14 -

Problem 1 (10 points) : Briefly explain the definitions of the following terms :

Solution

(a) radiation and radioactivity (2 points)

□ Radiation :

① 10CFR20 (Standards for Protection against Radiation) :

alpha & beta particles, gamma rays, X-rays, neutrons, high-speed electrons, high-speed protons, other atomic particles capable of producing ion. (Radiation, as used in this regulation, does not include non-ionizing radiation, such as radio- or microwaves, or visible, infrared, or ultraviolet light.)

② 한국 원자력법 시행령 :

알파선, 중양자선, 양자선, 베타선, 기타 중하전입자선, 중성자선,

감마선 및 엑스선, 5만 전자볼트 이상의 에너지를 가진 전자선

□ Radioactivity

: The emission of radiation by spontaneous nuclear transformation,

or property of emitting radiation by spontaneous nuclear transformation.

(b) transformation constant and half-life (2 points)

□ Transformation constant (□)

: The rate of spontaneous nuclear transformation of nuclide A is proportional to the number of nuclei A :[eqn: bold{dN over dt `=`-lambda `N}] (where, N = number of nuclei A, □= Transformation constant). The unit of this constant is sec-1, min-1 or hr-1, etc. Similarly, the term of decay constant or disintegration constant is used instead of this constant.

□ Half-life (T½ = ln2 /□)

: The time required to transform to one half of the original number of nuclei.

(c) transient and secular equilibriums of radioactivity (2 points)

The activity ratio of daughter to parent nuclide is ;

[eqn: bold{A_B over A_A `==`{lambda_B `N_B}over{lambda_A `N_A} `=`lambda_B {left[1-exp left{-(lambda_B -lambda_A )`t right} right]}over{lambda_B -lambda_A}}]

□ Transient equilibrium

When [eqn: bold{lambda_A `<`lambda_B}] (for shorter-lived daughters), as t→∞,

[eqn: bold{A_B over A_A `=`lambda_B over {lambda_B -lambda_A} `=`T_A over {T_A -T_B} `>`1}]  (constant).

□ Secular equilibrium

When [eqn: bold{lambda_A `<<`lambda_B}] (for very shorter-lived daughters), as t→∞, [eqn: bold{A_B over A_A `=` 1}].

(d) stochastic effects and deterministic effects of radiation (2 points)

□ Deterministic effects

: non-stochastic effects and proportional to the severity of the effects with threshold

□ Stochastic effects

: probabilistic effects with no threshold and proportional to probability of occurrences

(e) absorbed dose and dose equivalent (2 points)

□ Absorbed dose

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

□ Dose equivalent

① The absorbed dose modified by the biological effectiveness to human

due to radiation type and other modifying factors ; [eqn: bold{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

Problem 2 (10 points) : Compute the number of grams and the number of radioactive atoms contained in 1 mCi of :

Solution

The activity is defined by :[eqn: bold{A `=`N`lambda}]. Hence,[eqn: bold{N`=`A over lambda `=`A `left({T_1/2} over {ln2}right)}]

(a) Na-24 (T½ = 14.8 hr) (4 points)

□ The number of radioactive atoms

[eqn: N `=`(10^-3 `Ci)(3.7times10^10 `tps/Ci)left({14.8times3600`sec}over{ln2}right) `=`bold{2.84times10^12 `atoms}]

□ The number of grams of radioactive atoms

[eqn: m `=`left({2.84times10^12 `atoms}over{6.02times10^23 `atoms/mole} right)`(24 `g/mole) `=`bold{1.13times10^-10 `g `=`113`pg}]

(b) P-32 (T½ = 14.5 d) (3 points)

□ The number of radioactive atoms

[eqn: N `=`(10^-3 `Ci)(3.7times10^10 `tps/Ci)left({14.5times24times3600`sec}over{ln2}right) `=`bold{6.69times10^13 `atoms}]

