Chapter 25, Problem 61A

(a)

Interpretation Introduction

Interpretation:

The percentage of isotope remaining after 60 days is to be determined.

Concept Interpretation:

The process of radioactive decay is how an unstable atomic nucleus loses energy through radiation. A substance that has unstable nuclei is regarded as radioactive.

Answer to Problem 61A

Around $20\%$ .

Explanation of Solution

Finding the value of $60$ days on the x-axis is the first step in calculating the percentage of the radioisotope thorium- $234$ still present after decaying for $60$ days; the equivalent value for this percentage may then be found on the y-axis. This value on the y-axis is shown by the "about $20\%$ .

Therefore, about $20\%$ after a radioactive decay of $60$ days, just $20\%$ of the isotope thorium- $234$ is still present.

(b)

Interpretation Introduction

Interpretation:

Amount of radioactive samples remaining after 40 days is to be determined.

Concept Interpretation:

The process of radioactive decay is how an unstable atomic nucleus loses energy through radiation. A substance that has unstable nuclei is regarded as radioactive.

Answer to Problem 61A

Around $88g$ of the sample remains

Explanation of Solution

Finding the value of $40$ days on the x-axis is the first step in calculating the percentage of the radioisotope thorium- $234$ still present; the equivalent value for this percentage may then be found on the y-axis. This value on the y-axis is shown by the $88g$ of the sample.

Therefore, about $88g$ after a radioactive decay of $40$ days, just $88g$ of the isotope thorium- $234$ is still present.

(c)

Interpretation Introduction

Interpretation:

Time required to decay 44 g to 4.4 g is to be determined.

Concept Interpretation:

The process of radioactive decay is how an unstable atomic nucleus loses energy through radiation. A substance that has unstable nuclei is regarded as radioactive.

Answer to Problem 61A

Approximately $83$ days

Explanation of Solution

We must first determine the percentage of thorium- $234$ that remains in order to solve this issue.

  $\frac{4.4g}{44g}\times 100=10\%$

As a result, $10\%$ of thorium- $234$ is still present after decay.

Next, find the number of days it took for the decay by using the radioactive decay curve on the graph.

We can deduce from the graph that it took roughly $83$ days for $44g$ of thorium-234 to degrade into $4.4g$ of thorium- $234$ .

(d)

Interpretation Introduction

Interpretation:

Half-life is to be determined.

Concept Interpretation:

The process of radioactive decay is how an unstable atomic nucleus loses energy through radiation. A substance that has unstable nuclei is regarded as radioactive.

Answer to Problem 61A

Approximately $83$ days

Explanation of Solution

To address the issue, we will use the equation shown below:

  $A=A_0\times {\left(\frac{1}{2}\right)}^{t/T}$

where $(t)$ is the amount of time that has passed, $(A)$ is the amount of time that has passed, and $(A_0)$ is the half-life of the chemical $(T)$ (using the same units).

We have given:

  $\begin{array}{l}A=4.4g\\ A_0=44g\\ t=83days\\ 4.4g=44g\times {\left(\frac{1}{2}\right)}^{83/T}\\ \frac{4.4}{44}={\left(\frac{1}{2}\right)}^{83/T}\\ 0.1={0.5}^{83/T}\\ \log 0.1=\log ({0.5}^{83/T})\\ \log 0.1=\frac{83}{T}\times \log 0.5\\ T=(83)\times \frac{\log 0.5}{\log 0.1}\\ =25days\end{array}$

The half-life of thorium- $234$ is $25$ days.