A point source emitting 10 8 /sec with an average energy of 2 MeV is placed in t

ID: 3308177 • Letter: A

Question

A point source emitting 108 /sec with an average energy of 2 MeV is placed in the center of an 8.5 cm radius sphere of aluminum.

a) Give the appropriate vacuum particle fluence rate equation;

b) Then incorporate the attenuation piece to find the uncollided particle fluence rate equation;

c) Finally include the buildup piece and give the particle fluence rate equation including attenuation and buildup;

d) Determine the exposure rate on the surface of the sphere using the single-value buildup factor for isotropic point sources;

e) Determine the exposure rate on the surface of the sphere using the Taylor form of the buildup factor for isotropic point sources.

Linear Attenuation Coefficient for Al= 0.117 cm-1

Mass Attenuation Coefficient for Al = 0.0432 cm2/g

Build up for iscotropic point source= 1.75

Taylor exposure buildup for Al= (A- 16.981) (-a1- 0.04588) (a2- 0.00271)

Please show all work.

point source emission = 1.0E+8 /s

distance r= 8.5 cm

The emission is spread over a spher of radius 8.5 cm

fluence at 8.5 cm = 1.0E+8/4pi*8.52 =1.38E+6 /sq.cm/s in vaccum

b) linear attenuation of aluminum = 0.117 /cm

fluence rate with attenuation of aluminum = 1.38e+6 exp(-0.117 *8.5)

= 5.10e+5 /sq.cm/s

c) with buildup factor

Fluence rate = 5.1e+5*1.75 = 8.93e+5 /sq.cm/s

d)Energy of each phton E = 200 Mev

Energy fluency rate = 8.93e+5 * 200 E+6 = 1786e+11 ev /sq.cm/s

33.3 ev of energy absorbed produces 1- ion pair in air,

1 -ion pair yields 1.60e-19 C of elctrostatic charge

mass absorption co-efficient of air = 0.029 cm2/gm

1R = 2.54E-4 C /kg

dose rate = 1786e+11/33 * 1.6e-19 *1000*0.029 /2.54e-4

= 9.87e-2 R

= 98.7 mR

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