Chapter 11, Problem 11.27EP

(a)

Interpretation Introduction

Interpretation:

Half-life of the radionuclide has to be determined if after 5.4 days, 1/16 fraction of undecayed nuclide is present.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  All the radioactive nuclide do not undergo the decay at a same rate.  Some decay rapidly and others decay very slowly.  The nuclear stability can be quantitatively expressed by using the half-life.

The time required for half quantity of the radioactive substance to undergo decay is known as half-life.  It is represented as $t_{1/2}$.

The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,

$\left(\begin{array}{l}\text{Amount}\text{of}\text{radionuclide}\\ \text{undecayed}\text{after}n\text{half}\text{lives}\end{array}\right)\text{= }\left(\begin{array}{l}\text{Original}\text{amount}\\ \text{of}\text{radionuclide}\end{array}\right)\text{ x }\left(\frac{\text{1}}{{\text{2}}^n}\right)$

Answer to Problem 11.27EP

Half-life of the radionuclide is 1.4 days.

Explanation of Solution

Number of half-lives can be determined as shown below,

$\begin{array}{l}\left(\frac{\text{1}}{{\text{2}}^n}\right)=\frac{\text{1}}{16}\\ \left(\frac{\text{1}}{{\text{2}}^n}\right)=\frac{\text{1}}{{\text{2}}^{\text{4}}}\\ 2^n=2^4\end{array}$

As the bases are equal, the power can be equated.  This gives the number of half-lives that have elapsed as 4 half-lives.

In the problem statement it is given that the time is 5.4 days.  From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,

$\begin{array}{l}\text{5}\text{.4 days x }\left(\frac{\text{1 half-life}}{t_{1/2}}\right)\text{=}\text{4 half-lives}\\ t_{1/2}\text{=}\frac{\text{5}\text{.4 days}}{4}\\ =\text{1}\text{.4}\text{days}\end{array}$

Therefore, the half-life of the given sample is determined as 1.4 days.

Conclusion

Half-life of the given sample is determined.

(b)

Interpretation Introduction

Interpretation:

Half-life of the radionuclide has to be determined if after 5.4 days, 1/64 fraction of undecayed nuclide is present.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  All the radioactive nuclide do not undergo the decay at a same rate.  Some decay rapidly and others decay very slowly.  The nuclear stability can be quantitatively expressed by using the half-life.

The time required for half quantity of the radioactive substance to undergo decay is known as half-life.  It is represented as $t_{1/2}$.

The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,

$\left(\begin{array}{l}\text{Amount}\text{of}\text{radionuclide}\\ \text{undecayed}\text{after}n\text{half}\text{lives}\end{array}\right)\text{= }\left(\begin{array}{l}\text{Original}\text{amount}\\ \text{of}\text{radionuclide}\end{array}\right)\text{ x }\left(\frac{\text{1}}{{\text{2}}^n}\right)$

Answer to Problem 11.27EP

Half-life of the radionuclide is 0.90 day.

Explanation of Solution

Number of half-lives can be determined as shown below,

$\begin{array}{l}\left(\frac{\text{1}}{{\text{2}}^n}\right)=\frac{\text{1}}{64}\\ \left(\frac{\text{1}}{{\text{2}}^n}\right)=\frac{\text{1}}{{\text{2}}^{\text{6}}}\\ 2^n=2^6\end{array}$

As the bases are equal, the power can be equated.  This gives the number of half-lives that have elapsed as 6 half-lives.

In the problem statement it is given that the time is 5.4 days.  From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,

$\begin{array}{l}\text{5}\text{.4 days x }\left(\frac{\text{1 half-life}}{t_{1/2}}\right)\text{=}\text{6 half-lives}\\ t_{1/2}\text{=}\frac{\text{5}\text{.4 days}}{6}\\ =\text{0}\text{.90}\text{day}\end{array}$

Therefore, the half-life of the given sample is determined as 0.90 day.

