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

Concept: The law of radioactive decay states that the activity \( A \) after time \( t \) is given by \( A = A_0 \left( \frac{1}{2} \right)^n \), where \( n \) is the number of half-lives that have passed. \( n \) is calculated as the ratio of total time \( t \) to the half-life \( T_{1/2} \), i.e., \( n = \frac{t}{T_{1/2}} \).

Step 1: Determine the number of half-lives passed.
The activity dropped to \( 1/64 \) of the initial value: $$ \frac{A}{A_0} = \frac{1}{64} $$ Substitute this into the decay equation: $$ \left( \frac{1}{2} \right)^n = \frac{1}{64} $$ Express 64 as a power of 2: $$ \left( \frac{1}{2} \right)^n = \left( \frac{1}{2} \right)^6 $$ Therefore, \( n = 6 \).

Step 2: Calculate the half-life.
We have \( n = 6 \) half-lives elapsed in \( t = 30 \text{ seconds} \). $$ T_{1/2} = \frac{t}{n} = \frac{30 \text{ sec}}{6} = 5 \text{ sec} $$ $$\boxed{5 \text{ sec}}$$