What is the activity of a20ngsample${\mathbf{of}}^{\mathbf{92}}\mathbf{kr}$, which has a half-life of 1.84s?

1. Mass of ${\mathrm{the}}^{92}\mathrm{Kr}$sample,
2. The half-life ${\mathrm{of}}^{92}\mathrm{Kr}$sample, ${\mathrm{T}}_{1/2}=1.84\mathrm{s}$
3. The molar mass of ${\mathrm{the}}^{92}\mathrm{Kr}$sample, $\mathrm{A}=92\frac{\mathrm{g}}{\mathrm{mol}}$

Determine the number of atoms of a sample with a given mass can be calculated using the concept of stoichiometry. Again, with the given half-life, we get the time it takes to reduce to half its amount. This gives the disintegration constant which is the rate of decay of all the nuclei of the sample. Using the concept and given values, the activity of the sample which is the average number of disintegrations per second, or the decay rate of the sample and can be calculated. One Becquerel is disintegration per second of the sample.

The rate of decay is as follows:

$R=\lambda N$ ….. (i)

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

The disintegration constant as follows:

$\lambda =\frac{In2}{T_{{\displaystyle \frac{1}{2}}}}$ …. (ii)

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

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

$N=\frac{m}{A}{N}_A$ …… (iii)

Here,${\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23}\frac{\mathrm{atoms}}{\mathrm{mol}}$

The number of atoms of the Kryptonsample can be calculated using the given data in equation (iii) as follows:

$$\begin{gathered} \mathrm{N}=\frac{\left(20\times {10}^{-9}\mathrm{g}\right)}{\left(92{\displaystyle \frac{\mathrm{g}}{\mathrm{mol}}}\right)}\left(6.022\times {10}^{23}\frac{\mathrm{atoms}}{\mathrm{mol}}\right) \\ =1.3\times {10}^{14}\mathrm{atoms} \end{gathered}$$

Substitute equation (ii) in equation (i) with the above number of atoms value, and determine activity of the Kryptonsample as follows:

$$\begin{gathered} \mathrm{R}=\left(\frac{\mathrm{In}2}{{\mathrm{T}}_{{\displaystyle \frac{1}{2}}}}\right)\mathrm{N} \\ =\left(\frac{\mathrm{In}2}{1.84\mathrm{s}}\right)\left(1.3\times {10}^{14}\mathrm{atoms}\right) \\ =4.9\times {10}^{13}\mathrm{Bq} \end{gathered}$$

Hence, the activity of the sample is$4.9\times {10}^{13}\mathrm{Bq}$ .