Question

At \( t = 0 \), \( N_0 \) number of radioactive nuclei \( A \) start decaying into \( B \) with a decay constant \( \lambda_a \). The daughter nuclei \( B \) decay into nuclei \( C \) with a decay constant \( \lambda_b \). Then, the number of nuclei \( B \) at small time \( t \) (to the leading order) is:

• A. \( \lambda_a N_0 t \)
• B. \( (\lambda_a - \lambda_b)N_0 t \)
• C. \( (\lambda_a + \lambda_b)N_0 t \)
• D. \( \lambda_b N_0 t \)

Hint:

For small \( t \), ignore secondary decays when solving chain decay equations. The buildup of daughter nuclei depends only on the parent’s initial decay rate.

Answer 1

The Correct Option is A

Step 1: Define the decay processes.
At \( t = 0 \), only nuclei \( A \) are present. The rate of decay of \( A \) is given by \[ \frac{dN_A}{dt} = -\lambda_a N_A. \] Hence, at small \( t \), \[ N_A \approx N_0(1 - \lambda_a t). \] Step 2: Formation of nuclei \( B \).
Nuclei \( B \) are formed from \( A \) and decay with constant \( \lambda_b \): \[ \frac{dN_B}{dt} = \lambda_a N_A - \lambda_b N_B. \] At very small \( t \), \( N_B \) is initially negligible, so \( \lambda_b N_B \) can be ignored. Thus, \[ \frac{dN_B}{dt} \approx \lambda_a N_0. \] Step 3: Integrate for small \( t \).
\[ N_B \approx \lambda_a N_0 t. \] Step 4: Final Answer.
Therefore, \( N_B = \lambda_a N_0 t. \)