Question

A radioactive nucleus can decay by two different processes. The half-life for the first process is \(t_1\), and that for the second process is \(t_2\). The effective half-life \(\tau\) of the nucleus is given by:

• A. \(\dfrac{1}{\tau} = \dfrac{1}{t_1} + \dfrac{1}{t_2}\)
• B. \(\tau = t_1 + t_2\)
• C. \(\tau = t_1 - t_2\)
• D. \(\tau = t_1 t_2\)

Hint:

Independent decay channels add their decay constants; since \(t_{1/2}\propto 1/\lambda\), the reciprocals of the half-lives add.

Answer 1

The Correct Option is A

Reasoning through rates: Two parallel decay routes act like two independent "drains" on the same population of nuclei. Independent processes add in their rates, not their times.

Let $N$ be the number of nuclei. The first channel removes them at rate $\lambda_1 N$ and the second at rate $\lambda_2 N$, so
\[ -\frac{dN}{dt} = (\lambda_1 + \lambda_2)N = \lambda_{\text{eff}} N, \]
giving an effective decay constant $\lambda_{\text{eff}} = \lambda_1 + \lambda_2$.

Because half-life and decay constant are inversely proportional, $t_{1/2} \propto 1/\lambda$, the reciprocals of the half-lives add:
\[ \frac{1}{\tau} = \frac{1}{t_1} + \frac{1}{t_2}. \]
This is the harmonic-type combination, exactly like resistors in parallel or capacitors in series. The effective half-life is therefore shorter than either individual half-life, which makes physical sense since offering two ways to decay speeds up the process.
\[ \boxed{\dfrac{1}{\tau} = \dfrac{1}{t_1} + \dfrac{1}{t_2}} \]