Animated Solution for Physics - Atoms and Nuclei: A radioactive nucleus decays by two different processes. The half-life for the first process is 10 s and that for the second is 100 s. The effective half-life of the nucleus is close to

Imagine a radioactive nucleus, let's call it $A$, that has a choice. It doesn't just decay into one specific product, but it can simultaneously decay into two different products, $B$ and $C$. This is what we call parallel decay.

When multiple decay processes happen at the same time, their probabilities add up. Because the decay constant $\lambda$ represents the probability of decay per unit time, the effective decay constant ${\lambda }_{\text{eff}}$ is simply the sum of the individual decay constants.

Now, recall the fundamental relationship between the decay constant and half-life. The decay constant $\lambda$ is equal to the natural log of $2$ divided by the half-life $T_{1/2}$.

Let's substitute this relationship into our effective decay constant equation.

Notice how the term $\ln 2$ is common everywhere? We can safely cancel it out from both sides. This leaves us with a beautiful, simple relation that looks exactly like the formula for parallel resistors in electricity!

Let's bring in the values given in the question. The first half-life, $T_1$, is $10\text{ s}$, and the second half-life, $T_2$, is $100\text{ s}$. Let's plug these numbers right into our simplified equation.

Taking the common denominator as $100$, the numerator becomes $10+1$.

Finally, we just take the reciprocal to find the effective half-life.

Looking at our options, the closest value is $9\text{ s}$. That's our final answer! As a quick sanity check, notice that the effective half-life is shorter than both individual half-lives, which makes perfect physical sense because the nucleus has more pathways to decay.