Question

I need questions 1-6 answered

Constants and Conversion FactorsConstants and Conversion Factors

Γ for 137Cs=0.33 R m2/Ci hrΓ for 137Cs=0.33 R m2/Ci hr

C=0.00873 J/kg RC=0.00873 J/kg R

Mass attenuation coefficient of water=0.0326 cm2/gMass attenuation coefficient of water=0.0326 cm2/g

Mass attenuation coefficient of air=0.0293 cm2/gMass attenuation coefficient of air=0.0293 cm2/g

Attenuation coefficient of lead=1.23 cm−1Attenuation coefficient of lead=1.23 cm−1

For x-rays, 1Gy = 1Sv

1)What is the total exposure at a point 1 mm from a 100 mCimCi source of 137Cs137Cs after 3 hours?

2)Assuming a person is made entirely of water, what is the dose to the person at the same distance from the source as above?

3) If this person was shielded by 2 cmcm of lead at a distance of 2mm from the source, what will the new dose to the person be?

4)If the person is a member of the public and was exposed continuously to the dose in Question 3, how long will it take before this person reaches his yearly limit? Annual limit of exposure for the public is 1mSv. For x-rays 1 Gy=1 Sv1 Gy=1 Sv.

5)What is the time needed to reach the limit if there is no lead?

6)What is the time needed if there is lead, but a much stronger source of 5CiCi?

Dose calculation and radiation safety 2 3 4 5 6 1 Constants and Conversion Factors r for 137 Cs = 0.33 R m²/Ci hr C=0.00873 J/kg R Mass attenuation coefficient of water = 0.0326 cm²/g Mass attenuation coefficient of air = 0.0293 cm²/g Attenuation coefficient of lead = 1.23.cm-1 For x-rays, 1Gy = 1Sv.

Answer 1

1) what is the total exposure at a point 1 mm from 100 mCi source of 137Cs after 3 hours
half life of Cs-137 is 30.2 years = 30.2*365*24 hours = 264552 hours

activity after three hours is
A = 100mCi * e^(-3*ln(2)/264552)
A = 99.999213979342 mCi

distance from source , d = 1 mm

at time 0 < t < 3 hours
dosage is gamma*A*dt/(4*pi*d^2)
hence, total dosage is
integral( gamma*(100mCi * e^(-t*ln(2)/264552)*dt/(4*pi*d^2)) = 264552*gamma*(100mCi * (1 - e^(-t*ln(2)/264552))/(4*pi*d^2*ln(2))
hence,
total exposure in three hours is
E = 264552*gamma*(100mCi * (1 - e^(-t*ln(2)/264552))/(4*pi*d^2*ln(2))
E = 264552*0.33*(100 *10^-3 * (1 - e^(-3*ln(2)/264552))/(4*pi*0.001^2*ln(2))
E = 7878.138720987 R
E = 7878.138720987 R

but there is air in between
so,
E'= E*e^(-u*lambda*rho)
where lamdba = d
rho = density of air = 1.125 kg/m^3
u = attenuation coeff = 0.0293 cm^2/g
hence, E'= 0.99999670375543262*E = 7878.112752715 R

2) if the person is made of entirely water then
dosage = C*E'
dosage = 0.00873* 7878.112752715= 68.7759243312 J/kg
Dosage = 68.775924331 Gy

3) if its lead instead of air, and d = 2mm = 0.2 cm
E'= e^(-1.23*0.2)*E(0.001/0.002)^2
E' = 0.19548055623*E = 1540.02293924642 R
dosage = 0.00873*1540.02293924642 = 13.4444002 GY

4) annual limit is 1 Gy,
so annual limit is reached at this distance in less than 1 hour
1 = 0.00873*0.19548055623*264552*0.33*(100 *10^-3 * (1 - e^(-t*ln(2)/264552))/(4*pi*0.002^2*ln(2))
(1 - e^(-t*ln(2)/264552))= 2.3385*10^-6
t = 0.892562452657 hours
the person reaches yearly limit in 53.5537 min = 53 min 33.22 s