Chapter 19, Problem 34E

Interpretation Introduction

Interpretation: List of isotopes of krypton is given. The most stable and the hottest among them is to be stated. Time of decay of $87.5\%$ of each isotope is to be stated.

Concept introduction: Decay constant is the quantity that expresses the rate of decrease of number of atoms of a radioactive element per second. Half life of radioactive sample is defined as the time required for the number of nuclides to reach half of the original value.

The nuclides having longer half life are more stable while nuclides having shorter half life are less stable.

To determine: The most stable and the hottest isotope among the given isotopes of krypton; the time of decay for ${}^{73}\text{Kr}$; the time of decay for ${}^{74}\text{Kr}$; the time of decay for ${}^{76}\text{Kr}$ and the time of decay for ${}^{81}\text{Kr}$.

Answer to Problem 34E

Answer

The most stable isotope is ${}^{81}\text{Kr}$ and the hottest one is ${}^{73}\text{Kr}$

The time of decay for $87.5\%\text{of}{}^{73}\text{Kr}$ is $\underset{\_}{81.428s}$ .

The time of decay for $87.5\%\text{of}{}^{74}\text{Kr}$ is. $\underset{\_}{34.51s}$

The time of decay for $87.5\%\text{of}{}^{76}\text{Kr}$ is $\underset{\_}{44.41h}$

The time of decay for $87.5\%\text{of}{}^{81}\text{Kr}$ is $\underset{\_}{6.302\times 10^{5}years}$

Explanation of Solution

Explanation

The most stable isotope is ${}^{81}\text{Kr}$ and the hottest one is ${}^{73}\text{Kr}$

The nuclides having longer half life are more stable while nuclides having shorter half life are less stable. Thus the most stable isotope is ${}^{81}\text{Kr}$ and the hottest isotope is ${}^{73}\text{Kr}$.

The time of decay for ${}^{73}\text{Kr}$is $\underset{\_}{81.428s}$

Explanation

The decay constant can be calculated by the formula given below.

$\lambda =\frac{0.693}{{\text{t}}_{1/2}}$

Where

• $t_{1/2}$ is the half life of nuclide.
• $\text{λ}$ is the decay constant.

Substitute the value of $\lambda$ in the above expression.

$\begin{array}{c}\lambda =\frac{0.693}{27}{\text{s}}^{-\text{1}}\\ =0.0256{\text{s}}^{-\text{1}}\end{array}$

The fraction of isotope decayed is $\frac{87.5}{100}$.

The fraction remaining $\begin{array}{l}=1-\frac{87.5}{100}\\ =\frac{12.5}{100}\end{array}$

The time of decay can be calculated by the formula,

$t=\frac{2.303}{\lambda }\log \frac{n_0}{n}$

Where

• $n_0$ is the number of atoms initially present.
• $n$ is the number of atoms after time “t”.

Substitute the values of $\text{λ}$, ${\text{n}}_0$ and $n$ in the above expression.

$\begin{array}{c}t=\frac{2.303}{\lambda }\log \frac{n_0}{n}\\ t=\frac{2.303}{\text{λ}}\log \frac{100}{\text{12}\text{.5}}\\ t=\frac{2.303}{0.0256}\log \frac{100}{\text{12}\text{.5}}\\ t=\underset{\_}{81.428s}\end{array}$.

The time of decay for ${}^{74}\text{Kr}$is $\underset{\_}{34.51s}$.

Explanation

The decay constant is calculated by the formula,

$\lambda =\frac{0.693}{{\text{t}}_{1/2}}$

Where

• $t_{1/2}$ is the half life of nuclide.
• $\text{λ}$ is the decay constant.

Substitute the value of $t_{1/2}$ in the above expression.

$\lambda =\frac{0.693}{11.5}{\min }^{-\text{1}}$

The fraction of isotope decayed is $\frac{87.5}{100}$.

The fraction remaining $\begin{array}{l}=1-\frac{87.5}{100}\\ =\frac{12.5}{100}\end{array}$

The time of decay can be calculated by the formula,

$t=\frac{2.303}{\lambda }\log \frac{n_0}{n}$

Where

• $n_0$ is the number of atoms initially present.
• $n$ is the number of atoms after time “t”.

