Question

137 Cs is a common product of nuclear fission. Suppose an accident spills 550 mCi of 137 Cs in a lab room.

a. What mass of 137 Cs is spilled?

b. If the spill is not cleaned up, how long will it take until the radiation level drops to an acceptable level, for a room this size, of 25 mCi?

Short answer

Number of atom is,

$$\begin{gathered} N_0=\frac{R_0}{r} \\ Here{R}_{0}isinitilandrisfinaldecayrate \\ and, \\ r=\frac{In2}{t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ Here{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}ishalftime \end{gathered}$$

Step-by-step solution

Convert Ci to Bq

1Ci= 3.7x10¹⁰ Bq

so,

$$\begin{gathered} R_0=550mCi \\ =\left(550mCi\right)\left(\frac{{10}^{-3}Ci}{1mCi}\right)\left(\frac{3.7x{10}^{10}decay/sec}{1Ci}\right) \\ =20.35x{10}^{9}decay/sec \end{gathered}$$

And final decay rate of ¹³⁷Cs is calculated as follows,

Substitute 30 year for t_1/2 from Appendix C in above final rate equation.

$$\begin{gathered} r=7.32x{10}^{-10}decys \\ {N}_0=2.8x{10}^{19}atoms \end{gathered}$$

Here is the mass of uranium 6.32x10^-6kg