Chapter 20, Problem 25E

Interpretation Introduction

(a)

Interpretation:

The grams of the isotope left after $12$ minutes are to be stated.

Concept introduction:

The elements whose nucleus decays spontaneously are known as radioactive elements. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays. All the elements which have an atomic number greater than $83$ are radioactive. The

half life of a radioactive substance is defined as the time taken by half of its atoms to decay.

Answer to Problem 25E

The grams of the isotope left after $12$ minutes is $5.7\text{g}$.

Explanation of Solution

The formula to calculate the number of half life, $\text{n}$ is given below.

$\text{n}=\frac{\text{Time}}{\text{Half}\text{life}}$…(1)

The time is $12\min$.

The half life is $3.1\text{min}$.

Substitute the values of time and half life in equation (1).

$\begin{array}{rll}\text{n}& =& \frac{12\text{min}}{3.1\text{min}}\\ & =& 3.9\end{array}$

The formula to calculate amount of sample left after $\text{n}$ half life is given below.

$\text{R}=\text{S}\times {\left(\frac{\text{1}}{\text{2}}\right)}^{\text{n}}$…(2)

Where,

• $\text{R}$ is the amount of sample left.

• $\text{S}$ is the starting amount.

• $\text{n}$ is the number of half lives.

The starting amount, $\text{S}$ is $\text{84}\text{.6}\text{g}$.

The value of $\text{n}$ is $3.9$.

Substitute the value of $\text{n}$ and $\text{S}$ in equation (2).

$\begin{array}{rll}\text{R}& =& \text{84}\text{.6}\text{g}\times {\left(\frac{\text{1}}{\text{2}}\right)}^{\text{3}\text{.9}}\\ & =& \text{84}\text{.6}\text{g}\times 0.06699\\ & =& 5.7\text{g}\end{array}$

Therefore, the grams of the isotope left after $12$ minutes is $5.7\text{g}$.

Conclusion

The grams of the isotope left after $12$ minutes is $5.7\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The minutes in which the mass of ${}_{\text{81}}^{\text{208}}\text{Tl}$ will be $\text{3}\text{.48}\text{g}$ is to be stated.

Concept introduction:

The elements whose nucleus decays spontaneously are known as radioactive elements. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays. All the elements which have an atomic number greater than $83$ are radioactive. The

half life of a radioactive substance is defined as the time taken by half of its atoms to decay.

Answer to Problem 25E

The minutes in which the mass of ${}_{\text{81}}^{\text{208}}\text{Tl}$ will be $\text{3}\text{.48}\text{g}$ is $14\min$.

Explanation of Solution

The formula to calculate amount of sample left after $\text{n}$ half lives is given below.

$\text{R}=\text{S}\times {\left(\frac{\text{1}}{\text{2}}\right)}^{\text{n}}$…(2)

Where,

• $\text{R}$ is the amount of sample left.

• $\text{S}$ is the starting amount.

• $\text{n}$ is the number of half lives.

The starting amount, $\text{S}$ is $\text{84}\text{.6}\text{g}$.

The amount of sample left, $\text{R}$ is $\text{3}\text{.48}\text{g}$.

Substitute the value of $\text{R}$ and $\text{S}$ in equation (2).

$\begin{array}{rll}\text{3}\text{.48}\text{g}& =& \text{84}\text{.6}\text{g}\times {\left(\frac{\text{1}}{\text{2}}\right)}^{\text{n}}\\ {\left(\text{0}\text{.5}\right)}^{\text{n}}& =& 0.04113\\ \text{n}& =& \frac{\log \left(0.04113\right)}{\log \left(0.5\right)}\\ & =& 4.6\end{array}$

The formula to calculate the number of half lives, $\text{n}$ is given below.

$\text{n}=\frac{\text{Given}\text{time}}{\text{Half}\text{life}}$…(1)

The half-life is $3.1\min$.

The value of $\text{n}$ is $4.6$.

Substitute the values of time and half life in equation (1).

$\begin{array}{rll}4.6& =& \frac{\text{Time}}{3.1\text{min}}\\ \text{Time}& =& 3.1\min \times 4.6\\ & =& 14\min \end{array}$

Therefore, the minutes in which the mass of ${}_{\text{81}}^{\text{208}}\text{Tl}$ will be $\text{3}\text{.48}\text{g}$ is $14\min$.

Conclusion

The minutes in which the mass of ${}_{\text{81}}^{\text{208}}\text{Tl}$ will be $\text{3}\text{.48}\text{g}$ is $14\min$.