Calculation: radioactivity [atoms/ccm]

[NucE]

Homework Statement

The problem:
4000,000 gal of water remained in a contaminated building at 3 mile island. The principle sources of radioactivity were:
137Cs at 156E-6 Ci/cubic cm and 134Cs at 26E-6 Ci/ccm. How many atoms/ccm were in the water?

Homework Equations

137E-6*3.7E10 Ci= 5.77E6 Bq/ccm
A ctivity= R*N
R= Ln2/T1/2

The Attempt at a Solution

R= Ln2/30.07yrs*3600s = 6.4E-6/sec
A = R*N
N = 5.77E6(decays/sec-ccm)/6.4E-6(1/sec) = 9.02E11 atoms/ccm

The given answers:
Cs-137: 7.892E15 atoms/cm3
Cs-134: 9.045E15 atoms/cm3

 

[Astronuc]

R= Ln2/30.07yrs*3600s = 6.4E-6/sec

Be careful with conversions. There are 3600 s/hr, 24 hrs/day and 365.25 days/yr, or about 3.156 E7 s/yr.