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: Understanding the Question:
The activity (\( R \)) of a radioactive sample follows the law of radioactive decay. We need to find the decay constant \( \lambda \).

Step 2: Key Formula or Approach:
The activity at time \( t \) is given by:
\[ R = R_0 e^{-\lambda t} \] where \( R_0 \) is the initial activity.

Step 3: Detailed Explanation:
Given:
Initial activity, \( R_0 = 8000 \text{ dpm} \)
Activity at \( t = 4 \text{ min} \), \( R = 2000 \text{ dpm} \)
\[ 2000 = 8000 e^{-4\lambda} \] \[ \frac{2000}{8000} = e^{-4\lambda} \] \[ \frac{1}{4} = e^{-4\lambda} \implies 4 = e^{4\lambda} \] Taking natural log on both sides:
\[ \log_e 4 = 4\lambda \] \[ \log_e 2^2 = 4\lambda \] \[ 2 \log_e 2 = 4\lambda \] \[ \lambda = \frac{2 \log_e 2}{4} = 0.5 \log_e 2 \]

Step 4: Final Answer:
The decay constant is \( 0.5 \log_e 2 \).