Caesium after Chernobyl

Problem

In the 1986 Chernobyl accident, $\sim 70\ \text{kg}$ of caesium-137 ($T_{1/2} = 30{,}2$ years) were released.

• (a) Estimate the total initial activity released.
• (b) Estimate the residual activity in 2024, i.e. after $38$ years.
• (c) After how many years would the activity fall to a thousandth of the initial value?

Solution

Initial number of atoms. The molar mass of $^{137}\text{Cs}$ is $137\ \text{g/mol}$:

$N = \frac{70\,000}{137}\cdot 6{,}022\times 10^{23} \approx 3{,}08\times 10^{26}\ \text{atoms}$

Decay constant. Converting the half-life to seconds ($T_{1/2} = 30{,}2 \times 3{,}156\times 10^{7}\ \text{s} = 9{,}53\times 10^{8}\ \text{s}$):

$\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0{,}693}{9{,}53\times 10^{8}} \approx 7{,}27\times 10^{-10}\ \text{s}^{-1}$

(a) Initial activity. The activity is $A_0 = \lambda N$:

$A_0 = 7{,}27\times 10^{-10}\cdot 3{,}08\times 10^{26} \approx 2{,}24\times 10^{17}\ \text{Bq} = 224\ \text{PBq}$

(b) Activity in 2024 (after 38 years). The activity decays like the number of atoms, with factor $2^{-t/T_{1/2}}$:

$A = A_0\, 2^{-38/30{,}2} = A_0\, 2^{-1{,}258} \approx A_0\cdot 0{,}418 \approx 9{,}4\times 10^{16}\ \text{Bq} \approx 94\ \text{PBq}$

(c) Time to fall to a thousandth. Imposing $A/A_0 = 1/1000$:

$t = T_{1/2}\,\log_2(1000) = 30{,}2\cdot 9{,}97 \approx 301\ \text{years}$

$\ev{\text{(a) } \approx 2{,}24\times 10^{17}\ \text{Bq } (224\ \text{PBq});\quad \text{(b) } \approx 94\ \text{PBq};\quad \text{(c) } \approx 301\ \text{years}}$

Error in the textbook's answer key

The textbook’s answer key gives "$224\ \text{TBq}$" and "$94\ \text{TBq}$": this is a unit error. The calculation gives $A_0 \approx 2{,}24\times 10^{17}\ \text{Bq}$, i.e. $224\ \text{PBq}$ (petabecquerel, $10^{15}$), not $224\ \text{TBq}$ (terabecquerel, $10^{12}$). The discrepancy is a factor of $1000$: the digits "$224$" and "$94$" are correct, but the right prefix is P, not T.

Collegamenti

Argomenti: Fisica nucleare Concetti: Decadimento radioattivo