Chapter 20, Problem 100AP

Interpretation Introduction

Interpretation:

The energy released (in kJ) for the given decays and the heat released by the decay of ${}^{90}\text{Sr}$

are to be calculated.

Concept introduction:

The amount of a particular radioactive isotope left after time t is given as:

$\text{ln}\frac{N_t}{N_o}\text{=}-kt$

Here, $k$

is the rate constant for the radioactive decay, $N_t$ is the amount of it left after time t, and $N_0$ is the initial amount of the sample.

Nearly all radioactive decays are of first order and the rate constant is given as:

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

Here, $t_{1/2}$

is the half -life of the radioactive substance.

The energy during the decay of a radioactive substance is given as:

$\Delta E=(\Delta m){\text{c}}^{\text{2}}$

Here, $\Delta E$

is the amount of energy released, $\Delta m$

is the mass defect which occurs during the course of decay and c is the speed of the light.

The mass defect ($\Delta m$

) is given as the difference of total atomic mass of the product and the reactant in the balanced radioactive reaction.

Answer to Problem 100AP

Solution:

$\text{5}\text{.59 }\times {\text{ 10}}^{\text{-15}}\text{ J and 2}\text{.84 }\times {\text{ 10}}^{\text{-13}}\text{ J}$

$\text{0}\text{.024 mol }$.

$\text{4}\text{.26 }\times {\text{ 10}}^6\text{ kJ}$

Explanation of Solution

a) The energy released (in joules) in each of the following two decays

The decay of ${}^{90}\text{Sr}$ as follows:

${{}^{90}}_{\text{38}}\text{Sr }\to {}^{\text{90}}{}_{\text{39}}{\text{Y + }}^{\text{0}}{}_{-\text{1}}{\text{β (t}}_{\text{1/2}}=\text{ 28}\text{.1 year)}$

The decay of ${}^{90}\text{Y}$ as follows:

${{}^{90}}_{\text{39}}\text{Y}\to {}^{\text{90}}{}_{\text{40}}\text{Zr }{+}^{\text{0}}{}_{-\text{1}}{\text{β (t}}_{\text{1/2}}\text{ = 64 hr)}$

Mass of ${}^{90}\text{Sr}$

is $\text{89}\text{.907738 amu}$

Mass of ${}^{90}\text{Y}$

is $\text{89}\text{.907152 amu}$

Mass of electron is $\text{5}\text{.4857 }\times {\text{ 10}}^{\text{-4}}\text{ amu}$

Mass of ${}^{90}\text{Zr}$

is $\text{89}\text{.904703 amu}$

The balanced reaction for ${}^{90}\text{Sr}$

decay is given as follows:

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

Now, the mass defect $\Delta m$

is calculated as follows:

$\Delta m=\text{Mass of products}-\text{Mass of reactants}$

Substitute the values of masses in the above expression.

$\begin{array}{c}\Delta m={\text{[ Mass of }}^{90}\text{Y}+{\text{mass of e}}^{-}\text{]}-[{\text{Mass of }}^{90}\text{Sr]}\\ =\text{ [ 89}\text{.907152 amu + 5}.4857\times {\text{ 10}}^{\text{-4}}\text{ amu ]}-[89.907738\text{ amu ]}\\ =-\text{3}\text{.743 }\times {\text{ 10}}^{-\text{5}}\text{ amu}\end{array}$

Now, it is known that $1\text{ kg}$ is equivalent to $6.022\times {\text{ 10}}^{\text{26}}\text{amu}$

Thus, the conversion factor for this is $\frac{1\text{Kg}}{6.022\times {\text{ 10}}^{\text{26}}\text{ amu}}$

Hence, the mass defect can be converted to kilogram unit by using the above conversion factor as follows:

$\begin{array}{l}\Delta m=\left(-3.743\times {\text{ 10}}^{-\text{5}}\cancel{\text{amu}}\times \frac{1\text{kg}}{6.022\times {\text{ 10}}^{\text{26}}\cancel{\text{amu}}}\right)\\ \text{ = }-\text{6}\text{.216 }\times {\text{ 10}}^{-\text{32}}\text{kg}\end{array}$

Now, the energy during the decay of a radioactive substance is given as:

$\Delta E=(\Delta m){\text{c}}^{\text{2}}$

Substitute the values of speed of light and energy in the above expression.

