Question

When aboveground nuclear tests were conducted, the explosions shot radioactive dust into the upper atmosphere. Global air circulations then spread the dust worldwide before it settled out on ground and water. One such test was conducted in October 1976. What fraction of the ${}^{\mathbf{90}}\mathbf{Sr}$produced by that explosion still existed in October 2006? The half-life of ${}^{\mathbf{90}}\mathbf{Sr}\mathbf{}\mathbf{is}\mathbf{}\mathbf{29}\mathbf{}\mathbf{y}$.

Short answer

The fraction of the ${}^{90}\mathrm{Sr}$remaining explosion during the production is 0.49.

Step-by-step solution

The given data

a) The half-life of ${}^{90}\mathrm{Sr},{\mathrm{T}}_{1/2}29\mathrm{y}$,

b) Time of decay, t = 30y (1976 to 2006

Understanding the concept of decay  

Due to the production process, the occurrence of explosion due to the heating process also decays with the time being as per its life -span. Thus, using the decaying constant value of any substance in the equation of finding the undecayed material, determine the amount of explosion left behind after the initial decay.

Formulae:

The disintegration constant is given as follows:

$\mathrm{\lambda }=\frac{\mathrm{In}2}{{\mathrm{T}}_{{\displaystyle \frac{1}{2}}}}$ …… (i)

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

The undecayed sample remaining after a given timeis as follows:

$N=N_0{e}^{-\lambda t}$ ….. (ii)

Calculate the fraction of the remaining explosion of the Strontium isotope

Substituting equation (i) in equation (ii) and using given data, determine the fraction of the remaining explosion to that of the initial number of nuclei of the Strontium${}^{90}\mathrm{Sr}$ substance as follows:

$\frac{N}{N_0}=e^{-\frac{In2}{T_{{\displaystyle \frac{1}{2}}}}\left(t\right)}$

Substitute the values and solve as:

$$\begin{gathered} \frac{N}{N_0}=e^{-\frac{\mathrm{In}2}{29y}\left(30y\right)} \\ =0.49 \end{gathered}$$

Hence, the value of the fraction is 0.49.