Question

A radioactive element has rate of disintegration \(16,000\) disintegrations per minute at a particular instant. After four minutes, it becomes \(2000\) disintegrations per minute. The decay constant per minute is

• A. \(0.25log_e2\)
• B. \(0.50log_e3\)
• C. \(0.75log_e2\)
• D. \(0.8log_e3\)

Hint:

Activity falls exponentially: A = A0 e^(-lambda t).

Answer 1

The Correct Option is C

Step 1: Understand the concept
The rate of disintegration (activity) of a radioactive sample falls as \(A = A_0e^{-\lambda t}\).

Step 2: Substitute
\(A_0 = 16000\), \(A = 2000\) and \(t = 4\) min:
\[ \frac{2000}{16000} = \frac{1}{8} = e^{-4\lambda} \]

Step 3: Solve for \(\lambda\)
\[ 4\lambda = \ln 8 = 3\ln2 \Rightarrow \lambda = 0.75\ln 2\ \text{per minute} \]

Step 4: Check
Activity falls to \(\frac{1}{8}\) in three half-lives, so the half-life is \(\frac{4}{3}\) min and \(\lambda = \dfrac{\ln2}{4/3} = 0.75\ln2\). Option (C).

Final Answer:
The decay constant is 0.75 log_e 2 per minute. This is option (C). \[ \boxed{\text{(C) }0.75\log_e2} \]