$\begin{array}{c}\Delta E=\text{(}-\text{6}\text{.216 }\times {\text{ 10}}^{-\text{32}}\text{kg)(3}\times {\text{10}}^{\text{8}}{\text{m/sec)}}^{\text{2}}\\ \text{ =}-\text{5}\text{.59 }\times {\text{10}}^{-\text{15}}\text{J}\end{array}$

Thus, the energy released during the decay of ${}^{90}\text{Sr}$

is $-\text{5}\text{.59 }\times {\text{ 10}}^{\text{-15}}\text{J}$

The balanced reaction for ${}^{90}\text{Y}$

decay is given as follows:

${{}^{90}}_{\text{39}}\text{Y}\to {}^{\text{90}}{}_{\text{40}}\text{Zr}{+}^{\text{0}}{}_{-\text{1}}\text{β }$

Now, the mass defect $\Delta \text{m}$

is calculated as follows:

$\Delta m=\text{Mass of products}-\text{Mass of reactants}$

Substitute the values of masses in the above expression.

$\begin{array}{c}\Delta m={\text{[ Mass of }}^{90}{\text{Zr + Mass of e}}^{-}\text{]}-[{\text{Mass of }}^{90}\text{Sr]}\\ =\text{[ 89}\text{.904703 amu + 5}.4857\times {\text{10}}^{-\text{4}}\text{ amu ]}-[89.907152\text{ amu ]}\\ =-1.900\times {\text{10}}^{-\text{3}}\text{ amu}\end{array}$

Now, it is known that $1\text{ kg}$ is equivalent to $6.022\times {\text{ 10}}^{\text{26}}\text{amu}$

Thus, the conversion factor is $\frac{1\text{Kg}}{6.022\times {\text{ 10}}^{\text{26}}\text{ amu}}$

Hence, the mass defect can be converted to kilogram unit by using the above conversion factor as follows:

$\begin{array}{c}\Delta m=\left(-1.900\times {\text{10}}^{-\text{3}}\cancel{\text{amu}}\times \frac{1\text{kg}}{6.022\times {\text{ 10}}^{\text{26}}\cancel{\text{amu}}}\right)\\ =-3.156\times {\text{10}}^{-\text{30 }}\text{kg}\end{array}$

The energy during the decay of a radioactive substance is given as follows:

$\Delta E=(\Delta m){\text{c}}^{\text{2}}$

Substitute the values of energy and mass defect in the above expression.

$\begin{array}{c}\Delta E=\text{(}-3.156\times {\text{10}}^{-\text{30}}\text{kg)(3}\times {\text{10}}^{\text{8}}{\text{m/sec)}}^{\text{2}}\\ \text{ = }-2.84\times {\text{10}}^{-\text{13}}\text{J}\end{array}$

Thus, the energy released during the decay of ${}^{90}\text{Y}$

is $-2.84\times {\text{10}}^{-\text{13}}\text{J}$

b) The number of moles of ${}^{90}\text{Sr }$

that will decay in a year, starting with 1 mole of ${}^{90}\text{Sr }$.

Initial time, $\text{t = 0 min}$ moles of ${}^{90}\text{Sr}=1\text{ mol}$

Final time, $t=1\text{ year}$

${}^{\text{90}}\text{Sr , }{t}_{1/2}=\text{28}\text{.1 year}$

It is also known that the rate constant for the radioactive decay is given as follows:

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

Substitute the value of half-life.