□ The number of grams of radioactive atoms

[eqn: m `=`left({6.69times10^13 `atoms}over{6.02times10^23 `atoms/mole} right)`(32 `g/mole) `=`bold{3.56times10^-9 `g `=`3.56`ng}]

(c) Ra-226 (T½ = 1620 yr) (3 points)

□ The number of radioactive atoms

[eqn: N `=`(10^-3 `Ci)(3.7times10^10 `tps/Ci)left({1620times365times24times3600`sec}over{ln2}right) `=`bold{2.73times10^18 `atoms}]

□ The number of grams of radioactive atoms

[eqn: m `=`left({2.73times10^18 `atoms}over{6.02times10^23 `atoms/mole} right)`(226 `g/mole) `=`bold{1.03times10^-3 `g `=`1.03`mg}]

Problem 3 (10 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 4 (10 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) (5 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) (5 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 (10 points)

(a) What are the radiation risk factors for carcinogenesis and hereditary effects?

(b) Derive the annual limits of radiation exposure for radiation workers and general public?

Solution

(a) (5 points)

□ for carcinogenesis effects (fatal cancer) = 4.0×10-2 / Sv

□ for adverse hereditary effects = 1.0×10-2 / Sv

(b) (5 points)

ICRP-60 Recommendations are as follows :

	Radiation Workers	Public
Annual acceptable risk	8.0×10-4 / year	5.0×10-5 / year
Radiation risk factor	4.0×10-2 / Sv	5.0×10-2 / Sv

Therefore,

□ annual limit of radiation exposure for radiation workers is

[eqn: {8.0times10^-4 `/yr} over {4.0times10^-2 `/yr} `=`2.0times10^-2 `Sv/yr `=`bold{20`mSv/yr}]

□ annual limit of radiation exposure for general public is

[eqn: {5.0times10^-5 `/yr} over {5.0times10^-2 `/yr} `=`1.0times10^-3 `Sv/yr `=`bold{1`mSv/yr}]

Problem 6 (10 points) : 2 measurements taken. Gross for 10 minutes and background for 20 minutes for each measurement. The results of measurement are as follows :

Measurement	Gross Counts	Background Counts
1	230	22
2	229	21

Find the net count rate (cpm) with standard deviation.

Solution

Let

① gross count rate (including the background count rate) = [eqn: r_g `+`sigma_g] , and

② background count rate = [eqn: r_b `+`sigma_b] , then

③ net count rate = [eqn: r_n `+`sigma_n] ( where, [eqn: r_n `=`r_g `-`r_b] , [eqn: sigma_n `=`sqrt{sigma_g^2 `+`sigma_b^2}] )

In this problem,

[eqn: □``r_n `=`r_g `-`r_b ~=&{230+229}over{20} - {22+21}over{40} `=`22.95 `-`1.08 `=`21.87 ~cpm]

[eqn: □``sigma_n `=`sqrt{sigma_g^2 `+`sigma_b^2}`=`sqrt{{{r_g}over{t_g}}+{{r_b}over{t_b}}} `=`sqrt{{{22.95}over{10}}+{{1.08}over{20}}} `=`1.53 ~cpm]

Hence, net count rate with standard deviation is, [eqn: bold{r_n `+`sigma_n `=`(21.87 `+-`1.53)~cpm}]

Problem 7 (10 points) : List the release points of gaseous and liquid radioactive materials from a typical PWR-type nuclear power plant in effluents.

Solution

(1) Gaseous effluents (5 points)

□ Containment purge or ventilation exhausts

□ PAB vent exhausts

□ Gaseous waste processing system discharge (controllable)

□ Steam leakage from the secondary systems

□ Condenser air-ejector off-gas

□ Steam generator blowdown system vent

□ Radwaste and fuel building vent

(2) Liquid effluents (5 points) (all effluents are controllable)

□ BRS discharge (for H-3 control and water balance)

□ Dirty and miscellaneous wastes (containment sump, PAB floor drains, etc.)

