Question

A small quantity of solution containing Na24 radionuclide (half-life=15 h) of activity 1.0 microcurie is injected into the blood of a person. A sample of the blood of volume 1$cm^3$ taken after 5h 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 a person.
(1 curie =$3.7\times 10 ^{10}$disintegrations per second)

Answer 1

Hint: Use the expression for the radioactive decay and find the decay constant and final values of the molecules and substitute in the formula for the radioactive decay to get the required total volume of the blood present in the person.

Step by step solution:
Given that half life of the radionuclide $t_{\dfrac{1}{2}} = 15h$

Initial activity $N_o = 1 \mu$ curie

The activity of blood of volume $1cm^3$at 5hr $= 296 \dfrac{dis}{min}$

We know that the population decays exponentially at the rate of the decay constant.
The half-life is the time required for half of the original population of the radioactive atoms to decay

Hence the relationship between the half-life and the decay constant is given by

$\lambda = \dfrac{ln2}{t_{\dfrac{1}{2}}} = \dfrac{0.693}{t_{\dfrac{1}{2}}}$

Where $\lambda =$decay constant

Hence by substituting the given half time in the above equation we get,

$\lambda = \dfrac{0.693}{15}h^{-1} = 0.0462 h^{-1}$
As given the initial activity $N_o = 1\mu$ curie = $3.7\times 10^{10}$

Converting 1 microcurie to curie we get

Initial activity = $3.7\times 10^{10} \times 10^{-6} = 3.7 \times 10^{4} \dfrac{dis}{s}$
As given Activity of blood of volume $1cm^3$ at 5hr r = 296 $\dfrac{dis}{min}$
It can be written as r = $\dfrac{296}{60} \dfrac{dis}{s} = 4.93 \dfrac{dis}{s}$

According to radioactive decay law, the probability of a nucleus will decay is a constant i.e. independent of time but this probability may vary between different types of nuclei leading to the different decay rates.

The radioactive decay of a certain number of atoms is exponential in the time given by the equation
$N = N_o e^{-\lambda t}$

where N = total number of radioactive nuclei remaining after time t=5hr
Therefore the activity of the whole volume of blood at t = 5hr is given as

N = r V
Here V = volume

$V = \dfrac{N}{r} = \dfrac{N_o e^{-\lambda t}}{r} = \dfrac{3.7\times 10^{4} e^{-\lambda t }}{r} = 5.95\times 10^3 cm^3 = 5.95 L$

Note: In this kind of problem one has to be careful about the definition of half-life and carefully calculate the decay constant from the given details and check the units. One of the possible mistakes is that we don't see the units of the half and directly proceed with the problem which may give us wrong results.