Question

In an alpha decay, the daughter nucleus produced is itself unstable and undergoes further decay. If the number of parent and daughter nuclei at time \(t\) are \(N_{p}\) and \(N_{d}\) respectively and their decay constants are \(\lambda _{p}\) and \(\lambda _{d}\) respectively. What condition needs to be satisfied in order for \(N_{d}\) to remain constant?

• A. \(N_{p}N_{d}=\lambda _{d}\lambda _{p}\)
• B. \((N_{p})/(N_{d})=(\lambda _{p})/(\lambda _{d})\)
• C. \(2N_{p}\lambda _{d}=3N_{d}\lambda _{p}\)
• D. \(N_{p}\lambda _{p}=N_{d}\lambda _{d}\)

Hint:

The daughter is formed by parent decay and removed by its own decay; constant means these two rates are equal.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Daughter nuclei are produced at the rate at which parent nuclei decay and are lost at the rate at which daughter nuclei decay.

Step 2: Write the rate equation
\(\dfrac{dN_d}{dt}=\lambda_pN_p-\lambda_dN_d\)

Step 3: Condition for constant \(N_d\)
\(N_d\) stays constant when \(\dfrac{dN_d}{dt}=0\), so \(\lambda_pN_p=\lambda_dN_d\).

Step 4: Compare the options
This is option (D): the activities of parent and daughter are equal.

Final Answer:
The activities must be equal, that is Np times lambda p equals Nd times lambda d. \[ \boxed{\text{(D) }N_{p}\lambda_{p}=N_{d}\lambda_{d}} \]