One of the dangers of radioactive fallout from a nuclear bomb is its ${}^{\mathbf{90}}\mathit{S}\mathit{r}$, which decays with a 29-yearshalf-life. Because it has chemical properties much like those of calcium, the strontium, if ingested by a cow, becomes concentrated in the cow’s milk. Some of the ${}^{\mathbf{90}}\mathit{S}\mathit{r}$ends up in the bones of whoever, drinks the milk. The energetic electrons emitted in the beta decay of ${}^{\mathbf{90}}\mathit{S}\mathit{r}$damage the bone marrow and thus impair the production of red blood cells. Abomb produces approximately 400gof ${}^{\mathbf{90}}\mathit{S}\mathit{r}$. If the fallout spreads uniformly over aarea $\mathbf{2000}\mathbf{}\mathit{k}{\mathit{m}}^{\mathbf{2}}$, what ground area would hold an amount of radioactivity equal to the “allowed” limit for one person, which is 74000 counts/s?

a) Half-life of${}^{40}Sr$ ,$T_{\frac{1}{2}}=29years$

b) Mass of the bomb produces ${}^{40}Sr$of mass,$M=400g$

c) Area of activity,$A=2000k{m}^2$

d) Activity of the radioactivity decay,$R=74000\frac{counts}{s}$

Here, the count rate in the area is the decay rate of the isotope. Since spreading is uniform in the given area, the rate of decay is directly proportional to the abundance through an area of the isotope in the ground region.

The disintegration constant as follows:

$\lambda =\frac{\ln 2}{T_{{\displaystyle \frac{1}{2}}}}$ …… (i)

Here,$T_{\frac{1}{2}}$is the half-life of the substance.

The rate of decay of a radioactivity is as follows:

$R=\lambda N$ …… (ii)

The number of atoms in a given mass of a substance is as follows:

$N=\frac{M}{m}{N}_A$ …… (iii)

$R=74000\frac{counts}{s}$Since the spreading is assumed uniform, the count rate is given using equations (ii) and (iii) as follows:

$R=\lambda \left(\frac{M}{m}\right)N_A\left(\frac{a}{A}\right)\left(\because f=abundanceoftheradioactivity=\frac{a}{A}\right)$

Here,is the mass of ${}^{40}Sr$ produced, is the mass of a single nucleus, Ais the area over which fall out occurs, andis the area in question.

Now, substituting the value of disintegration constant from equation (i) in the above expression, the ground area acan be given as:

$$\begin{gathered} a=A\left(\frac{m}{M}\right)\left(\frac{R}{\lambda }\right)N_A \\ =\left(\frac{A}{N_A}\right)\left(\frac{m}{M}\right)\left(\frac{R{T}_{{\displaystyle \frac{1}{2}}}}{\ln 2}\right) \\ Themolarmassof{}^{40}SrisM=90\frac{g}{mol} \end{gathered}$$

With the given data in the problem, we can get the ground area that would hold an amount of radioactivity equal to “allowed” limit for one person as follows:

$$\begin{gathered} a=\frac{\left(2000\times {10}^6{m}^2\right)}{\left(6.022\times {10}^{23}mo{l}^{-1}\right)}\left(\frac{90{\displaystyle \frac{g}{mol}}}{400g}\right)\left[\frac{\left(74000{\displaystyle \frac{counts}{s}}\right)\left(29years\right)\left(3.15\times {10}^7{\displaystyle \frac{s}{y}}\right)}{\ln 2}\right] \\ =7.3\times {10}^{-2}{m}^{-2} \\ =730c{m}^2 \end{gathered}$$

Hence, the value of the ground area is$730c{m}^2$ .