Chapter 21, Problem 21.57QA

Interpretation Introduction

To:

a. Write a balanced nuclear equation describing the decay of $Sr$

b. Calculate the number of atoms of  $Sr$ present in a 200 mL glass of milk with 1.25 Bq/L of  $Sr$ radioactivity.

c. Explain why strontium-90 is more concentrated in milk than other foods such as grains, fruits or vegetables.

Answer to Problem 21.57QA

Solution:

a. $Sr \to Y+ {\rm B}$

b. $3.28 \times {10}^8$ atoms.

c. Milk is rich in calcium and since strontium and calcium belong to the same group, they have similar chemical properties.

Explanation of Solution

a) The average atomic mass of Sr is 87.62. This means that the mass number is greater than the average atomic mass, Thus, $Sr$ is a neutron rich nuclide and undergoes ${\rm B}$ decay.

In the decay reaction, the sum of the mass numbers (superscripts) of the nuclei on the left side of the reaction must be equal to those on the right side. Similarly, the sum of the atomic numbers (subscripts) of the nuclei on the left side of the reaction must be equal to those on the right side.

The reaction is shown below.

$Sr \to Y+ {\rm B}$

b)

1) Concept:

The radioactivity of the strontium-90 is given for per liter of the milk, which can give us the amount of radioactivity present in the given quantity 200 mL. Half-life of strontium-90 is provided from which we can find out the decay rate constant by the relation $k= \frac{0.693}{t_{1/2}}$; where the unit of k is $decays events/(atom.s)$

Radioactivity, the number of decayed atoms and the decay rate constant are related as $A=kN$; where $A$ is the radioactivity in decay events/s, $k$ is the decay rate constant and $N$ is the number of atoms going decay process.

2) Formulae:

i) $k= \frac{0.693}{t_{1/2}}$

ii) $A=kN$

3)  Given:

i) Half-life of Strontium-90 $t_{1/2}=28.8 years$

ii) Radioactivity of strontium-90 $=1.25 Bq/L$

iii) Quantity of the sample $=200 mL$

4) Calculations:

The half-life of $Sr$ is in years, which we have to convert it into seconds. The conversion is done as:

$28.8 years \times \frac{365 days}{1 year} \times \frac{24 hours}{ 1 day} \times \frac{60 min}{1 hour} \times \frac{60 s}{1 min}=9.0823 \times {10}^8 s$

Now, the rate constant of the reaction is:

$k= \frac{0.693}{t_{1/2}}=\frac{0.693}{9.08 \times {10}^8 s}=7.63 \times {10}^{-10} decay events/(atom.s)$

The radioactivity of the $Sr$ is given as $1.25 Bq/L$. Hence radioactivity of the $200 mL$ of the milk sample will be:

$A= 200 mL \times \frac{1 L}{1000 mL } \times \frac{1.25 Bq}{1 L}$

$A =0.25 Bq of radioactivity$

$A =0.25 \frac{decay events}{s}$

The relation between the radioactivity and the number of atoms that can decay is given by:

$A=kN$

Thus the number of atoms decayed can be calculated as:

$N= \frac{A}{k}= \frac{\frac{0.25 decay events}{s}}{7.63 \times {10}^{-10} \frac{decay events}{atom . s}}$

$N= 3.28 \times {10}^8 decay atoms$

$200 mL$ of the milk sample contains $3.28 \times {10}^8$ atoms of $Sr$

c) Milk is rich in calcium.  $Sr$ is a group 2 element and has similar chemical properties as that of calcium. Hence, the milk will absorb higher quantity of $Sr$ because of the same chemical reactivity and thus the concentration of the $Sr$ is higher in milk than the other food products.

Conclusion:

Strontium undergoes  ${\rm B}$ decay and is found in milk in higher proportion as compared to other foods.