-131}]

①

[eqn: U_{Thyroid ; I-131} ~≡& int_{0}^{50y} A_{thyroid`;`I-131} (t) `dt ~~["# of transformations"] ## =& int_{0}^{50y} A_{thyroid`;`I-131} (t) `dt `=` int_{0}^{50y}A_{thyroid`;`I-131}(0) e^{ -lambda_E (thyroid`;`I-131)`t } `dt ## =&{A_{thyroid`;`I-131}(0)} over { lambda_E (thyroid`;`I-131)} `=`{0.2times10^6 `Bq}over{1.05times10^-6 `sec^-1 } ## =&1.91times10^11 ``transformations]

② [eqn: SEE(Thyroid←Thyroid`)_{I-131} `=`1.14times10^-2 Mev / g cdot t]

Hence,

[eqn: H_50 (Thyroid←Thyroid)_{I-131} `=&(1.6times10^-10 )times(1.91times10^11 )times(1.14times10^-2 ) ## =&3.48times10^-1 `Sv `=`bold{348`mSv}]

Problem 3 (20 points) : One of major thermal neutron interactions with the soft tissue in the human body is (n,□) reaction of hydrogen ; H1 (n,□) H2. The accompanied □-ray has the energy of 2.23 MeV. Find the dose-equivalent rate (mSv/hr) of a man of 70 kg who is exposed by the thermal neutron flux of 4×106 n/sec·□.

Data : # of hydrogen atoms per kg of tissue = 5.98×1025 #/kg

□n□ = (n,□) X-section of hydrogen for thermal neutrons = 3.3×10-25 □

absorbed fraction of 2.23 MeV gamma within a total body = 0.278

Solution

For H1 (n,□) reaction ; the specific activity of □-emitter in the body is given by : [eqn: q `=` phi `N`sigma_{n gamma} ~~[Bq/kg]]

where, [eqn: phi] = thermal neutron flux, [eqn: [n/sec cdot cm^2 ]]

[eqn: N] = # of hydrogen atoms per kg of tissue, [eqn: 5.98times10^25 ~"#"/kg]

[eqn: sigma_{n gamma}] = (n,□) cross-section for hydrogen, [eqn: 3.3times10^-25 ~cm^2]

[eqn: q `=` (4times10^6 `n/sec cdot cm^2 )(5.98times10^25 `"#"/kg)(3.3times10^-25 `cm^2 ) `=`7.89times10^7 ~Bq/kg]

Hence, the absorbed dose rate due to 2.23 MeV photons is :

[eqn: {dot D}_{n gamma} `=&q`(Bq/kg)`E`(MeV)`AF`(tot larrow tot)`(1.6times10^-13 `J/MeV) ## =&(7.89times10^7 `Bq/kg)`(2.23`MeV)`(0.278)`(1.6times10^-13 `J/MeV) ## =& 7.83times10^-6 `Gy/sec]

The dose-equivalent rate is :

[eqn: {dot H}_{n gamma} `=& {dot D}_{n gamma} `w_gamma `N_gamma ## =& (7.83times10^-6 `Gy/sec)(1)(1) `=`7.83times10^-6 `Sv/sec `=`bold{28.19`mSv/hr}]