Substitute the values of $\text{λ}$, ${\text{n}}_0$ and $n$ in the above expression.

$\begin{array}{c}t=\frac{2.303}{\lambda }\log \frac{n_0}{n}\\ t=\frac{2.303}{\text{λ}}\log \frac{100}{\text{12}\text{.5}}\\ t=\frac{2.303\times 11.5}{0.693}\log 8\\ t=\underset{\_}{34.51min}\end{array}$

The time of decay for ${}^{76}\text{Kr}$ is $\underset{\_}{44.41h}$

Explanation

The decay constant can be calculated by the formula given below.

$\lambda =\frac{0.693}{{\text{t}}_{1/2}}$

Where

• $t_{1/2}$ is the half life of nuclide.
• $\text{λ}$ is the decay constant.

Substitute the value of $t_{1/2}$ in the above expression.

$\lambda =\frac{0.693}{14.8}{\text{h}}^{-\text{1}}$

The fraction of isotope decayed is $\frac{87.5}{100}$.

The fraction remaining $\begin{array}{l}=1-\frac{87.5}{100}\\ =\frac{12.5}{100}\end{array}$

The time of decay can be calculated by the formula,

$t=\frac{2.303}{\lambda }\log \frac{n_0}{n}$

Where

• $n_0$ is the number of atoms initially present.
• $n$ is the number of atoms after time “t”.

Substitute the values of $\text{λ}$, ${\text{n}}_0$ and $n$ in the above expression.

$\begin{array}{c}t=\frac{2.303}{\lambda }\log \frac{n_0}{n}\\ t=\frac{2.303}{\text{λ}}\log \frac{100}{\text{12}\text{.5}}\\ t=\frac{2.303\times 14.8}{0.693}\log 8\\ t=\underset{\_}{44.41h}\end{array}$

The time of decay for ${}^{81}\text{Kr}$ is $\underset{\_}{6.302\times 10^{5}years}$.

Explanation

The decay constant can be calculated by the formula given below.

$\lambda =\frac{0.693}{{\text{t}}_{1/2}}$

Where

• $t_{1/2}$ is the half life of nuclide.
• $\text{λ}$ is the decay constant.

Substitute the value of $t_{1/2}$ in the above expression.

$\lambda =\frac{0.693}{2.1\times {10}^5}\text{year}$

The fraction of isotope decayed is $\frac{87.5}{100}$.

The fraction remaining $\begin{array}{l}=1-\frac{87.5}{100}\\ =\frac{12.5}{100}\end{array}$

The time of decay can be calculated by the formula,

$t=\frac{2.303}{\lambda }\log \frac{n_0}{n}$

Where

• $n_0$ is the number of atoms initially present.
• $n$ is the number of atoms after time “t”.

Substitute the values of $\text{λ}$, ${\text{n}}_0$ and $n$ in the above expression.

$\begin{array}{c}t=\frac{2.303}{\lambda }\log \frac{n_0}{n}\\ t=\frac{2.303\times 2.1\times {10}^5}{0.693}\log \frac{100}{\text{12}\text{.5}}\\ t=\frac{2.303\times 2.1\times {10}^5}{0.693}\log 8\\ t=\underset{\_}{6.302\times 10^{5}years}\end{array}$

Conclusion

Conclusion

The most stable isotope is ${}^{81}\text{Kr}$ and the hottest one is ${}^{73}\text{Kr}$

The time of decay for $87.5\%\text{of}{}^{73}\text{Kr}$ is $\underset{\_}{81.428s}$ .

The time of decay for $87.5\%\text{of}{}^{74}\text{Kr}$ is. $\underset{\_}{34.51s}$

The time of decay for $87.5\%\text{of}{}^{76}\text{Kr}$ is $\underset{\_}{44.41h}$

The time of decay for $87.5\%\text{of}{}^{81}\text{Kr}$ is $\underset{\_}{6.302\times 10^{5}years}$