Question

Calculate the mass of a sample of (initially pure) ${}^{\mathbf{40}}\mathbf{K}$that has an initial decay rate of$\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}$disintegrations/s. The isotope has a half-life of $\mathbf{1}\mathbf{.}\mathbf{28}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{y}$.

Short answer

The mass of a sample of${}^{40}\mathrm{K}$ is $0.66\mathrm{g}$.

Step-by-step solution

The given data

a) Initial decay rate of ${}^{40}\mathrm{K},{\mathrm{R}}_0=1.70\times {10}^5\mathrm{disintegration}/\mathrm{s}$,

b) Half-life of the isotope,$T_{1/2}=1.28\times {10}^9\mathrm{or}4.04\times {10}^{16}\mathrm{s}$

c) Molar mass of the sample,$A=40\mathrm{g}/\mathrm{mol}$

Understanding the concept of decay  and mass

The radioactive decay is due to the loss of the elementary particles from an unstable nucleus to convert them into a more stable one. From the concept of the decay rate, we can get the number of undecayed nuclei. Now, using this in the equation of finding the number of nuclei then determine unknown mass of the sample using its molar mass value and Avogadro number.

Formulae:

The rate of decay is as follows:

$R=\left(\frac{\mathrm{In}2}{T_{1/2}}\right)N$ ……. (i)

Here, $\lambda$ is the disintegration constant, Nis the number of undecayed nuclei.

$T_{1/2}$is the half-life of the substance, the number of atoms in a given mass of an atom.

$$\begin{gathered} N=\frac{m}{A}{N}_A \\ \mathrm{Here},{\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23}\frac{\mathrm{atoms}}{\mathrm{mol}} \end{gathered}$$ …… (ii)

Calculate the mass of the potassium sample

Substituting value of number of undecayed nuclei from the equation (i) in equation (ii), determine the mass of the potassium sample as follows:

$m=\left(\frac{R{T}_{1/2}}{\mathrm{In}2}\right)\left(\frac{A}{N_A}\right)$

Substitute the values and solve as:

$$\begin{gathered} m=\left(\frac{\left(1.70\times {10}^5{\displaystyle \frac{\mathrm{disintegration}}{s}}\right)\left(4.04\times {10}^{16}\mathrm{s}\right)}{\mathrm{In}2}\right)\left(\frac{40{\displaystyle \frac{\mathrm{g}}{\mathrm{mol}}}}{6.022\times {10}^{23}{\displaystyle \frac{\mathrm{atoms}}{\mathrm{mol}}}}\right) \\ =0.66\mathrm{g} \end{gathered}$$

Hence, the value of the mass is 0.66g.