Animated Solution for Physics - Atoms and Nuclei: The activity of a radioactive sample falls from $700{\text{ s}}^{-1}$ to $500{\text{ s}}^{-1}$ in $30\text{ min}$. Its half-life is close to

Imagine you are observing a radioactive sample in a laboratory. At the very moment you start your stopwatch ($t=0$), the Geiger counter registers an activity of $700{\text{ s}}^{-1}$. This means exactly 700 nuclei are disintegrating every single second.

As time ticks on, the unstable nuclei deplete, and the rate of decay naturally slows down. Exactly $30\text{ minutes}$ later, you check the counter again, and the activity has dropped to $500{\text{ s}}^{-1}$. Our mission is to determine the half-life ($T_{1/2}$) of this mysterious sample—the time it takes for its activity to drop to exactly half of its initial value.

The fundamental law of radioactive decay states that the activity $A$ of a sample at any given time $t$ follows a strict exponential curve. We can express this mathematically as:

Here, $A_0$ is the initial activity, and $T_{1/2}$ is the half-life. This form of the equation is incredibly powerful because it directly incorporates the half-life, bypassing the need to calculate the decay constant $\lambda$ first.

Let's substitute the raw data from our experiment into this master equation:

To isolate our unknown variable $T_{1/2}$, we first need to clean up the equation. Dividing both sides by $700$ gives us:

Dealing with fractions less than one can be mentally taxing, so let's take the reciprocal of both sides. This flips the fractions and changes the base on the right side from $1/2$ to $2$:

Since $7/5$ is exactly $1.4$, our equation simplifies beautifully to:

To solve for an exponent, we must invoke the power of logarithms. Taking the natural logarithm ($\ln$) of both sides allows us to bring the exponent down:

Rearranging this to solve for $T_{1/2}$ yields:

In a competitive exam like JEE, you are expected to know that $\ln (2)\approx 0.693$. To find $\ln (1.4)$, we can use logarithm properties: $\ln (1.4)=\ln (14/10)=\ln (7/5)=\ln (7)-\ln (5)$. Knowing that $\ln (7)\approx 1.946$ and $\ln (5)\approx 1.609$, we get $\ln (1.4)\approx 0.337$.

Plugging these values in:

Looking at our options, $62\text{ minutes}$ is the undeniable correct answer.

What if you forgot the value of $\ln (7)$ under exam pressure? Physics rewards intuition! Look closely at the number $1.4$. It is tantalizingly close to $1.414$, which is the well-known value of $\sqrt{2}$.

If we boldly approximate $1.4\approx \sqrt{2}=2^{1/2}$, our exponential equation becomes:

Since the bases are now identical, we can simply equate the exponents:

Solving this gives $T_{1/2}\approx 60\text{ minutes}$. While not perfectly exact, $60$ is close enough to $62$ to confidently select option (a) when the other choices ($52,66,72$) are significantly further away. This is the art of competitive physics!