Q. 5.78 - An Introduction to General, Organic, and Biological Chemistry

Question

Cesium-$137$, used in cancer treatment, has a half-life of $30 yr. (5.2,5.4)$

a. Write the balanced nuclear equation for the beta decay of cesium-$137$.

b. How many milligrams of a $16-mg$ sample of cesium-$137$ remain after $90yr$?

c. How many years are required for $28\mathrm{mg}$ of cesium- $137$ to decay to $3.5\mathrm{mg}$ of cesium- $137$ ?

Step-by-Step Solution

Answer

Option a

     The following is the whole nuclear reaction:

${}_{55}{}^{137}\mathrm{Cs}\to {}_{55}{}^{137}\mathrm{Ba}+{}_{-1}{}^{0}e$

Option b

     After $90 years,2.0 mg Cs-137$  still exists

Option c

     As a result, it takes $90 years$ for a sample to decay from $28 mg to 3.5 mg.$

Step 1: Given Information (a)

(a) A neutron changes to a proton when a beta particle $\left(\beta \right)$ is emitted from a nucleus. The beta particle $\left({}_{-0}{}^{1}e\right)$ has one negative charge.

The mass number remains constant when a beta particle is emitted from a nucleus, while the atomic number increases by one.

 $\left({}_{-0}{}^{1}e\right)$is the symbol for the beta particle.

Step 2: Given Explanation (Option a)

     The following is an incomplete nuclear equation for beta decay:

${}_{55}{}^{137}\mathrm{Cs}\to ?+{}_{-1}{}^{0}e$

Find the missing mass number in the equation above:

     Cesium has a mass number of $137$ , which is equal to the sum of the mass numbers of the new nucleus and a beta particle.

$Mass number of cesium = mass number of new nucleus + mass number of beta partical$

Rearrange the equation above.

$Mass number of new nucleus = mass number of cesium - mass number of beta particle$

Substitute the value in the equation above.

$$\begin{gathered} Mass number of new nucleus =137-0 \\ =137 \end{gathered}$$

Step 3: Calculation of Option a

      In the equation above, find the missing atomic number:

      Cesium has an atomic number of $55$, which is the result of adding the atomic numbers of a new nucleus and a beta particle.

$\mathrm{Atomic} \mathrm{number} \mathrm{of} \mathrm{cesium} =( \mathrm{atomic} \mathrm{number} \mathrm{of} \mathrm{new} \mathrm{nucleus} ) - ( \mathrm{atomic} \mathrm{number} \mathrm{of} \mathrm{beta} \mathrm{particle}) :$

Rearrange the equation above:

$Atomic number of new nucleus =(atomic number of cesium )+( atomic number of beta particle )$

Substitute the following value in the equation:

$Atomic number of new nucleus =55+1=56$

Step 4: Conclusion of Option a

Determine the new element's symbol:

The atomic number $56$ for barium is $Ba$ in the periodic table.

The nucleus of this isotope of $Ba$ is denoted as ${}_{55}{}^{137}Ba$

The following is the whole nuclear reaction:

${}_{55}{}^{137}\mathrm{Cs} \to {}_{55}{}^{137}\mathrm{Ba} +{}_{-0}{}^1\mathrm{e}$

Step 5: Given Information (Option b)

(b)      Half-life: A radioisotope's half-life is the time it takes for half of a sample to decay.

The half-life of cesium is $30 y$.

The starting dose of cesium-$137$ is $16 mg$

Make a plan to figure out the unknown quantity.

days miligrams of $Cs-137$ remaining half-lives half-life number of miligrams of $Cs-137$ remaining half-lives

Step 6: Given Explanation ( Option b)

     Calculate the amount of sample that remains after 90 years using the conversion factor.

$$\begin{gathered} half- life of Cs-137=30 years \\ \frac{half -life of Cs-137}{30 y} and \frac{30 y}{half- life of Cs-137} \end{gathered}$$

To begin, calculate the number of half-lives in the period that has passed.

$$\begin{gathered} Number of half-lives=90 years\times \frac{1 half -life of Cs-137}{30 years} \\ =3 half-lives of Cs-137 \end{gathered}$$

Calculate how much of the sample decays in three half-lives and how much  $\mathrm{Cs}$

 remains in milligrammes.

$16 \mathrm{mg} \mathrm{Cs}-137\stackrel{1 \mathrm{half}-\mathrm{life}}{\to } 8.0 \mathrm{mg} \mathrm{Cs}-137\stackrel{2 \mathrm{half}-\mathrm{lives}}{\to } 4.0 \mathrm{mg} \mathrm{Cs}-137\stackrel{3 \mathrm{half}-\mathrm{lives}}{\to } 2.0 \mathrm{mg} \mathrm{Cs}-137$

After $90 \mathrm{years}, 2.0 \mathrm{mg} \mathrm{Cs}-137$ still exists

Step 7: Given Information (Option c)

(c)      Half-life: A radioisotope's half-life is the time it takes for half of a sample to decay.

The starting dose of $\mathrm{Cs}-137 \mathrm{is} 28 \mathrm{mg}$$\mathrm{Cs}-137$.

The amount of  left is $3.5 mg$.

$half life of Cs-137=30 years$

The following is the half-life conversion factor:

$$\begin{gathered} \mathrm{half}-\mathrm{life} \mathrm{of} \mathrm{Cs}-137=30\mathrm{years} \\ \frac{half-life of Cs-137}{30 y} and \frac{30 y}{half-life of Cs-137} \end{gathered}$$

Step 8: Given Explanation (Option c)

     Calculate the number of half-lives required to reduce the sample size from $28 \mathrm{to} 3.5 \mathrm{mg}.$

$28 \mathrm{mg} \mathrm{Cs}-137\stackrel{1 \mathrm{half}-\mathrm{life}}{\to } 14 \mathrm{mg} \mathrm{Cs}-137\stackrel{2 \mathrm{half}-\mathrm{lives}}{\to } 7 \mathrm{mg} \mathrm{Cs}-137\stackrel{3 \mathrm{half}-\mathrm{lives}}{\to } 3.5 \mathrm{mg} \mathrm{Cs}-137$

Three half-lives are required to reduce the sample from$28 \mathrm{to} 3.5 \mathrm{mg}$.

Calculate the number of years necessary using the conversion factor.

$$\begin{gathered} \mathrm{half}-\mathrm{life} \mathrm{of} \mathrm{Cs}-137=30 \mathrm{years} \\ 3 \mathrm{half}-\mathrm{lives} \mathrm{of} \mathrm{Cs}-137\times \frac{30 \mathrm{years}}{\mathrm{half}-\mathrm{life} \mathrm{of} \mathrm{Cs}-137} = 90.0 \mathrm{y} \end{gathered}$$

As a result, it takes $90 years$ for a sample to decay from $28 mg to 3.5 mg.$