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September 30, 2014

September 30, 2014

Posted by **anonymous** on Tuesday, February 5, 2008 at 9:46pm.

1. Set up the appropriate integral to represent the radiation dosage delivered by the absorbed iodine in eight days.

2. Use the fact that eight days is the half-life of the isotope to find the initial radiation intensity in millrems/hour.

3. To the nearest 10 millirems, how much of the total radiation is delivered in six weeks?

I got this:

D = int dt Io e^-.0866 t from t=0 to t = 8

but what do I do after that to get B and C?

- calculus -- PLEASE HELP! -
**Damon**, Tuesday, February 5, 2008 at 10:20pmNow I did this yesterday :)

However I did it in days and for part 2 you need it in hours.

To do it in hours, get 8 days in hours

8*24 = 192 hours half life

so

I = Io e^-kt now t in hours

.5 = e^-192 t

ln .5 = -.693 = -192 k

k = .00361

so

I = Io e^-.00361 t

Now do your integral from t = 0 to t = 8 days

D = int I dt = int Io e^-.00361 t from t = 0 to t = 192

or

D = (Io/.00361)(1/2)= 138 Io

so

10^7 millirem = 10^4 rem = 138 Io

so

Io = 10,000/138 = 72.5 rem/hr = 72.5*10^3 millirem/hr

now in 6 weeks

6 weeks = 7*24*6 = 1008 hr

D = (Io /.00361)(1-e^-(.00631*1008))

D = (72.5*10^3/.00361)(1 essentially)

D = 20083*10^3 millirems = 20*10^6 millirems essentially or about twice the total dose we got in one half life logically enough

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