Question

The radioactivity of a certain radioactive element drops to 1/64 of its initial value in 30 sec. Its half-life is:

• A. \( 2 \text{ sec} \)
• B. \( 4 \text{ sec} \)
• C. \( 5 \text{ sec} \)
• D. \( 6 \text{ sec} \)

Hint:

Remember that \( 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64 \). Knowing powers of 2 simplifies half-life problems significantly.

Answer 1

The Correct Option is C

Step 1: Understand half-life decay.
After each half-life, the activity falls to half its previous value. After $n$ half-lives the activity is $A_0 \left(\tfrac{1}{2}\right)^n$.
Step 2: Write the given drop as a fraction.
The activity falls to $\tfrac{1}{64}$ of the start, so \[ \left(\tfrac{1}{2}\right)^n = \frac{1}{64} \]
Step 3: Express 64 as a power of 2.
Since $64 = 2^6$, we have $\tfrac{1}{64} = \left(\tfrac{1}{2}\right)^6$. Matching exponents gives $n = 6$ half-lives.
Step 4: Relate half-lives to total time.
Those $6$ half-lives all happened within the given $30$ seconds.
Step 5: Divide to get one half-life.
\[ T_{1/2} = \frac{30 \text{ s}}{6} = 5 \text{ s} \]
Step 6: State the answer.
So each half-life lasts $5$ seconds, which is option 3.
\[ \boxed{5 \text{ sec}} \]