Question

The rate of disintegration of a radioactive element at any time is proportional to its mass at that time, where \(k\) \((k > 0)\) is the constant of proportionality. The time during which an original mass of 1.5 gm will disintegrate to a mass of 0.5 gm is \(\ldots\)

• A. \(klog3\)
• B. \(\frac{1}{k}log3\)
• C. \(klog5\)
• D. \(\frac{1}{k}log5\)

Hint:

Solve dm/dt = -km to get m = m0 e^(-kt), then set m = m0/3.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The rate of disintegration is proportional to the mass present, and the mass decreases with time. So the differential equation has a negative sign.

Step 2: Key Formula or Approach:
\[ \frac{dm}{dt} = -km \Rightarrow \frac{dm}{m} = -k\,dt \Rightarrow \log m = -kt + C \]

Step 3: Detailed Explanation:
At \(t = 0\), \(m = 1.5\) g, so \(C = \log 1.5\). Then
\[ \log\frac{m}{1.5} = -kt \]
We want the time when \(m = 0.5\) g:
\[ \log\frac{0.5}{1.5} = -kt \Rightarrow \log\frac13 = -kt \Rightarrow -\log 3 = -kt \]
\[ t = \frac{1}{k}\log 3 \]
Option (A) has \(k\) multiplied instead of divided. Options (C) and (D) involve \(\log 5\), which would be correct for a drop to one fifth of the mass.

Final Answer:
The time required is \(\dfrac{1}{k}\log 3\), option (B). \[ \boxed{\frac{1}{k}\log 3 \text{ (B)}} \]