$\begin{array}{c}k\text{ =}\left(\frac{0.693}{\text{28}\text{.1 years}}\right)\\ =0.0247{\text{ year}}^{-\text{1}}\end{array}$

Now, number of moles of ${}^{90}\text{Sr }$

decaying in one year can be calculated as follows:

$\text{ln}\frac{N_t}{N_o}=-kt$

Here, $N_o$ is initial number of moles while $N_t$

is the final number of moles.

Substitute the values of $N_o$, $k$ and $t$

in the above expression.

$\begin{array}{c}\text{ln}\frac{N_t}{1\text{ mol}}\text{=}-\text{(0}{\text{.0247 year}}^{-\text{1}})\times \text{1 year}\\ \text{ln}\frac{N_t}{1\text{ mol}}\text{ =}-\text{(0}\text{.0247})\\ \frac{N_t}{1\text{ mol}}=e^{-\text{ (0}\text{.0247})}\\ N_t\text{ = 0}{\text{.9756 mol }}^{\text{90}}\text{Sr }\end{array}$

Thus, the number of moles that has decayed in one year is calculated as follows:

$\begin{array}{c}n=1\text{ mol}-\text{0}\text{.9756 mol }\\ \text{= 0}\text{.024 mol}\end{array}$

So, moles of ${}^{90}\text{Sr}$

decayed $\text{= 0}\text{.024 mol}$

c) The amount of heat released (in kJ) corresponding to the $\text{ 0}\text{.0244 mol}$ of ${}^{90}\text{Sr}$ decayed to ${}^{90}\text{Zr}$.

It is known that half- life of ${}^{90}\text{Y}$

is much shorter than that of ${}^{90}\text{Sr}$. Hence, it can be assumed that all ${}^{90}\text{Y}$

formed from ${}^{90}\text{Sr}$

is converted to ${}^{90}\text{Zr}$. Now, the energy changes observed in part (a) are for the individual nuclei.

Now, it is known that one mole of particles is equivalent to Avogadro’s number.

Hence, the conversion factor is $\frac{6.022\times {\text{ 10}}^{\text{23}}\text{ nuclei}}{1\text{ mol}}$

Thus, by using the above conversion, the number of nuclei decayed in one year can be calculated as follows:

$\left(\text{0}\text{.0244 }\cancel{\text{mol}}\times \frac{6.022\times {\text{ 10}}^{\text{23}}\text{ nuclei}}{1\cancel{\text{mol}}}\right)=\text{1}\text{.47 }\times {\text{ 10}}^{\text{22}}\text{ nuclei}$

Now, the total energy released during the decay given in part (a) is as follows:

$\text{(5}\text{.59 }\times {\text{ 10}}^{-\text{15}}\text{J) + ( 2}\text{.84 }\times {\text{ 10}}^{-\text{13}}\text{J )}=\text{2}\text{.90}\times {\text{10}}^{-13}\text{ J}$

This energy corresponds to the energy released by one nuclei.

Hence, the conversion factor is $\frac{\text{2}\text{.90 }\times {\text{ 10}}^{-13}\text{ J}}{1\text{ nuclei}}$

Moreover, it is also known that one kilojoules is equivalent to thousand joules.

Hence, the conversion factor is $\frac{\text{1 kJ}}{1000\text{ J}}$

Thus, by using the above conversion factor, the energy released in one year can be calculated as follows:

$\left((1.47\times {\text{ 10}}^{\text{22}}\cancel{\text{nuclei}}\text{) }\times \frac{\text{2}\text{.90 }\times {\text{ 10}}^{-13}\cancel{\text{J}}}{1\cancel{\text{nuclei}}}\times \frac{\text{1 kJ}}{1000\cancel{\text{J}}}\right)\text{ = 4}\text{.26}\times {\text{10}}^{\text{6}}\text{ kJ}$

Hence, the heat released by the decay of ${}^{90}\text{Sr}$ in one year is $\text{4}\text{.26 }\times {\text{ 10}}^{\text{6}}\text{ kJ}$.