What is the half-life period of a radioactive material if its activity drops to $$\dfrac{1}{16^{th}}$$ of its initial value of $$30$$ years?

We need to find the half-life of a radioactive material whose activity drops to $$\frac{1}{16}$$ of its initial value in 30 years.

Activity at time $$t$$: $$A = A_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$$

where $$T_{1/2}$$ is the half-life.

$$\frac{A}{A_0} = \frac{1}{16} = \left(\frac{1}{2}\right)^{t/T_{1/2}}$$

Since $$\frac{1}{16} = \left(\frac{1}{2}\right)^4$$:

$$\left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^{t/T_{1/2}}$$

$$\frac{t}{T_{1/2}} = 4$$

$$T_{1/2} = \frac{t}{4} = \frac{30}{4} = 7.5 \text{ years}$$

The correct answer is Option C: $$7.5$$ years.