5 -

Problem 1 (20 points) : Briefly explain the definitions of the following terms.

a) radiation and radioactivity (10 points)

Solution

① Radiation :

ⓐ 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.)

ⓑ 한국 원자력법 시행령

알파선, 중양자선, 양자선, 베타선, 기타 중하전입자선, 중성자선, 감마선 및 엑스선, 5만 전자볼트 이상의 에너지를 가진 전자선

② Radioactivity :

The emission of radiation by spontaneous nuclear transformation, or property of emitting radiation by spontaneous nuclear transformation.

b) transformation constant and half-life (10 points)

Solution

① Transformation constant (□) :

The rate of spontaneous nuclear transformation of nuclide A is proportional to the number of nuclei A : [eqn: 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.

② Half-life (T½ = ln2 /□) :

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

Problem 2 (20 points) : Find the followings when a 1□ flask is filled with pure 222Rn (T1/2 = 3.82 days) gas. Assume the temperature is 20℃ and the pressure is 76 mmHg).

a) transformation constant of 222Rn (5 points)

Solution

[eqn: □~lambda ~[day^-1 ]`=`ln2 over {3.82``days} ``=``bold{1.81times10^-1 ``day^-1}]

[eqn: □~lambda ~[hr^-1 ]`=`ln2 over {3.82times24``hr} ``=``bold{7.56times10^-3 ``hr^-1}]

[eqn: □~lambda ~[sec^-1 ]`=`ln2 over {3.82times24 times3600``sec} ``=``bold{2.10times10^-6 ``sec^-1}]

b) number of nuclei per liter of pure 222Rn (5 points)

Solution

The amount of 222Rn in the flask in moles becomes :

X moles/□ : 76 mmHg = 1 mole/22.4□ : 760mmHg

[eqn: therefore~X `=&4.46times10^-3 `moles/□ ## =&(4.46times10^-3 `moles)(6.02times10^23 ``"#"/mole)/□`=`bold{2.69times10^21 ``"#" /□}]

c) activity of 222Rn in curies (5 points)

Solution

[eqn: A_Rn (t=0)`=`(2.69times10^21 )(2.10times10^-6 )`=`5.65times10^15 `Bq `=`bold{1.53times10^5 `Ci}]

d) activity of 222Rn in curies after 5 days cooling (5 points)

Solution

[eqn: A_Rn (t=5`days)`=`(1.53times10^5 `Ci)`exp left({-{ln2 over 3.82}times5}right) `=`bold{6.18times10^4 `Ci}]

Problem 3 (20 points) : Calculate the equilibrium concentration of 222Rn (T1/2 = 3.82 days) in a sealed room of 500□ where 1 Ci of 226Ra (T1/2 = 1,620 yrs) is stored. Find the ventilation exhaust rate (□/hour) in order to keep the concentration of 222Rn in the room under 4×10-6 Ci/□.

Solution

Let [eqn: A `=`Ra^226] & [eqn: B `=`Rn^222]

The rate of change in the number of [eqn: Rn^222] : [eqn: dN_B over dt `=`lambda_A `N_A `-`lambda_B `N_B `-`R`N_B]

where, [eqn: R] = room ventilation exhaust rate [volumes/hr]

a) In a sealed room without ventillation, Rn222 (Radon gas) is in secular equilibrium with Ra226. If Rn222 is in secular equilibrium with Ra226, then

[eqn: dN_B over dt `=`lambda_A `N_A `-`lambda_B `N_B `-`R`N_B `=`0 ~~and~~R `=`0]

In this case, Rn222 activity in the room becomes ; [eqn: A_B `=`lambda_B `N_B `=`lambda_A `N_A `=`A_A `=`1`Ci]

Hence, the concentration in the room becomes ;

[eqn: C_B `=`A_B over V `=`{1`Ci} over {500`m^3} `=`bold{2times 10^-3 `Ci/m^3}]

□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□

b) The number of [eqn: Rn^222] in the room is given by :

[eqn: N_B `(t`=`inf) `=`{lambda_A `N_A}over{(lambda_B `+`R)} `=`{A_A}over{(lambda_B `+`R)}]   [eqn: left(because ~dN_B over dt `=`0right)]

In this case, the activity concentration of [eqn: Rn^222] in the room in an equilibrium state is given by :

[eqn: C_B `(t`=`inf) `=`{A_B `(t`=`inf)} over V `=`{lambda_B `N_B `(t`=`inf)} over V `=`{lambda_B `A_A}over{(lambda_B `+`R)`V}]

Since [eqn: lambda_B] << [eqn: R] [eqn: left(~because~lambda_B `=`ln2 over {3.82times24`hrs} `=`7.56times10^-3 `hr^-1 ~right)], the concentration of Rn222 in the room easily saturates. In this case, the ventilation rate becomes :

[eqn: RV `=`{lambda_B `A_A}over {C_B `(t=inf)} `=`{(7.56times10^-3 hr^-1 )(1`Ci)}over{4times10^-6 Ci/m^3 }`=`bold{1.89times10^3 ~[m^3 /hr]}]

Problem 4 (20 points) : 131I is an important fission product in assessing the health effects following a postulated accident in the nuclear power plant. The thermal power level of the reactor is 330 MWt. Find the equilibrium activity of 131I in the core in curies. Assume the fission yield of 131I is 3%, and neglect the neutron absorption reaction of 131I within the reactor. The half-life of 131I is 8 days.

