Chapter 19, Problem 19.2KSP

Interpretation Introduction

Interpretation:

The original mass of sample of ${}^{\text{61}}\text{Cu}$ has to be determined 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,

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

• Half-life of radioactive decay is the time required for a radioactive sample to decay to one half of the atomic nucleus.

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}$

Answer to Problem 19.2KSP

The correct answer is option (e) $8.78\times {10}^{4}mg$

Explanation of Solution

Reason for correct option

To find: The original mass of sample of ${}^{\text{61}}\text{Cu}$

The original mass of sample of ${}^{\text{61}}\text{Cu}$ is $8.78\times {10}^{4}mg$

Original mass of ${}^{61}Cu$ is calculated

Given,

$\begin{array}{c}\text{Half-life}\text{=}\text{3}\text{.35}\text{h}\\ \text{Time}\text{=}\text{24}\text{h}\\ \text{Mass}\text{of}\text{sample}\text{after}\text{24}\text{h}\text{=}\text{612}\text{.8}m\text{g}\end{array}$

Half-life of a first order reaction is ${\text{t}}_{\text{1/2}}=\frac{0.693}{k}$

Hence,

The rate constant of the reaction is,

$\begin{array}{c}\text{k}\text{=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\\ \text{=}\frac{\text{0}\text{.693}}{\text{3}\text{.35}\text{h}}\\ \text{=}\text{0}\text{.207}{\text{h}}^{-1}\end{array}$

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 original mass of sample of ${}^{\text{61}}\text{Cu}$ is,

$\begin{array}{c}\text{ln}\frac{{\text{[A]}}_{\text{t}}}{{\text{[A]}}_{\text{0}}}\text{=}-\text{kt}\\ \text{ln}\frac{\text{612}\text{.8}\text{mg}}{{\text{[A]}}_{\text{0}}}\text{=}-\left(\text{0}\text{.207}{\text{h}}^{-1}\right)\left(24h\right)\\ \text{ln}\frac{\text{612}\text{.8}\text{mg}}{{\text{[A]}}_{\text{0}}}\text{=}-\left(4.968\right)\\ \frac{\text{612}\text{.8}\text{mg}}{{\text{[A]}}_{\text{0}}}=6.957\times {10}^{-3}\\ {\text{[A]}}_{\text{0}}=8.78\times {10}^{4}mg\end{array}$

The original mass of sample of ${}^{\text{61}}\text{Cu}$ is $8.78\times {10}^{4}mg$

Hence the correct option for the original mass of sample of ${}^{\text{61}}\text{Cu}$ is option e.

Reasons for incorrect options:

Writes all original mass of sample of ${}^{\text{61}}\text{Cu}$ given in the options except option e.

1. a) $\text{85}\text{.5}\text{mg}$
2. b) $\text{4}\text{.39}\text{×}\text{10}{}^{\text{3}}\text{mg}$
3. c) $\text{736}\text{mg}$
4. d) $\text{6}\text{.40}\text{×}\text{10}{}^{\text{3}}\text{mg}$

The original mass of sample of ${}^{\text{61}}\text{Cu}$ is found as $8.78\times {10}^{4}mg$

So, this value of original mass of sample of ${}^{\text{61}}\text{Cu}$ is not matching with any of the above given original mass of sample of ${}^{\text{61}}\text{Cu}$ in remaining options.

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

Conclusion

The correct option for original mass of sample of ${}^{\text{61}}\text{Cu}$ is identified from the given options. The original mass of sample of ${}^{\text{61}}\text{Cu}$ is determined by using the data’s given for the radioactive decay.