Conclusion

Half-life of the given sample is determined.

(c)

Interpretation Introduction

Interpretation:

Half-life of the radionuclide has to be determined if after 5.4 days, 1/256 fraction of undecayed nuclide is present.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  All the radioactive nuclide do not undergo the decay at a same rate.  Some decay rapidly and others decay very slowly.  The nuclear stability can be quantitatively expressed by using the half-life.

The time required for half quantity of the radioactive substance to undergo decay is known as half-life.  It is represented as $t_{1/2}$.

The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,

$\left(\begin{array}{l}\text{Amount}\text{of}\text{radionuclide}\\ \text{undecayed}\text{after}n\text{half}\text{lives}\end{array}\right)\text{= }\left(\begin{array}{l}\text{Original}\text{amount}\\ \text{of}\text{radionuclide}\end{array}\right)\text{ x }\left(\frac{\text{1}}{{\text{2}}^n}\right)$

Answer to Problem 11.27EP

Half-life of the radionuclide is 0.68 day.

Explanation of Solution

Number of half-lives can be determined as shown below,

$\begin{array}{l}\left(\frac{\text{1}}{{\text{2}}^n}\right)=\frac{\text{1}}{256}\\ \left(\frac{\text{1}}{{\text{2}}^n}\right)=\frac{\text{1}}{{\text{2}}^{\text{8}}}\\ 2^n=2^8\end{array}$

As the bases are equal, the power can be equated.  This gives the number of half-lives that have elapsed as 8 half-lives.

In the problem statement it is given that the time is 5.4 days.  From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,

$\begin{array}{l}\text{5}\text{.4 days x }\left(\frac{\text{1 half-life}}{t_{1/2}}\right)\text{=}8\text{ half-lives}\\ t_{1/2}\text{=}\frac{\text{5}\text{.4 days}}{8}\\ =\text{0}\text{.68}\text{day}\end{array}$

Therefore, the half-life of the given sample is determined as 0.68 day.

Conclusion

Half-life of the given sample is determined.

(d)

Interpretation Introduction

Interpretation:

Half-life of the radionuclide has to be determined if after 5.4 days, 1/1024 fraction of undecayed nuclide is present.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  All the radioactive nuclide do not undergo the decay at a same rate.  Some decay rapidly and others decay very slowly.  The nuclear stability can be quantitatively expressed by using the half-life.

The time required for half quantity of the radioactive substance to undergo decay is known as half-life.  It is represented as $t_{1/2}$.

The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,

$\left(\begin{array}{l}\text{Amount}\text{of}\text{radionuclide}\\ \text{undecayed}\text{after}n\text{half}\text{lives}\end{array}\right)\text{= }\left(\begin{array}{l}\text{Original}\text{amount}\\ \text{of}\text{radionuclide}\end{array}\right)\text{ x }\left(\frac{\text{1}}{{\text{2}}^n}\right)$

Answer to Problem 11.27EP

Half-life of the radionuclide is 0.54 day.

Explanation of Solution

Number of half-lives can be determined as shown below,

$\begin{array}{l}\left(\frac{\text{1}}{{\text{2}}^n}\right)=\frac{\text{1}}{1024}\\ \left(\frac{\text{1}}{{\text{2}}^n}\right)=\frac{\text{1}}{{\text{2}}^{10}}\\ 2^n=2^{10}\end{array}$

As the bases are equal, the power can be equated.  This gives the number of half-lives that have elapsed as 10 half-lives.

In the problem statement it is given that the time is 5.4 days.  From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,

$\begin{array}{l}\text{5}\text{.4 days x }\left(\frac{\text{1 half-life}}{t_{1/2}}\right)\text{=}10\text{ half-lives}\\ t_{1/2}\text{=}\frac{\text{5}\text{.4 days}}{10}\\ =0.\text{54}\text{day}\end{array}$

Therefore, the half-life of the given sample is determined as 0.54 day.

Conclusion

Half-life of the given sample is determined.