Chapter 14, Problem 14.98PAE

(a)

Interpretation Introduction

Interpretation: The number of ${}^{\text{26}}\text{Al}$ nuclei decay per second in a 1 mol sample of ${}^{\text{26}}\text{Al}$ should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

$k=\frac{0.693}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Where, $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is half-life

k is the decay constant

Answer to Problem 14.98PAE

Solution:

$3.063\times {10}^{-14}$ decays per second

Explanation of Solution

Calculating decay constant for $Al$ as follows:

$\begin{array}{c}k=\frac{0.693}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ k=\frac{0.693}{7.17\times {10}^5\text{ yr}}\\ =9.66\times {10}^{-7}{\text{ yr}}^{-1}\end{array}$

Now, calculate the number of decays per second in 1 mole of the $Al$

$\begin{array}{c}k=9.66\times {10}^{-7}{\text{ yr}}^{-1}\left(\frac{1\text{ yr}}{365\text{ day}}\right)\left(\frac{1\text{ day}}{24\text{ h}}\right)\left(\frac{1\text{ h}}{60\min }\right)\left(\frac{1\text{ min}}{60s}\right)\\ =3.063\times {10}^{-14}{\text{ s}}^{-1}\end{array}$

(b)

Interpretation Introduction

Interpretation: Energy released in a single ${}^{2\text{6}}\text{Al}$ decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

${\text{E=Δmc}}^{\text{2}}$

Here, $\text{Δm}$ is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

Answer to Problem 14.98PAE

Solution:

$3.49MeV$

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

$\begin{array}{l}=\left|\left(Massof Mg+Massof positron\right)-Massof Al\right|\\ =\left|\left(25.982593+0.0005486\right)-25.986892\right|\\ =3.75\times {10}^{-3}\text{ amu}\end{array}$

Since, 1amu =931.7 MeV

Thus,

$\begin{array}{c}3.75\times {10}^{-3}\text{ amu}=3.75\times {10}^{-3}\times 931.7\text{ MeV}\\ \text{=3}\text{.5 MeV}\end{array}$

Thus, energy released will be $\text{3}\text{.5 MeV}$.

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are $6.023\times {10}^{23}$ number of particles.

Answer to Problem 14.98PAE

Solution:

$3.86\times {10}^{12}MeV$

Explanation of Solution

Converting decay per second to decay per minute as follows:

$\begin{array}{c}k=3.063\times {10}^{-14}{\text{ s}}^{-1}\left(\frac{60\text{ s}}{1\text{ min}}\right)\\ =1.84\times {10}^{-12}{\text{ min}}^{-1}\end{array}$

In 1 mole there are $6.023\times {10}^{23}$

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

$\begin{array}{c}Heat=\text{3}\text{.5 MeV}\times 6.023\times {10}^{23}{\text{ mol}}^{-1}\times 1.837\times {10}^{-12}{\text{ min}}^{-1}\\ =3.87\times {10}^{12}MeV{\text{ mol}}^{-1}{\text{ min}}^{-1}\end{array}$