Chapter 21, Problem 21.55QP

(a)

Interpretation Introduction

Interpretation: The information regarding the elevation of Strontium $-90$ in cow’s milk after the Chernobyl accident is given. Various questions based on the given radioactivity are to be answered.

Concept introduction: The time taken by any species to get reduced to half of its original concentration is called half-life of that species.

To determine: The balanced equation for the decay of ${}^{90}\text{Sr}$.

Answer to Problem 21.55QP

Solution

The balanced equation for the decay process is shown as,

${}^{{}_{38}^{90}}\text{Sr}{\to }_{39}^{90}\text{Y}{+}_{-1}^0\text{e}$

Explanation of Solution

Explanation

There occurs beta decay in case of ${}^{90}\text{Sr}$. Therefore, the final product has one more extra proton than the reactant. The balanced equation for the decay process is shown as,

${}^{{}_{38}^{90}}\text{Sr}{\to }_{39}^{90}\text{Y}{+}_{-1}^0\text{e}$

(b)

Interpretation Introduction

To determine: The number of atoms present in $200\text{mL}$ of glass of milk.

Answer to Problem 21.55QP

Solution

The number of atoms present in $200\text{mL}$ of glass of milk is $\underset{\_}{3.125\times 10^{8}atom}$.

Explanation of Solution

Explanation

Given

The half life of Strontium $-90$ is $28.8\text{years}$.

The radioactivity is $1.25\text{Bq/L}$.

The volume of milk is $200\text{mL}$

The conversion of years into days is done as,

$1\text{year}=365\text{days}$

Hence the conversion of $28.8\text{years}$ into days is done as,

$\begin{array}{c}28.8\text{years}=28.8\times 365\text{days}\\ =10512\text{days}\end{array}$

The conversion of days into $\text{h}$ is done as,

$1\text{day}=24\text{h}$

Hence the conversion of $10512\text{days}$ into $\text{h}$ is done as,

$\begin{array}{c}10512\text{days}=10512\times 24\text{h}\\ =252,288\text{h}\end{array}$

The conversion of $\text{h}$ into $\text{s}$ is done as,

$1\text{h}=3600\text{s}$

Hence the conversion of $252,288\text{h}$ into $\text{s}$ is done as,

$\begin{array}{c}252,288\text{h}=252,288\times 3600\text{s}\\ =\text{9}\text{.08}\times {\text{10}}^8\text{s}\end{array}$

The half-life of first order reaction is given as,

$t_{1/2}=\frac{0.693}{\lambda }$ (1)

Where,

• $t_{1/2}$ is the half-life period.
• $\lambda$ is the decay constant of the reaction.

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

$\begin{array}{c}{t}_{1/2}=\frac{0.693}{\lambda }\\ \lambda =\frac{0.693}{\text{9}\text{.08}\times {\text{10}}^8\text{s}}\\ =0.08\times {10}^{-8}{\text{s}}^{-1}\end{array}$

This shows about $0.08\times {10}^{-8}{\text{disintegration atom}}^{-1}{\text{s}}^{-1}$.

The relation between Becquerel and ${\text{disintegration s}}^{-1}$ is given as,

$1\text{Bq}=1{\text{disintegration s}}^{-1}$

Hence, the conversion of $0.08\times {10}^{-8}{\text{disintegration s}}^{-1}$ to $\text{Bq}$ is done as,

$0.08\times {10}^{-8}{\text{disintegration s}}^{-1}=0.08\times {10}^{-8}\text{Bq}$

The conversion of $\text{mL}$ to $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence, the conversion of $200\text{mL}$ to $\text{L}$ is done as,

$\begin{array}{c}200\text{mL}=200\times 0.001\text{L}\\ =\text{0}\text{.2}\text{L}\end{array}$

The activity in $1\text{L}$ of glass of milk $=1.25\text{Bq}$

The activity in $0.2\text{L}$ of glass of milk $\begin{array}{c}=0.2\times 1.25\text{Bq}\\ =\text{0}\text{.25}\text{Bq}\end{array}$

The activity $0.08\times {10}^{-8}\text{Bq}$ is present in $=1\text{atom}$

The activity $1\text{Bq}$ is present in $=\frac{1}{0.08\times {10}^{-8}}\text{atom}$

The activity $0.25\text{Bq}$ is present in $\begin{array}{c}=0.25\times \frac{1}{0.08\times {10}^{-8}}\text{atom}\\ =\underset{\_}{3.125\times 10^{8}atom}\end{array}$

Therefore, the number of atoms present in $200\text{mL}$ of glass of milk is $\underset{\_}{3.125\times 10^{8}atom}$.

(c)

Interpretation Introduction

To determine: The reason behind the more concentration of Strontium $-90$ in milk rather than in grains, fruits and vegetables.

Answer to Problem 21.55QP

Solution

The similarity in the chemistry of Strontium and Calcium makes Strontium more concentrated in milk rather than in grains, fruits and vegetables.

Explanation of Solution

Explanation

As Strontium is member of Group $2$ alkaline earth metals and Calcium also belongs to the same group. It means that both should have the similar chemistry. As Calcium is present in milk, it ultimately leads to the more presence of Strontium $-90$ in milk rather than in grains, fruits and vegetables.

Conclusion

1. a. The balanced equation for the decay process is shown as,
2. b. ${}^{{}_{38}^{90}}\text{Sr}{\to }_{39}^{90}\text{Y}{+}_{-1}^0\text{e}$
3. c. The number of atoms present in $200\text{mL}$ of glass of milk is $\underset{\_}{3.125\times 10^{8}atom}$.
4. d. The similarity in the chemistry of Strontium and Calcium makes Strontium more concentrated in milk rather than in grains, fruits and vegetables