JEE Challenger
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PhyHardAdv 2017 • Paper 1

Estimation of Total Blood Volume Using Radioactive Iodine Decay

An isotope of Iodine, 131I{}^{131}\text{I}, undergoes β\beta-decay to a Xenon isotope with a half-life T1/2=8 daysT_{1/2} = 8\text{ days}. A tracer solution containing 131I{}^{131}\text{I} with an initial activity of 2.4×105 Bq2.4 \times 10^5\text{ Bq} is administered into the bloodstream of a patient. The injected tracer is known to distribute evenly throughout the circulatory system in under 30 minutes30\text{ minutes}. After a duration of 11.5 hours11.5\text{ hours}, a 2.5 ml2.5\text{ ml} sample of blood is collected, displaying an activity of 115 Bq115\text{ Bq}. Calculate the approximate total blood volume of the patient in liters. (Given: ex≈1+xe^x \approx 1 + x for ∣x∣≪1|x| \ll 1 and ln⁡2≈0.7\ln 2 \approx 0.7).

Official Numerical Answer5

Topics & Concepts

NucleiRadioactivity

Step-by-Step Solution

To find the total blood volume of the patient, we track the decay of the radioactive isotope 131I^{131}\text{I} and its distribution throughout the circulatory system over time.

Step 1: Calculate the Decay Constant (λ\lambda)

The half-life of 131I^{131}\text{I} is given as: T1/2=8 days=8×24 hours=192 hoursT_{1/2} = 8 \text{ days} = 8 \times 24 \text{ hours} = 192 \text{ hours}

The decay constant λ\lambda is defined as: λ=ln⁡2T1/2\lambda = \frac{\ln 2}{T_{1/2}}

Given ln⁡2≈0.7\ln 2 \approx 0.7: λ=0.7192 hours−1\lambda = \frac{0.7}{192} \text{ hours}^{-1}

Step 2: Calculate the Activity of the Tracer After t=11.5 hourst = 11.5 \text{ hours}

The activity A(t)A(t) of a radioactive substance at time tt is given by: A(t)=A0e−λtA(t) = A_0 e^{-\lambda t}

where A0=2.4×105 BqA_0 = 2.4 \times 10^5 \text{ Bq} is the initial activity.

First, let's calculate the exponent λt\lambda t: λt=(0.7192)×11.5=8.05192≈0.04193\lambda t = \left(\frac{0.7}{192}\right) \times 11.5 = \frac{8.05}{192} \approx 0.04193

Since ∣λt∣≪1|\lambda t| \ll 1, we use the approximation e−x≈1−xe^{-x} \approx 1 - x: e−λt≈1−λt=1−8.05192=183.95192e^{-\lambda t} \approx 1 - \lambda t = 1 - \frac{8.05}{192} = \frac{183.95}{192}

Now, calculating the total activity A(t)A(t) remaining in the body after 11.5 hours11.5 \text{ hours}: A(t)=2.4×105×183.95192A(t) = 2.4 \times 10^5 \times \frac{183.95}{192} A(t)=240000192×183.95=1250×183.95=229937.5 BqA(t) = \frac{240000}{192} \times 183.95 = 1250 \times 183.95 = 229937.5 \text{ Bq}

Step 3: Calculate the Total Blood Volume (VV)

After 11.5 hours11.5 \text{ hours}, the tracer is uniformly distributed throughout the total blood volume VV.

The activity per unit volume (activity density) is: Activity Density=A(t)V\text{Activity Density} = \frac{A(t)}{V}

A sample of volume v=2.5 ml=2.5×10−3 Lv = 2.5 \text{ ml} = 2.5 \times 10^{-3} \text{ L} is drawn, which shows an activity of Asample=115 BqA_{\text{sample}} = 115 \text{ Bq}.

Thus: Asample=(A(t)V)×vA_{\text{sample}} = \left(\frac{A(t)}{V}\right) \times v

Rearranging the formula to solve for the total volume VV: V=A(t)⋅vAsampleV = \frac{A(t) \cdot v}{A_{\text{sample}}}

Substitute the calculated values into the equation: V=229937.5 Bq×2.5×10−3 L115 BqV = \frac{229937.5 \text{ Bq} \times 2.5 \times 10^{-3} \text{ L}}{115 \text{ Bq}} V=229937.5115×2.5×10−3 LV = \frac{229937.5}{115} \times 2.5 \times 10^{-3} \text{ L} V=1999.4565×2.5×10−3 L≈4.9986 LV = 1999.4565 \times 2.5 \times 10^{-3} \text{ L} \approx 4.9986 \text{ L}

Rounding to the nearest whole integer: V≈5 litersV \approx 5 \text{ liters}

Final Answer:

The approximate total blood volume of the patient is 55 liters.

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