Chapter 21, Problem 21.90AP

(a)

Interpretation Introduction

Interpretation: Cancer therapy using Rhenium $-188$ has been given to being effective in treating patients with colorectal cancer. The stated questions corresponding to Rhenium $-188$ are to be answered.

Concept introduction: The physical phenomenon that involves the spontaneous emission of radiations is known as radioactivity.

To determine: The symbol of Rhenium $-188$ and the number of protons, electrons and neutrons in it.

Answer to Problem 21.90AP

Solution

The symbol of Rhenium $-188$ is ${}^{188}\mathrm{Re}$. The number of protons is $75$, the number of electrons is also $75$ and the number of neutrons is $113$.

Explanation of Solution

Explanation

The symbol for Rhenium $-188$ is ${}^{188}\mathrm{Re}$. The atomic number of Rhenium is $75$ and the number of protons is equal to the atomic number, therefore, the number of protons is $75$. As the species is neutral and in a neutral species the number of protons is equal to the number of electrons, therefore the number of electrons is also $75$. The number of neutrons is given as,

$\text{Mass number}=\text{Number of protons}+\text{Number of neutrons}$

Substitute the value of mass number and number of protons in the above equation,

$\begin{array}{c}\text{Mass number}=\text{Number of protons}+\text{Number of neutrons}\\ \text{188}=\text{75}+\text{Number of neutrons}\\ \text{Number of neutrons}=113\end{array}$

(b)

Interpretation Introduction

To determine: The possibility of the lower number of neutrons in the isotopes of ${}^{188}\mathrm{Re}$.

Answer to Problem 21.90AP

Solution

The various isotopes of Rhenium are ${}^{182}\mathrm{Re}$, ${}^{183}\mathrm{Re}$, ${}^{184}\mathrm{Re}$, ${}^{185}\mathrm{Re}$, ${}^{186}\mathrm{Re}$, ${}^{187}\mathrm{Re}$, ad these have lower number of neutrons in the isotopes of ${}^{188}\mathrm{Re}$.

Explanation of Solution

Explanation

Those species that have the same atomic number but different mass number are called isotopes. The various isotopes of Rhenium are ${}^{182}\mathrm{Re}$, ${}^{183}\mathrm{Re}$, ${}^{184}\mathrm{Re}$, ${}^{185}\mathrm{Re}$, ${}^{186}\mathrm{Re}$, ${}^{187}\mathrm{Re}$, as all these are isotopes it means that they have the same atomic number of $75$ but as the mass number is different and is less than Rhenium $-188$, it means that the number of neutrons in them is less in comparison to ${}^{188}\mathrm{Re}$.

(c)

Interpretation Introduction

To determine: The percentage of the Rhenium that remains after binding to the antibody.

Answer to Problem 21.90AP

Solution

The percentage of the Rhenium that remains after binding to the antibody is $\underset{\_}{98\%}$.

Explanation of Solution

Explanation

Given

The half-life of Rhenium is $17\text{h}$.

The time period is $30\min$.

The conversion of $\text{h}$ into $\min$ is given as,

$1\text{h}=60\min$

Hence, the conversion of $17\text{h}$ into $\min$ is given as,

$\begin{array}{c}17\text{h}=17\times 60\min \\ =1020\min \end{array}$

The number of half lives is given as,

$n=\frac{t}{t_{1/2}}$

Where,

• $n$ is the number of half-lives.
• $t$ is the total time period for the reaction.
• $t_{1/2}$ is the half-life.

Substitute the value of $t$ and $t_{1/2}$ in the above equation as,

$\begin{array}{c}n=\frac{t}{t_{1/2}}\\ =\frac{30\min }{1020\min }\\ =0.0294\end{array}$

The ratio of the final and initial amount of the radioactive substance remaining after a given time is calculated as,

$\frac{\text{Final}\text{amount}}{\text{Initial}\text{amount}}={\left(0.5\right)}^n$

Substitute the value of $n$ in the above equation as,

$\begin{array}{c}\frac{\text{Final}\text{amount}}{\text{Initial}\text{amount}}={\left(0.5\right)}^n\\ \frac{\text{Final}\text{amount}}{\text{Initial}\text{amount}}={\left(0.5\right)}^{0.0294}\\ =0.98\end{array}$

Therefore, the percentage of the Rhenium that remains after binding to the antibody $\begin{array}{c}=0.98\times 100\\ =\underset{\_}{98\%}\end{array}$

(d)

Interpretation Introduction

To determine: The reason behind the non-usage of alpha emitters in the treatment of cancer.

Answer to Problem 21.90AP

Solution

The damage caused by alpha particles to the various tissues is the reason that alpha emitters are not effective in treatment of cancer.

Explanation of Solution

Explanation

The energy of alpha particles is higher than that of the beta particles. They even damage the tissue and also cause ionization in a small volume. As the energy of beta-particles is lower than that of the alpha particles, therefore they do not cause much damage to the tissue. Therefore, alpha emitters are not effective in treatment of cancer.

(e)

Interpretation Introduction

To determine: The two most abundant isotopes of Rhenium, their natural abundances and the path taken by radioactive isotope for the decay process it undergoes.

Answer to Problem 21.90AP

Solution

The two most abundant isotopes of Rhenium are ${}^{185}\mathrm{Re}$ and ${}^{187}\mathrm{Re}$ ad their natural abundances are $37.4\%$ for ${}^{185}\mathrm{Re}$ and $62.6\%$ for ${}^{187}\mathrm{Re}$.

Radioactive isotopes try to achieve the atomic number of the element that is present next to it in the periodic table.

Explanation of Solution

Explanation

The two most abundant isotopes of Rhenium are ${}^{185}\mathrm{Re}$ and ${}^{187}\mathrm{Re}$ ad their natural abundances are $37.4\%$ for ${}^{185}\mathrm{Re}$ and $62.6\%$ for ${}^{187}\mathrm{Re}$.

There is either more or less number of neutrons present in the radioactive isotopes and these are highly unstable. Therefore, the decay occurs either by the emission of alpha particles or beta particles and the pathway they adopt is such that they try to achieve the atomic number of the element that is present next to it in the periodic table. The element that is present next to Rhenium in periodic table is $\text{Os}$.

Conclusion

1. a. The symbol of Rhenium $-188$ is ${}^{188}\mathrm{Re}$. The number of protons is $75$, the number of electrons is also $75$ and the number of neutrons is $113$.
2. b. The various isotopes of Rhenium are ${}^{182}\mathrm{Re}$, ${}^{183}\mathrm{Re}$, ${}^{184}\mathrm{Re}$, ${}^{185}\mathrm{Re}$, ${}^{186}\mathrm{Re}$, ${}^{187}\mathrm{Re}$,ad these have lower number of neutrons in the isotopes of ${}^{188}\mathrm{Re}$.
3. c. The percentage of the Rhenium that remains after binding to the antibody is $\underset{\_}{98\%}$.
4. d. The damage caused by alpha particles to the various tissues is the reason that alpha emitters are not effective in treatment of cancer.
5. e. The two most abundant isotopes of Rhenium are ${}^{185}\mathrm{Re}$ and ${}^{187}\mathrm{Re}$ ad their natural abundances are $37.4\%$ for ${}^{185}\mathrm{Re}$ and $62.6\%$ for ${}^{187}\mathrm{Re}$.

   Radioactive isotopes try to achieve the atomic number of the element that is present next to it in the periodic table