Question

The radioactivity of a certain material drops to \(\frac{1}{16}\) of the initial value in 2 hours. The half-life of this radionuclide is

• A. 10 min
• B. 20 min
• C. 30 min
• D. 40 min

Hint:

If activity becomes \(\frac{1}{16}\), it means 4 half-lives have passed because \(1/16 = (1/2)^4\).

Answer 1

The Correct Option is C

Step 1: Use radioactive decay formula.
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \]
Step 2: Compare with given fraction.
Given:
\[ \frac{N}{N_0} = \frac{1}{16} \]
But:
\[ \frac{1}{16} = \left(\frac{1}{2}\right)^4 \]
So number of half-lives \(n = 4\).
Step 3: Total time is 2 hours.
\[ 2 \text{ hours} = 120 \text{ min} \]
Step 4: Find half-life.
\[ T_{1/2} = \frac{120}{4} = 30 \text{ min} \]
Final Answer:
\[ \boxed{30 \text{ min}} \]