Question

A radioactive element has rate of disintegration 9000 disintegration per minute at a particular instant. After two minutes it becomes 3000 disintegration per minute. The decay constant per minute is

• A. \(0.5\mathrm{log}_{\mathrm{e}}3\)
• B. \(0.2\mathrm{log}_{\mathrm{e}}3\)
• C. \(0.5\mathrm{log}_{\mathrm{e}}2\)
• D. \(0.2\mathrm{log}_{\mathrm{e}}2\)

Hint:

The disintegration rate follows the exponential decay law. Take the ratio of the two rates and use logarithms.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The rate of disintegration is proportional to the number of nuclei, so \(R = R_0 e^{-\lambda t}\).

Step 2: Substitution
\(R_0 = 9000\), \(R = 3000\), \(t = 2\text{ min}\).
\(\dfrac{R_0}{R} = e^{\lambda t}\), so \(3 = e^{2\lambda}\).

Step 3: Calculation
Taking natural logarithms, \(2\lambda = \log_e 3\).
\(\lambda = 0.5\log_e 3\text{ per minute}\).

Final Answer:
The decay constant is \(0.5\log_e 3\) per minute. \[ \boxed{\text{(A) }0.5\log_{e}3} \]