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Problem 1 (20 points) : BF3 counter is a typical neutron counter (thermal neutron) which is based upon B10(n,□)Li7 reaction using BF3 gas in the ion chamber. Alpha particles are counted instead of neutrons. When a thermal neutron counting rate is 500 cpm, what is the thermal neutron flux? Assuming the efficiency of 1 □-count per reaction.

Data :

□ Filled with 100% enriched B10F3 gas at p = 10 cmHg, and T = 20℃ in a detector volume of 80□.

□ The average absorption X-section of B10 at 20℃ is 3,553 b.

Solution

The number of B10 atoms in the chamber,

[eqn: N `=` {{80cm^3 times 10^-3 l/cm^3} over {22.4`l}} left(10 over 76 right) left(273 over {273+20}right)(6.02times10^23 )(1) `=`2.64times10^20 `atoms]

The average absorption X-section of B10 (1/v absorber) is 3,553b.

Hence, the total reaction rate becomes,

[eqn: N`sigma_a `phi `=` (2.64times10^20 )(3553times10^-24 )`phi `=`{500 `counts} over {60`sec}] [eqn: (``because~1 ``"reaction" `=`1``alpha-count``)]

[eqn: therefore ~phi `=` bold{8.88 ~"neutrons"/cm^2 cdot sec}]

Problem 2 (20 points) : 5 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
3	228	20
4	227	19
5	234	23

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+228+227+234}over{50} - {22+21+20+19+23}over{100} ## =&229.6 `-`10.5 `=`219.1 ~cpm]

[eqn: □``sigma_n `=`sqrt{sigma_g^2 `+`sigma_b^2}`=`sqrt{{{r_g}over{t_g}}+{{r_b}over{t_b}}} `=`sqrt{{{229.6}over{50}}+{{10.5}over{100}}} `=`2.2 ~cpm]

Hence, net count rate with standard deviation is, [eqn: bold{r_n `+`sigma_n `=`(219.1 `+-`2.2)~cpm}]

Problem 3 (20 points) : In case of air-submersion doses by radioactive noble gases, explain that the external radiation is completely controlling compared to lung or absorbed doses.

Solution

All the radioisotopes of noble gases emit either photons or □-particles of considerable energy.  For skin and tissues near the surface of the body, [eqn: g_E `=`0.5]

[eqn: {dot H_E} over {dot H_L} `=`{k`g_E `M_L}over{V_L `g_L `rho_A} ~□~ 130 over g_L]  [eqn: (`` because~g_L ``<``1 ``)]

[eqn: {dot H_E} over {dot H_A} `=`{rho_T `k`g_E}over{rho_A `delta `g_A} ~>~ 200]     [eqn: left(`` rho ``<``2`,~{{rho_T}over{rho_A} `□`800}`,~ g_A ``<``1 ``right)]

Hence, for noble gases, the external radiation is completely controlling compared to lung or absorbed doses.

※ Nomenclature

□ [eqn: k] : relative mass stopping power of tissue (∼1)

□ [eqn: g_A] : geometrical factor determined by the dimensions of a person & the range of the radiations

□ [eqn: g_E] : geometrical factor to allow for shielding by overlying tissue

□ [eqn: g_L] : geometrical factor

□ [eqn: M_L] : mass of the lungs (1,000 g)

□ [eqn: V_L] : average volume of air contained in the lungs (∼3×10-3 □)

□ [eqn: rho_A] : density of air (∼1,300g/□)

□ [eqn: rho_T] : density of tissue (106 g/□)

□ [eqn: delta] : solubility of gas in tissue (increases with gas atomic weight)

Problem 4 (10 points) : For □-emitters uniformly distributed in the bone, what are the largest and smallest absorbed fractions among AF(BS←trabecular bone), AF(BS←cortical bone), AF(RM←trabecular bone), and AF(RM←cortical bone). Explain the reasons of your choice.

Solution

For □-emitters uniformly distributed in the bone, the absorbed fractions, AF(T←S) are

□ AF(BS←trabecular bone) = 0.025

□ AF(BS←cortical bone) = 0.006∼0.02

□ AF(RM←trabecular bone) = 0.05 : Largest

□ AF(RM←cortical bone) = 0 : Smallest

Problem 5 (10 points) : What are the two types of bone for a human and their major biological effects of radiation?

Solution

The biological effects to the bone are on :

1) (5 points)

red bone marrow (RM) which is the haematopoietic stem cells of marrow within the trabecular bone (1,500 g), which induces "leukemia"

2) (5 points)

bone surfaces (BS) which is the osteogenic tissue on endosteal surfaces and certain epithelial cells close to bone surface (10 mm from the relevant bone surfaces) in the trabecular and cortical bones (= 120 g), which induces "osteosarcoma (bone cancer)". The masses of bone surface of two types of bone are equal (60 grams each).

Problem 6 (20 points) : In an ion chamber which is filled with air at STP conditions, the current appeared in the ammeter is 0.02mA. Find the exposure rate (Xunit/sec).

Data : The volume of the chamber is 4 □.

Solution

□ air condition in the chamber : STP (20℃, 760 mmHg)

□ external current appeared in the ammeter : 0.02 mA = 2×10-5 C/sec

□ volume of the chamber : 4□

Therefore, the exposure (# of i.p. produced in the STP air by □- or X-radiation) rate is,

[eqn: dot X `=`{{2times10^-5 `C/sec}over{4cm^3 times(1.293times10^-6 `g/cm^3 `)}}left({273+20}over{273} right)left(760 over 760 right) `=`bold{4.15~Xunit/sec}]