Question

A radioactive element with $6\times10^{5}$ atoms initially is left with $0.75\times10^{5}$ undecayed atoms in $48$ years. The half-life is:

• A. 16 years
• B. 24 years
• C. 12 years
• D. 6 years
• E. 18 years

Hint:

$0.75$ is $1/8$ of $6$. Since $1/8 = (1/2)^3$, exactly 3 half-lives have passed.

Answer 1

The Correct Option is A

Step 1: Concept
Use the relation $N = N_0(1/2)^n$, where $n$ is the number of half-lives.

Step 2: Analysis
$0.75\times10^5 = 6\times10^5(1/2)^n \implies 0.75/6 = (1/2)^n \implies 1/8 = (1/2)^n$.
Therefore, $n = 3$ half-lives.

Step 3: Calculation
Total time $T = n \times T_{1/2} \implies 48 = 3 \times T_{1/2}$.
$T_{1/2} = 16$ years. Final Answer: (A)