Chapter 20, Problem 20.69QP

Interpretation Introduction

Interpretation:

It should be calculated the radioactivity in millicuries of 15.6 mg of ${}^{\text{90}}\text{Sr}$ and it should be explained the reason for the dangerous behavior of this isotope.

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.
• These emitted radiations may be alpha radiations( $\text{α}$), beta radiations( $\text{β}$) or gamma radiations( $\text{γ}$)
• These unstable nuclei are the nuclei with more than 83 protons and which do not lie within the belt of stability.
• Radioactive decay is in the first order kinetics. Rate of radioactive decay at a time t is,

$\begin{array}{l}Rateofdecayatatime\text{R = k N}\\ \text{k-}\text{First}\text{order}\text{rate}{\text{constant and its unit is t}}^{-1}\\ \text{N-}\text{number}\text{of}\text{radioactive}\text{nuclei}\text{present}\text{at}\text{time}\text{'t}\end{array}$

Suppose the number of radioactive nuclei at time zero is ${\text{N}}_{\text{0}}$ and at a time t is ${\text{N}}_{\text{t}}$,

$ln\frac{N_t}{N_0}=-kt$

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

  Half-life of the radiation,  $t_{1/2}=\frac{0.693}{k}$

  Half-life and rate constant for radioactive isotopes vary greatly from nucleus to nucleus.

To determine: The value of k

Answer to Problem 20.69QP

• Because of both Ca and Sr belongs to group 2A, radioactive strontium that has been ingested into the human body becomes concentrated in bones and can damage blood cell production
• $Radioactivityinmillicuries(mCi)=2.12\times 10^{3}mCi$

Explanation of Solution

Explanation

Half-life of radioactive decay is the time required for a radioactive sample to decay to one half of the atomic nucleus. From the half-life of the reaction, the value of k can be easily determined.

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

Here the half lie of strontium 90 is 29.1years.

So, the value of k in ${\text{s}}^{\text{-1}}$,

$\begin{array}{l}\text{k}\text{=}\left(\frac{\text{0}\text{.693}}{\text{29}\text{.1 years}}\text{×}\frac{\text{1}\text{yr}}{\text{365}\text{d}}\text{×}\frac{\text{1}\text{d}}{\text{24}\text{h}}\text{×}\frac{\text{1}\text{h}}{\text{3600s}}\right)\\ \text{=}\text{7}\text{.55}\text{×}{\text{10}}^{\text{-10}}{\text{s}}^{\text{-1}}\end{array}$

To determine: radioactivity in millicuries of 15.6 mg of ${}^{\text{90}}\text{Sr}$

$\text{Radioactivity in millicuries (mCi)}\text{=}\text{2}\text{.12}{\text{× 10}}^{\text{3}}\text{mCi}$

Radioactive decay is in the first order kinetics. Rate of radioactive decay at a time t is,

$\begin{array}{l}Rateofdecayatatime\text{R = k N}\\ \text{k-}\text{First}\text{order}\text{rate}{\text{constant and its unit is t}}^{-1}\\ \text{N-}\text{number}\text{of}\text{radioactive}\text{nuclei}\text{present}\text{at}\text{time}\text{'t}\end{array}$

The value of N can be determined as follows, given mass of ${}^{\text{90}}\text{Sr}$ is 15.6 mg

$\text{15}\text{.6}\text{mg}\text{=}\frac{\text{15}\text{.6}}{\text{1000}}\text{=}\text{0}\text{.0156}$ g

So, $\begin{array}{l}\text{N=}\left(\text{0}\text{.0156}\text{g}{}^{\text{90}}\text{Sr}\text{×}\frac{\text{1}\text{mol}}{\text{90}\text{g}{}^{\text{90}}\text{Sr}}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{atom}}{\text{1}\text{mol}}\right)\\ \text{=}\text{1}\text{.04}{\text{×10}}^{\text{20}}\text{atoms}\end{array}$

From the value of k and N, R can be calculated as follows;

Here in the case of ${}^{\text{90}}\text{Sr}$, $\text{k=}\text{7}\text{.55×}{\text{10}}^{\text{-10}}{\text{s}}^{\text{-1}}$ and $\text{N=}\text{1}\text{.04}\text{×}{\text{10}}^{\text{20}}\text{atoms}$

$\begin{array}{l}\text{R = k}\text{N}\\ \text{=}\text{7}\text{.55×}{\text{10}}^{\text{-10}}{\text{s}}^{\text{-1}}\text{×}\text{1}\text{.04}\text{×}{\text{10}}^{\text{20}}\text{atoms}\\ \text{=}\text{7}\text{.85}\text{×}{\text{10}}^{\text{10}}\text{atoms/second}\text{=}\text{7}\text{.85}{\text{×10}}^{\text{10}}\text{disintegration/s}\end{array}$

$\begin{array}{l}\text{A}\text{curie}\text{is}\text{defined}\text{as}\text{3}\text{.70}\text{×}{\text{10}}^{\text{10}}\text{disintegrations}\text{persecond}\\ \text{So,theactivity}\text{in}\text{curies}\text{is:}\\ \text{Radioactivity}\text{in}\text{millicuries}\text{(mCi)}\text{=}\text{(7}\text{.85}{\text{×10}}^{\text{10}}\text{disintegration/s)}\left(\frac{\text{1}\text{Ci}}{\text{3}\text{.70}\text{×}{\text{10}}^{\text{10}}\text{disintegrations}\text{/second}}\right)\\ \text{=}\text{2}\text{.12}\text{Ci}\text{=}\text{2}\text{.12}{\text{×10}}^{\text{3}}\text{m}\text{C}\text{i}\end{array}$

Conclusion

It is calculated the radioactivity in millicuries of 15.6 mg of ${}^{\text{90}}\text{Sr}$ and it is  explained the reason for the dangerous behavior of this isotope.