Animated Solution for Physics - Atoms and Nuclei: The half-life of ${\text{Au}}^{198}$ is $2.7\text{ days}$. The activity of $1.50\text{ mg}$ of ${\text{Au}}^{198}$ if its atomic weight is $198{\text{ g mol}}^{-1}$ is ($N_A=6\times 10^{23}\text{ / mol}$)

The problem of calculating the radioactivity of a sample is a beautiful intersection of the macroscopic world we can measure and the quantum world we cannot see. Let's dive into the decay of Gold-198!

Imagine you are holding a tiny, glowing vial containing exactly $1.50\text{ mg}$ of the radioactive isotope ${\text{Au}}^{198}$. Even though the mass is incredibly small, the number of unstable atoms inside is astronomically large. Our mission is to find its activity, which is the rate at which these atoms are disintegrating.

Activity

The fundamental law of radioactive decay tells us that the activity $A$ is directly proportional to the number of undecayed nuclei $N$ present in the sample. The constant of proportionality is the decay constant $\lambda$.

This gives us our master equation:

To solve the problem, we need to find both $\lambda$ and $N$ independently and then bring them together.

First, let's tackle the decay constant $\lambda$. We know it is related to the half-life $T_{1/2}$ by the relation:

Here is where many students fall into a trap! The half-life is given as $2.7\text{ days}$. To get the standard SI unit of activity (Becquerels, or disintegrations per second), we must convert this time into seconds.

Next, we need the total number of atoms $N$. We can find this using the mole concept. The number of moles is the given mass divided by the molar mass. Multiplying this by Avogadro's number $N_A$ gives the total number of atoms. Remember to convert the mass from milligrams to grams!

Now, we substitute our expressions for $\lambda$ and $N$ back into the master equation.

This massive expression gives us the activity in Becquerels. However, if we look at the options, they are all in Curies (Ci). One Curie is defined as $3.7\times 10^{10}$ disintegrations per second. Therefore, we must divide our entire result by this conversion factor.

Now comes the heavy lifting. If you carefully crunch these numbers, the powers of ten simplify nicely.

Looking at the options, the closest matching value is $357\text{ Ci}$. In competitive exams like JEE, slight variations can occur due to the approximations used for constants like $\ln 2$ or $N_A$. Always trust your calculation and choose the closest option!