Question

For every 1000 disintegrations of a particular isotope, the following number of photons - at the given energies - are observed to be emitted: 51 photons at 80 keV, 51 photons at 300 keV, 853 photons at 400 keV, 69 photons at 600 keV, and 16 photons at 800 keV. a) Compute the Gamma Ray Constant in R-m2 /h-Ci for this isotope. You may find the data in the attached table useful. Hint: Remember that 1 cm2 /gm ≠ 1 m2 /kg! (15 pts) ?? = 193.8 ∑ ???? ???? (??????) � ?????? ?? � ??????,?? ?? b) Compute the exposure rate in R/h in air for at a distance of 2.5 m, from a 3 Ci source of this same isotope. Assume the source irradiates isotropically. (10 pts) Note that for any arbitrary activity “A” in Ci, and distance “d” in m, ???? ???? = ???? ??2

Answer 1

Per 1000 photons

energy

80 * 51 = 4080 kev ( = 0.0024 m2/kg- mass absorption co-efficient)

300*51 = 15300 kev ( = 0.0029 - mass absorption co-efficient)

400*853 = 341200 kev ( = 0.0029 - mass absorption co-efficient)

600*69 = 41400 kev ( = 0.0029 - mass absorption co-efficient)

800* 16 = 12800 kev ( = 0.0029 - mass absorption co-efficient)

energy aborbed per 1000 photons

= 4080*0.0024+ 0.0029(15300 + 341200 + 41400 + 12800 )

= 1200.82 kev-m2/kg

This energy is emitted in an imaginary sphere of 4r2

w of air = 33 ev/ion pair

charge associated with one ion pair = 1.602e-19 C/ion pair

the exposure

X = (1200.82/33)*1.62e-19 C

= 58.95 e-16 C-m2/1000phtons-kg

per Ci per Hr (1Ci = 3.7e+10 dis/s)

X = 58.95e-16 *3.7e+7*3600 C-m2/Ci-kg

    = 7.852e-4 C-m2/Ci-kg

This charge is liberated in an area of 4 sq.m considering an imaginary sphere of 1 sq.m surounding the source

1R = 2.58e-4 C/kg

X = 7.582e-4/4*3.14*2.58e-4 R-m2/Ci-hr

= 0.234 R-m2/Ci-hr - gamara exposure constant.

b)

exposure at 2.5 m from 3 Ci for 1 hr

= 0.234 *3 /(2.5)2  

= 0.112 R