A small quantity of solution containing Na^24 radio nuclide (half-life = 15 h) of activity 1.0 micro

Q: Why do we use $R = R_0 e^{-\lambda t}$ instead of just dividing the initial activity by the sample activity?

Because the radioactive isotope is continuously decaying! The activity of the Sodium-24 injected into the body doesn't stay constant. Over the 5 hours it takes for the blood to mix and the sample to be drawn, a significant portion of the isotope has decayed. If we used the initial activity, we would overestimate the blood volume.

Q: How did we get the decay constant $\lambda$?

The decay constant $\lambda$ is related to the half-life $t_{1/2}$ by the formula $\lambda = \frac{\ln 2}{t_{1/2}}$. Since the half-life is given as 15 hours, we calculate $\lambda = \frac{0.693}{15} \text{ h}^{-1}$.

Q: Why did we convert the sample activity to disintegrations per second?

Consistency is key in physics! The initial activity was given in microcuries, and the conversion factor provided ($1 \text{ Ci} = 3.7 \times 10^{10} \text{ dps}$) gives us the rate in disintegrations per second. To compare the total activity with the sample activity, they must be in the exact same units.

Q: What does the assumption 'mixes uniformly' mean mathematically?

It means the concentration of the radioactive isotope is the same everywhere in the blood. Mathematically, this implies that the ratio of Activity to Volume is constant: $\frac{\text{Total Activity}}{\text{Total Volume}} = \frac{\text{Sample Activity}}{\text{Sample Volume}}$.

Q: Is 5.95 Liters a realistic blood volume for a human?

Yes! The average adult human body contains about 4.5 to 5.5 liters of blood, so 5.95 liters is a very realistic and medically sound value. This tracer method is actually a real technique used in medicine to measure blood volume!

Question

A small quantity of solution containing $	ext{Na}^{24}$ radio nuclide (half-life = $15	ext{ h}$) of activity $1.0	ext{ microcurie}$ is injected into the blood of a person. A sample of the blood of volume $1	ext{ cm}^3$ taken after $5	ext{ h}$ shows an activity of $296$ disintegrations per minute. Determine the total volume of the blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person.
($1	ext{ curie} = 3.7 	imes 10^{10}$ disintegrations per second)

Sigma Percentile · Accepted Answer

Imagine you have a large tank of water, and you want to know its exact volume without draining it. How would you do it? One brilliant way is to use a tracer! You add a known amount of a detectable substance, let it mix completely, and then take a tiny sample. By measuring the concentration of the substance in your small sample, you can easily calculate the total volume of the tank. This is exactly what we are doing in this problem, but instead of a water tank, it's the human circulatory system, and instead of dye, we are using a radioactive isotope, Sodium-24!

### Analyzing the Setup

We are given a small quantity of $	ext{Na}^{24}$ with a half-life of $15 	ext{ h}$. Its initial activity is $1.0 	ext{ microcurie}$. This is injected into a person's blood. After $5 	ext{ h}$, a $1 	ext{ cm}^3$ sample of blood is drawn, and its activity is measured to be $296$ disintegrations per minute. We need to find the total volume of blood in the person's body.

### The Master Equation

First, let's get all our units consistent. The standard unit for activity is disintegrations per second (dps).

The initial activity $R_0$ is $1.0 \mu	ext{Ci}$.
$$R_0 = 10^{-6} 	imes 3.7 	imes 10^{10} = 3.7 	imes 10^4 	ext{ dps}$$

The activity of our $1 	ext{ cm}^3$ sample after $5 	ext{ h}$, let's call it $r$, is $296 	ext{ dpm}$.
$$r = \frac{296}{60} = 4.93 	ext{ dps}$$

We also need the decay constant $\lambda$.
$$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{15} = 0.0462 	ext{ h}^{-1}$$

### Final Calculation

Now, what is the total activity $R$ of the entire blood volume after $5 	ext{ h}$? According to the radioactive decay law:
$$R = R_0 e^{-\lambda t}$$

Since the radioactive solution mixes uniformly, the ratio of the total volume $V$ to the sample volume ($1 	ext{ cm}^3$) must be equal to the ratio of the total activity $R$ to the sample activity $r$.
$$V = \frac{R}{r} = \frac{R_0 e^{-\lambda t}}{r}$$

Let's plug in the numbers!
$$V = \frac{3.7 	imes 10^4}{4.93} e^{-0.0462 	imes 5}$$
$$V = 7505 	imes e^{-0.231}$$
$$V = 7505 	imes 0.7937 = 5957 	ext{ cm}^3$$

Converting this to liters, we get $5.95 	ext{ L}$.

And there we have it! By just taking a tiny $1 	ext{ cm}^3$ sample of blood, we were able to determine that the person has about $5.95$ liters of blood in their body. Isn't physics amazing?

Visualized Solution

$λ = \frac{l n 2}{t _{1/2}}$

$R_{0} = 1 μ Ci$

$r = \frac{296}{60} dps$

$V = \frac{R}{r}$

$V = \frac{3.7 \times 1 0 ^{4}}{4.93} e^{- 0.0462 \times 5}$

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=t1/2​ln2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R0​=1μCi

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=60296​ dps

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=rR​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=4.933.7×104​e−0.0462×5