Chapter 30, Problem 42P

To determine

(a)

The activity immediately after 782 µg of ${}_{53}{}^{131}I$ was ingested by a patient.

Answer to Problem 42P

Solution:

The activityimmediately after 782 µg of ${}_{53}{}^{131}I$ was ingested by a patient is 3.59×10¹² decays/s.

Explanation of Solution

Given info:

782 µg of ${}_{53}{}^{131}I$ are ingested by a patient.

Formula used:

The activity ($R$)is defined as:

$R=\lambda N_o{e}^{-\lambda t}$ (1)

Where:

• $\lambda$ is called the decay constant

• $N_o$ is called the number of parent nuclei present at time t=0

• $t$ is the time elapsed

The decay ($\lambda$ )constant can be obtained using:

$\lambda =\frac{Ln(2)}{T_{1/2}}$ (2)

Where, $T_{1/2}$ is called the half-life of isotope. In this problem, the half-life ($T_{1/2}$)for iodine isotope ${}_{53}{}^{131}I$ is 8.0207 days.

Calculation:

Convert the half-life($T_{1/2}$)in seconds:

$T_{1/2}=8.0207days.\left(\frac{24h}{1\text{ day}}\right).\left(\frac{3600s}{1h}\right)=6.93x{10}^{5}s$ (3)

Replacing (3)into equation (2)we get:

$\lambda =\frac{Ln(2)}{6.93\times {10}^{5}s}=1.00\times {10}^{-6}{s}^{-1}$ (4)

Now, the atomic mass for ${}_{53}{}^{131}I$ is 131 g, therefore the initial number of parent nuclei ($N_0$)is:

$N_o=\left(\frac{6.02x{10}^{23}nuclei}{131g}\right).(782x{10}^{-6}g)=3.59x{10}^{18}nuclei$ (5)

To determine

(b)

The activity 1.50 h later when the thyroid is being tested.

Answer to Problem 42P

Solution:

The activity 1.50 h later when the thyroid is being tested is 3.57x10¹² decays/s.

Explanation of Solution

Calculation:

Convert the time elapsed in seconds as follows:

$t=(1.5h)\times \left(\frac{3600s}{1\text{ h}}\right)=5400s$ (8)

To determine

(c) To determine:

The activity 3.0 months later.

Answer to Problem 42P

Solution:

The activity 3.0 months later is 1.16×10⁹ decays/s.

Explanation of Solution

Calculation:

Convert the time elapsed in seconds as follows:

$t=3.0\text{ months}\text{.}\left(\frac{\text{31 days}}{\text{1 month}}\right).\left(\frac{24\text{ h}}{1\text{ day}}\right).\left(\frac{3600\text{ s}}{1h}\right)=8.04\times {10}^6\text{s}$ (11)