131I is an isotope of Iodine that decays into an isotope of Xenon with a half-life of 8 days. A smal

Q: Why can't we just use the initial activity of $2.4 \times 10^5$ Bq to find the volume?

Because the tracer is radioactive! During the 11.5 hours it takes for the serum to mix and the sample to be drawn, a portion of the Iodine-131 has decayed into Xenon. If you use the initial activity, you will overestimate the total volume because you are assuming more active tracer is present than there actually is.

Q: Why did we write $e^{-0.042}$ as $\frac{1}{1.042}$ instead of $1 - 0.042$?

The problem explicitly states to use the approximation $e^x \approx 1 + x$. While $e^{-x} \approx 1 - x$ is also a valid Taylor expansion for small $x$, using $e^{-x} = \frac{1}{e^x} \approx \frac{1}{1+x}$ often yields a slightly more accurate result in these specific types of calculations and perfectly aligns with the provided hint.

Q: What does 'activity' actually mean in this context?

Activity refers to the number of radioactive decays happening per second. The unit Becquerel (Bq) is defined as one decay per second. So, an activity of 115 Bq means that in that 2.5 ml sample, 115 Iodine atoms are decaying into Xenon every single second.

Q: Why do we assume the concentration is the same everywhere?

The problem states that the injected serum gets 'distributed uniformly in the blood stream'. This is a crucial assumption in the isotope dilution method. It means the tracer is perfectly mixed, so any random sample you pull will have the exact same proportion of tracer to blood as the entire body.

Q: Is 5 liters a realistic blood volume for a human?

Yes, absolutely! The average adult human body contains approximately 4.5 to 5.5 liters of blood. This makes our calculated answer of 5 liters highly realistic and serves as a good real-world sanity check for our math.

Question

$^{131}	ext{I}$ is an isotope of Iodine that decays into an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with $^{131}	ext{I}$ is injected into the blood of a person. The activity of the amount of $^{131}	ext{I}$ injected was $2.4 	imes 10^5$ Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 h, 2.5 ml of blood is drawn from the person's body, and gives an activity of 115 Bq. The total volume of blood in the person's body, in litres is approximately (you may use $e^x \approx 1 + x$ for $|x| \ll 1$ and $\ln 2 \approx 0.7$).

Sigma Percentile · Accepted Answer

Imagine you are tasked with finding the exact volume of water in a massive, irregularly shaped underground reservoir. You cannot simply drain it into measuring cups. Instead, you pour in a known quantity of a harmless dye, wait for it to mix completely, and then extract a tiny, 10-milliliter sample. By measuring the concentration of the dye in that small sample, you can mathematically deduce the total volume of the entire reservoir.

This elegant concept is known as the **isotope dilution method**, and it is precisely what is happening in this problem, albeit with a radioactive tracer in the human bloodstream.

### Analyzing the Setup

We are given an initial injected activity of $R_0 = 2.4 	imes 10^5 	ext{ Bq}$ of Iodine-131. This tracer mixes uniformly into the patient's total blood volume, which we will call $V$.

After a time $t = 11.5 	ext{ hours}$, a small sample of volume $v = 2.5 	ext{ ml}$ is drawn, and its activity is measured to be $r = 115 	ext{ Bq}$.

If this were a simple dye, we could just equate the concentrations immediately. However, Iodine-131 is radioactive, meaning its total activity is constantly decreasing according to the **radioactive decay law**:

$$ R(t) = R_0 e^{-\lambda t} $$

Before we can compare the sample to the total, we must determine exactly how much total activity $R(t)$ remains in the body at the exact moment the sample is taken.

### The Master Equation

First, let's calculate the exponent $\lambda t$. We know the decay constant $\lambda = \frac{\ln 2}{T_{1/2}}$. The half-life $T_{1/2}$ is 8 days, and the elapsed time $t$ is 11.5 hours. We must ensure our units match, so we convert the hours into days:

$$ \lambda t = \frac{\ln 2}{8 	ext{ days}} 	imes \left( \frac{11.5}{24} 	ext{ days} ight) $$

Using the given approximation $\ln 2 \approx 0.7$, we get:

$$ \lambda t = \frac{0.7 	imes 11.5}{8 	imes 24} = \frac{8.05}{192} \approx 0.042 $$

Now, we substitute this back into our decay equation to find the total remaining activity. The problem generously provides the approximation $e^x \approx 1 + x$ for small values of $x$. Therefore, $e^{-0.042} = \frac{1}{e^{0.042}} \approx \frac{1}{1 + 0.042} = \frac{1}{1.042}$.

$$ R(t) = \frac{2.4 	imes 10^5}{1.042} 	ext{ Bq} $$

### Final Calculation

Because the serum is uniformly distributed, the activity per unit volume (the concentration) must be identical in the sample and in the entire body. We can set up the following ratio:

$$ \frac{R(t)}{V} = \frac{r}{v} $$

Substituting our known values:

$$ \frac{1}{V} \left( \frac{2.4 	imes 10^5}{1.042} ight) = \frac{115}{2.5} $$

$$ \frac{1}{V} \left( \frac{2.4 	imes 10^5}{1.042} ight) = 46 $$

Now, we simply rearrange to solve for the total blood volume $V$:

$$ V = \frac{2.4 	imes 10^5}{46 	imes 1.042} = \frac{240000}{47.932} \approx 5007 	ext{ ml} $$

Converting milliliters to liters, we find that the total volume of blood in the person's body is approximately **5 Litres**.

Enter Numerical Value:

Visualized Solution

$Isotope Dilution Method$

$Radioactive Decay Law$

$Calculating the Exponent λ t$

$Evaluating λ t$

$Total Activity After 11.5 h$

$Activity Concentration$

$Calculating Total Volume V$

$Final Result$

$The Way Forward$

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

The activity of a freshly prepared radioactive sample is $1 0^{10}$ disintegrations per second, whose mean life is $1 0^{9}$ s. The mass of an atom of this radioisotope is $1 0^{- 25}$ kg. The mass (in mg) of the radioactive sample is

The activity of a radioactive sample falls from $700 s^{- 1}$ to $500 s^{- 1}$ in $30 min$ . Its half-life is close to

The half-life of $^{215} At$ is $100 μ s$ . The time taken for the activity of a sample of $^{215} At$ to decay to $\frac{1}{16} th$ of its initial value is

A radioactive sample $S_{1}$ having an activity of $5 μ Ci$ has twice the number of nuclei as another sample $S_{2}$ which has an activity of $10 μ Ci$ . The half lives of $S_{1}$ and $S_{2}$ can be

At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 s the number of undecayed nuclei reduces to 12.5%. Calculate (a) mean life of the nuclei, (b) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number.

After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is

A radioactive sample is undergoing $α$ -decay. At any time $t_{1}$ , its activity is $A$ and another time $t_{2}$ , the activity is $\frac{A}{5}$ . What is the average life time for the sample?

The half-life of $^{131} I$ is 8 days. Given a sample of $^{131} I$ at time $t = 0$ , we can assert that

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isotope Dilution Method

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radioactive Decay Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Exponent λt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating λt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Activity After 11.5 h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Activity Concentration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Total Volume V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Result

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward