Question

What is the half-life period of a radioactive material if its activity drops to 1/16 th of its initial value in 30 years?

• A. 9.5 years
• B. 8.5 years
• C. 7.5 years
• D. 10.5 years

Answer 1

The Correct Option is C

To find the half-life period of the radioactive material, we will use the basic principle of radioactive decay that relates the remaining quantity of a substance to its initial amount over time, governed by the formula:

\(N = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}\) 

Where:

• \(N\) is the final quantity of the substance.
• \(N_0\) is the initial quantity of the substance.
• \(t\) is the time period.
• \(T_{1/2}\) is the half-life of the substance.

According to the problem, the activity drops to \(\frac{1}{16}\) of its initial value in 30 years. We will plug these values into the formula:

\(\frac{1}{16} = \left(\frac{1}{2}\right)^{30/T_{1/2}}\)

Note that \(\frac{1}{16} = \left(\frac{1}{2}\right)^4\). Hence, we set up the equation:

\(4 = \frac{30}{T_{1/2}}\)

Solving for the half-life \(T_{1/2}\) gives:

\(T_{1/2} = \frac{30}{4} = 7.5\text{ years}\)

Thus, the half-life period of the radioactive material is 7.5 years which matches with the given option.

Answer 2

The correct answer is (C) : 7.5 years
\(∵A=\frac{A0}{\frac{t}{2^{T_{1/2}}}}\)
\(⇒2^{\frac{t}{T_{1/2}}}\)
\(=\frac{A_0}{A}=16\)
\(⇒\frac{t}{T_{1/2}}=4\)
\(⇒\frac{30}{T_{1/2}}=4\)
\(⇒T_{1/2}=\frac{30}{4}\)
=7.5 years