Data : 1 fission/sec = thermal power of 3×10-17 MWt.

Solution

U235 (Fission) → I131 (T½ = 8 days) → …

The rate of change in the number of nuclide [eqn: i] at [eqn: under r] in the reactor core is given by :

[eqn: {dN_i `(under r , t)} over dt `=`gamma_i int_E dE`Sigma_f `(under r , E)`phi`(under r , E) `-`lambda_i `N_i `(under r , t)]  where, [eqn: gamma_i] = Fission yield

Solving this 1st order differential equation, the number of nuclide [eqn: i] becomes :

[eqn: N_i `(under r , t) `=`{gamma_i int_E dE`Sigma_f `(under r , E)`phi`(under r , E)} over lambda_i left(1 `-`e^{-lambda_i t}`right)]

By integrating over the core volume, the total activity in the core becomes :

[eqn: A_i `(t) `=`int_V d under r `lambda_i `N_i`(under r , t) `=`gamma_i `int_V d under r `int_E dE`Sigma_f `(under r , E)`phi`(under r , E) left(1 `-`e^{-lambda_i t}`right)]

where, [eqn: int_V d under r `int_E dE`Sigma_f `(under r , E)`phi`(under r , E)] = the total fission rate in the reactor

Hence, for a 330 MWt reactor, the total fission rate becomes :

[eqn: int_V d under r `int_E dE`Sigma_f `(under r , E)`phi`(under r , E) `=`(330`MW`t)left({1`"fissions/sec"}over {3times10^-17 `MW`t}right) `=`1.1times10^19 `fissions/sec]

The activity equation becomes : [eqn: A_i `(t) `=`gamma_i `(1.1times10^19 )` left(1 `-`e^{-lambda_i t}`right) ~[Bq]]

Therefore, the equilibrium activity of [eqn: I^131] becomes :

[eqn: A_i `(t~rarrow~inf) `=&gamma_i `(1.1times10^19 ) `=`0.03times(1.1times10^19 ) ## =&3.3times10^17 ~[Bq] `=`bold{8.92times10^6 `[Ci] `=`8.92 `[MCi]}]

Problem 5 (20 points) : 99mTc (half-life = 6.00 hrs) is usually used for liver-scanning for a medical diagnostic purpose. 99mTc is generated as the daughter product of 99Mo (half-life = 66.0 hrs) following its beta transformation. If 1.0 Mbq 99mTc is needed for a liver-scanning, and if 2 days elapse between shipment of the Tc-generator (99Mo) and the use of 99mTc in the test, how many micro-curie of 99Mo must be shipped?

Solution

Mo99 (□-transformation with T½ = 66 hrs) → Tc99m (T½ = 6 hrs) →

Let [eqn: A `=`Mo^99] & [eqn: B `=`Tc^99m]

The rate of change in the number of [eqn: Mo^99] : [eqn: dN_A over dt `=`-lambda_A N_A ~~cdots~~①]

The rate of change in the number of [eqn: Tc^99m] : [eqn: dN_B over dt `=`lambda_A N_A `-`lambda_B N_B ~~cdots~~②]

Solving Eq. ① □[eqn: N_A (t) `=`N_A (0) exp(-lambda_A `t) ~~cdots~~③]

Solving Eq. ② ([eqn: N_B (0) `=`0]) □[eqn: N_B (t)`=`{lambda_A N_A (0)}over{lambda_B - lambda_A}left[ exp(-lambda_A t) - exp(-lambda_B t) right]]

In activity expression,

□ [eqn: A_B (t)`=`lambda_B N_B (t) `=`{lambda_B} over {lambda_B - lambda_A} A_A (0) left[ exp(-lambda_A t) - exp(-lambda_B t) right]]

□ [eqn: therefore ~ {A_A (0) } `=` {(lambda_B - lambda_A ) {A_B (t)}} over {{lambda_B } left[ exp(-lambda_A t) - exp(-lambda_B t) right]}]

Now, [eqn: lambda_A `=`ln2 over {66`hr} `=`1.05times10^-2 ``[hr^-1 ]], [eqn: lambda_B `=`ln2 over {6`hr} `=`1.155times10^-1 ``[hr^-1 ]]

After 2 days cool,[eqn: A_B (t=2days) `=`1`MBq], hence

[eqn: therefore ~ {A_A (0) } `=& {{(1.155times10^-1 - 1.05 times10^-2 ) times{1}} over {{(1.155times10^-1 )} left[ e^{(-1.05 times10^-2 ) times 48} - e^{(-1.155times10^-1 ) times 48} right]}}MBq ### =& 1.51`MBq `=`4.08times10^-5 `Ci `=`bold{40.8`muCi}]