□ S/G blowdown treatment system discharge

□ Chemical wastes (laboratory, sampling) small volume

□ Condensate demineralizer regeneration system discharges

□ Turbine building floor drain sump

□ Detergent wastes (decontamination); based on experiences

□ Clean waste (equipment and RCDT drains)

※ Acronyms

□ PAB : Primary Auxiliary Building

□ BRS : Boron Recycle System

□ RCDT : Reactor Coolant Drain Tank

Problem 8 (20 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) (2 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) (4 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) (4 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) (2 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) (4 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) (4 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 9 (10 points) : The average primary coolant leakage into the auxiliary building is 80 kg/day.

1)  If there is no ventilation in the building,

find the equilibrium concentration of Xe-131m (Ci/m3) in the building air.

2)  Find the required ventilation rate

to keep the concentration of Xe-131m under DAC (2×107 Bq/m3).

3)  Find the annual release rate of Xe-131m to the environment via this ventilation exhaust.

Data :

□ Volume of the auxiliary building = 500 m3

□ Xe-131m concentration in the primary coolant during power operation = 23 □Ci/g

□ Half-life of Xe-131m = 11.8 days

Solution

(1) (4 points)

The auxiliary building encloses the several systems which handle the primary coolant. Typical systems are CVCS (chemical and volume control system) and BRS (boron recycle system).

The rate of change in airborne activity of nuclide i in the building is given by:

[eqn: dP_i over dt `=`l_pp `C_i `-`lambda_i `P_i `-`R_v `P_i]

where, [eqn: P_i] = activity of nuclide i in the PAB, [□Ci]

[eqn: l_pp] = primary leak rate into the auxiliary building [eqn: =` {(80kg)(1000g/kg)} over 24hr/day `=`3333`g/hr]

[eqn: C_i] = activity of nuclide i in the reactor coolant

[eqn: =` (23 mu Ci/g)(3.7times10^4 Bq/mu Ci) `=`8.51times10^5 Bq/g]

[eqn: lambda_i `=`ln2 over {(11.8days)(24hrs/day)} `=`2.45times10^-3 /hr]

[eqn: R_v ~left(`= F_v over V_PAB `right)] = vent exhaust rate of the auxiliary building, [vol/hr]

[eqn: P_v] = vent exhaust rate of the auxiliary building, [m3/hr]

[eqn: V_PAB] = volume of auxiliary building = 500 m3

In an equilibrium state, the airborne concentration of radioactivity in the auxiliary building atmosphere is : [eqn: P_i `=`{l_pp `C_i} over {(lambda_i + R_v )V_PAB}]

If there is no ventilation, the concentration becomes :

[eqn: P_i `=`{l_pp `C_i} over {lambda_i `V_PAB} `=` {(3333g/hr)(8.51times10^5 Bq/g)} over {(2.45times10^-3 /hr)(500m^3 )} `=`bold{2.32times10^9 Bq/m^3}]

(2) (3 points)

To keep the concentration of Xe-131m under DAC, the required ventilation rate becomes :

[eqn: P_i `=`{l_pp `C_i} over {(lambda_i + R_v )V_PAB} `□`DAC]

[eqn: R_v `V_PAB ~□& {l_pp `C_i} over {DAC} `-`lambda_i `V_PAB ## &=`` {(3333g/hr)(8.51times10^5 Bq/g)} over {2times10^7 Bq/m^3} `-`(2.45times10^-3 /hr)(500m^3 ) ## &=``1.41times10^2 `m^3 /hr]

[eqn: bold{therefore~R_v `V_PAB ~□~1.41times10^2 `m^3 /hr `=`3.37times10^3 `m^3 /day}]

(3) (3 points)

All the leaked Xe-131m will be eventually vented out to the environment. Hence, the annual release of Xe-131m becomes :

[eqn: R_i ~=&l_pp `C_i `T ## =&(3333g/hr)(8.51times10^5 Bq/g)(365days/yr)(24hrs/day) ## =&bold{2.49times10^13 ``Bq/year `=`6.72times10^2 ``Ci/year}]