Animated Solution for Physics - Atoms and Nuclei: A radioactive material decays by simultaneous emission of two particles with half-lives of $1400\text{ yr}$ and $700\text{ yr}$, respectively. What will be the time after the which one-third of the material remains ? [Take, $\ln 3=1.1$]

Imagine a radioactive nucleus that is highly unstable. Instead of following a single predictable path to stability, it has a choice. It can decay by emitting one type of particle to become a new element, or it can emit a completely different particle to become another element. This is known as simultaneous decay or branching decay.

When a substance decays through multiple independent paths simultaneously, the overall rate at which the parent nuclei disappear is simply the sum of the rates of the individual processes. Mathematically, the decay constants add up:

In our problem, the radioactive material decays via two paths with half-lives $T_1=1400\text{ yr}$ and $T_2=700\text{ yr}$. We know that the decay constant $\lambda$ is related to the half-life by the equation $\lambda =\frac{\ln 2}{T_{1/2}}$.

Let's calculate the individual decay constants:

Now, we find the effective decay constant by adding them together. To make the addition easier, we can use a common denominator of $1400$:

The fundamental law of radioactive decay states that the number of undecayed nuclei $N(t)$ at any time $t$ is given by:

We are asked to find the time $t$ when exactly one-third of the original material remains. This means we set $N(t)=\frac{N_0}{3}$:

The initial amount $N_0$ beautifully cancels out from both sides, leaving us with a pure exponential equation:

To bring the variable $t$ down from the exponent, we take the natural logarithm ($\ln$) on both sides. Remember the logarithmic property $\ln (1/x)=-\ln x$:

Now, we substitute our expression for ${\lambda }_{eff}$ back into the equation:

The problem provides the approximation $\ln 3=1.1$. We also use the standard approximation $\ln 2\approx 0.693$. Plugging these values in, we get:

Solving for $t$:

Rounding to the nearest option, we find that it takes approximately $740\text{ years}$ for the material to reduce to one-third of its initial amount.