Question

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as : (where $\lambda$ is the decay constant)

• A. $\frac{\ln(3/2)}{\lambda}$
• B. $\frac{1}{2} \frac{\ln 2}{\lambda}$
• C. $\frac{2 \ln 2}{\lambda}$
• D. $\frac{\ln 2}{\lambda}$

Hint:

Remember that the decay equation relates the amount remaining. Always subtract the decayed amount from the initial amount before using the formula.
Use log properties: \(\ln a - \ln b = \ln(a/b)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Radioactive decay follows the law \(N = N_0 e^{-\lambda t}\), where \(N\) is the number of nuclei remaining at time \(t\).
If a certain fraction has decayed, the remaining fraction is \(1 - \text{decayed fraction}\).
Step 2: Key Formula or Approach:
Time taken to reach \(N\) nuclei from \(N_0\) is: \[t = \frac{1}{\lambda} \ln\left( \frac{N_0}{N} \right)\]
Step 3: Detailed Explanation:
Instance 1: A quarter (\(1/4\)) of the nuclei have decayed.
Remaining nuclei \(N_1 = N_0 - \frac{1}{4}N_0 = \frac{3}{4}N_0\).
Time \(t_1 = \frac{1}{\lambda} \ln\left( \frac{N_0}{3/4 N_0} \right) = \frac{1}{\lambda} \ln\left( \frac{4}{3} \right)\).

Instance 2: Half (\(1/2\)) of the nuclei have decayed.
Remaining nuclei \(N_2 = N_0 - \frac{1}{2}N_0 = \frac{1}{2}N_0\).
Time \(t_2 = \frac{1}{\lambda} \ln\left( \frac{N_0}{1/2 N_0} \right) = \frac{1}{\lambda} \ln(2)\).

The time gap \(\Delta t\) is:
\[ \Delta t = t_2 - t_1 = \frac{1}{\lambda} \left[ \ln 2 - \ln\left( \frac{4}{3} \right) \right] \]
\[ \Delta t = \frac{1}{\lambda} \ln\left( \frac{2}{4/3} \right) = \frac{1}{\lambda} \ln\left( \frac{2 \times 3}{4} \right) = \frac{1}{\lambda} \ln\left( \frac{6}{4} \right) = \frac{\ln(3/2)}{\lambda} \]
Step 4: Final Answer:
The time gap is \(\frac{\ln(3/2)}{\lambda}\).