Question

Question: The nuclide ${}^{\mathbf{19}}\mathbf{O}$prepared by neutron irradiation of ${}^{\mathbf{19}}\mathbf{F}$has a half-life of 29 seconds.

1. How many ${}^{\mathbf{19}}\mathbf{O}$are in a freshly prepared sample if its decay rate is $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$?
2. After $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{min}$how many ${}^{\mathbf{19}}\mathbf{O}$atoms remain?

Short answer

1. Number of ${}^{19}\mathrm{O}$atoms present in a freshly prepared sample $=1.05\times {10}^6\mathrm{atoms}$
2. Number of ${}^{19}\mathrm{O}$atoms present after$2.00\min$$=6.034\times {10}^4\mathrm{atoms}$

Step-by-step solution

Calculating the rate constant for the decay process 

The rate constant of the decay process can be calculated from the half-life in the following way:

$$\begin{gathered} \mathrm{k}=\frac{0.693}{29}{\mathrm{s}}^{-1} \\ =0.0238{\mathrm{s}}^{-1} \end{gathered}$$

The rate constant for the radioactive decay process is$0.0238{\mathrm{s}}^{-1}$.

Calculating the number of atoms of O19present in a freshly prepared sample

The decay rate is defined as the number of atoms of a radioactive sample disintegrating per unit time.

$$\begin{gathered} \mathrm{A}=\mathrm{kN} \\ \mathrm{N}=\frac{\mathrm{A}}{\mathrm{k}} \\ \mathrm{N}=\frac{2.5\times {10}^4{\mathrm{s}}^{-1}}{0.0238{\mathrm{s}}^{-1}} \\ \mathrm{N}=1.05\times {10}^6\mathrm{atoms} \end{gathered}$$

The number of atoms of localid="1660714264418" ${}^{19}\mathrm{O}$present in a freshly prepared sample$=1.05\times {10}^6\mathrm{atoms}$

Number of atoms of O19present after 2.00 min

A radioactive decay process is a first-order reaction.

Let $\mathrm{N}$be the number of atoms present after $2.00\min$and ${\mathrm{N}}_0$be the initial number of atoms present in the sample.

Therefore,

$$\begin{gathered} \mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{kt}} \\ =1.05\times {10}^6\times {{\mathrm{e}}^{-}}^{0.0238{\mathrm{s}}^{-1}\times 120\mathrm{s}} \\ =6.03\times {10}^4\mathrm{atoms} \end{gathered}$$

Number of data-custom-editor="chemistry" ${}^{19}\mathrm{O}$atoms present after$2.00\min$$=6.03\times {10}^4\mathrm{atoms}$