Question

A 70 kg laboratory worker is exposed to a 1.48 GBq 27 Co sou body has a cross sectional area of 1.5 m2 and is normally about 4 60 rce. Assume the person's .0 m from the source for hours per day. 27 Co emits y rays of energy 1.33 MeV and 1.17 MeV in quick succession. Approximately 50% of the rays interact with the body and deposit their energy. The rest just pass through (a) (b) What effective dose does that laboratory worker receive? Comment on the dose the laboratory worker received when considering the normal allowed annual dose is 1.0 mSv. the person' s A 70 kg laboratory worker is exposed to a 1.48 GBq 2 Co source. Ass body has a cross sectional area of 1.5 m2 and is normally about 4.0 m from 4.0 hours per day, succession. Approximately 50% of the rays interac body has a cross so a 1.48 GBa Co source Assume 27 Co emits y rays of energy 1.33 MeV and 1.17 MeV in quick t with the body and deposit their energy. The rest just pass through. (a) (b) What effective dose does that laboratory worker receive? Comment on the dose the laboratory worker received when con allowed annual dose is 1.0 mSv. sidering the normal

Answer 1

there are three types of Dose 1.Radiation dose

2.Equivalent Dose

3 Effective Dose

4 Exposure rate

Radiation dose can be defined as the amount of energy absorbed by the body per unit mass and Unit is Grey

1 Grey=1 joule of energy absorbed per kilogram mass of the absorber.

Equivalent dose is defines as amount of energy absorbed  per unit mass for a particular organ in body unit is Sievert(S)

and Effective Dose is the amount of energy absorbed  the per unit mass by the entire body. Unit is Sieverts

mass of the Worker=70kg

Activity of the radioactive Cobalt given=1.48GBq

1Gbq=10^9 disintegration per second of the radio active material

there are two energy of gamma radiation i.e 1.33MeV,1.77MeV

mean energy=(1.33+1.77)/2=1.25Mev

therefore Dose=d=1.25/70=0.018MeV/Kg

Equivalent Dose=H

allradiationtype

where Wr is called radiation weight factor and for gamma radiation it is =1.0

and D=Dose

H=1*0.018= 0.018Mev/Kg

Rate of Exposure of radiation

$R=(C*A)/d^{2}$

where C=exposure constant=13.2 for cobalt

A=activity=1.48GBq

d=distance from the source to the body=4.0m

R=(13.2*1.48)/16=1.221GBq/m^2

Area or worker exposed=m^2

therefore effective dose=

' E- W**H*exposurerate*Areaofexposurexper centageofbodyexposed* . alltissuetype time

where W^t is called Tissue Weight factor as different tissue have different succeptibility for same type of radiation in this case I assumed skin,bone,brain which W^t=0.01

calculated on image

Research work Date En Setem ma See me lh alloweeh Worler fre Com Page 157

Date Research work 1992,624-Ser est WoT 产4ャ er im Labo lar