Chapter 19, Problem 19.3KSP

Interpretation Introduction

Interpretation:

The remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ has to be identified from the given option.

Concept introduction:

• Unstable nuclei emit radiation spontaneously to become stable nuclei by losing energy. This process of emission of radiation by unstable nuclei is known as radioactive decay.
• Radioactive decay is in the first order kinetics. Rate of radioactive decay at a time t is,

$\begin{array}{c}\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=}-\text{kt}\\ \text{where,}\text{k=}\text{First}\text{order}\text{rate}\text{constant}\\ {\text{N}}_{\text{0}}\text{=}\text{number of radioactive nuclei at time zero}\\ {\text{N}}_{\text{t}}\text{=}\text{number of radioactive nuclei at a time,}\text{t}\end{array}$

Answer to Problem 19.3KSP

The correct option is option (b) $2.42g$

Explanation of Solution

Reason for correct option

To find: The remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$

The remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ is $2.42g$

The remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ is calculated

Given,

$\begin{array}{c}\text{Rate constant,k}\text{=}\text{0}\text{.0533}{\text{d}}^{-1}\\ \text{Time}\text{=}\text{24}\text{h}\text{=}\text{1d}\\ \text{Original}\text{mass}\text{of}\text{sample}\text{=}\text{2}\text{.55}\text{g}\end{array}$

Suppose the number of radioactive nuclei at time zero is ${\text{N}}_{\text{0}}$ and at a time t is ${\text{N}}_{\text{t}}$,

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=}-\text{kt}$

Radioactive nuclei at time zero and time t are the initial concentration and the final concentration.

So the equation can also be written as,

$\text{ln}\frac{{\text{[A]}}_{\text{t}}}{{\text{[A]}}_{\text{0}}}\text{=}-\text{kt}$

Therefore,

The remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ is,

$\begin{array}{c}\text{ln}\frac{{\text{[A]}}_{\text{t}}}{{\text{[A]}}_{\text{0}}}\text{=}-\text{kt}\\ \text{ln}\frac{{\text{[A]}}_{\text{t}}}{\text{2}\text{.55}\text{g}}\text{=}-\left(\text{0}\text{.0533}{\text{d}}^{-1}\right)\left(1d\right)\\ \text{ln}\frac{{\text{[A]}}_{\text{t}}}{\text{2}\text{.55}\text{g}}\text{=}-\left(\text{0}\text{.0533}\right)\\ \frac{{\text{[A]}}_{\text{t}}}{\text{2}\text{.55}\text{g}}=0.948\\ {\text{[A]}}_{\text{t}}=2.42g\end{array}$

The remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ is $2.42g$

Hence the correct option for the remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ is option b

Reasons for incorrect options:

Write all the remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ given in the options except option b.

(a) $\text{0}\text{.948}\text{g}$

(c) $\text{0}\text{.136}\text{g}$

(d) $\text{0}\text{.710}\text{g}$

(e) $\text{0}\text{.873}\text{g}$

The remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ in the given reaction is found as $2.42g$.

So, this value of remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ is not matching with any of the above given remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ in remaining options.

Hence the incorrect options are a, c, d and e.

Conclusion

The correct option for remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ is identified from the given options. The remaining mass of sample of ${}^{\text{126}}\text{I}$ after $24h$ is determined by using the data’s given for the radioactive decay.