Question

Two radioactive substances \(A\) and \(B\) have decay constant \(5\lambda\) and \(\lambda\) respectively. At \(t=0\), they have the same number of nuclei. The ratio of number of nuclei of \(A\) to those of \(B\) will be \(\left(\frac{1}{e}\right)^{2}\) after a time interval

• A. \(\frac{1}{4\lambda }\)
• B. \(4\lambda\)
• C. \(2\lambda\)
• D. \(\frac{1}{2\lambda }\)

Hint:

Use N = N0 e^(-lambda t) for each substance and set the ratio equal to e^(-2).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Radioactive decay follows \(N=N_0e^{-\lambda t}\).
Both substances start with the same \(N_0\).

Step 2: Write the ratio
\(N_A=N_0e^{-5\lambda t}\) and \(N_B=N_0e^{-\lambda t}\), so
\(\dfrac{N_A}{N_B}=e^{-5\lambda t+\lambda t}=e^{-4\lambda t}\)

Step 3: Calculation
The ratio is given as \(\left(\dfrac{1}{e}\right)^{2}=e^{-2}\), so \(4\lambda t=2\).

Step 4: Compare the options
\(t=\dfrac{1}{2\lambda}\), which is option (D).

Final Answer:
The required time is 1/(2 lambda). \[ \boxed{\text{(D) }\dfrac{1}{2\lambda}} \]