Q. 5.56 - An Introduction to General, Organic, and Biological Chemistry

Question

Use the following decay curve for iodine$-131$ to answer problems a to c: $(5.4)$

a. Complete the values for the mass of radioactive iodine$- 131$ on the vertical axis.

b. Complete the number of days on the horizontal axis.

c. What is the half-life, in days, of iodine$-131$?

Step-by-Step Solution

Answer

The values for the mass radio active on vertical axis has derived and number of days is calculated

(part a) The mass od iodine $-131$ in $mg.$

(part b) The horizontal axis is $24$ days and $32$ days.

(part c) The half-life of Jod $131$ is $8$ days.

Step 1- Given information (part a)

(part a) The decay curve of iodine $-131$ is as follows.

In a given graph, the horizontal axis represents the number of days and the vertical axis represents the mass of iodine $-131$in $mg$.

Step 2-Given information (part b)

(part b) You need to find the missing mass, represented by a long line on the vertical axis. Represent the line as $a, b, c$as follows:

In the graph above, the line ' $a$'is at the midpoint of the distance between $0 mg$ and  $80 mg$.

Therefore, the line' $a$'represents $40 mg$. '$b$'is between $0 mg$ and $20 mg$. Therefore, the line'$b$'represents $10 mg$. Then the line segment'$c$' is at the midpoint of the segment between $0 mg$ and $10 mg$. Therefore, the line '$c$' represents $5 mg$.

Let us draw the decay curve with the missing values of the mass of $1-131$.

Step 3- Explanation (part a)

Therefore, the missing values for the vertical mass $1-131$are from $80 mg$ to $0 mg$, $40 mg$,  $10 mg$ and $5 m$.

(part a) Complete the number of days on the horizontal axis 

Step 4- Explanation (part b)

(part b) Use the graph to determine the half life of time. One unit of the horizontal line represents $8$ days. The two units represent 16 days. This means $8 \times 2=16$ days. Therefore, $3$ units on the horizontal axis correspond to $8 \times 3=24$ days. Similarly, the $4$ units on the horizontal axis correspond to $8 \times 4=32$ days.

Therefore, the decay curve for $1-131$ with the closed value for days looks like this:

Therefore, the number of missing days from $16$ on the horizontal axis is $24$ days and $32$ days.

Step 5- Explanation (part c)

(part c) From the figure above, we can see that the initial mass of $1-131$ on the baseline is $80mg$. During the  $8$ days, the remaining mass of $1-131$will be $40 mg$. At  $16$-day intervals, the remaining $1-131$ mass  is $20mg$.

$\text{ At zero days }80\mathrm{mgI}-131\stackrel{\text{ After }8\text{ days }}{⟶}40\mathrm{mgI}-131\stackrel{\text{ After }16\text{ days }}{⟶}20\mathrm{mgI}-131$

$\stackrel{\text{ After }24\text{ days }}{⟶}10\mathrm{mgI}-131\stackrel{\text{ Afer }32\text{ days }}{⟶}5\mathrm{mgI}-131$

From the above sequence, we can see that after the interval of $8$ days, the mass is reduced to half of the previous mass. Therefore, the half-life of $1-131$ is $8$ days. Therefore, we can show that the sequence can be displayed as:

$\stackrel{3\text{ hallelives }}{⟶}10\mathrm{mgI}-131\stackrel{4\text{ halflives }}{⟶}5\mathrm{mgI}-131$

Therefore, the half-life of Jod $131$ is $8$ days.