Question

A radioactive element has rate of 8000 disintegrations per minute. After four minutes it becomes 2000 disintegrations per minute. The decay constant per minute is

• A. $0.8 \log_{e} 2$
• B. $0.6 \log_{e} 2$
• C. $0.5 \log_{e} 2$
• D. $0.2 \log_{e} 2$

Hint:

Every half-life, the activity drops by half. Here it dropped to $1/4$ ($2$ half-lives) in $4$ mins, so $T_{1/2} = 2$ mins. $\lambda = \log_e 2 / T_{1/2}$.

Answer 1

The Correct Option is C

Step 1: Formula
Activity $A = A_0 e^{-\lambda t}$.

Step 2: Analysis
$2000 = 8000 e^{-4\lambda} \implies 1/4 = e^{-4\lambda}$.
$4 = e^{4\lambda} \implies \log_e 4 = 4\lambda$.

Step 3: Calculation
$2 \log_e 2 = 4\lambda$
$\lambda = \frac{2}{4} \log_e 2 = 0.5 \log_e 2$.

Step 4: Conclusion
Hence, the decay constant is $0.5 \log_{e} 2$ per minute. Final Answer: (C)