Question

A radioactive nucleus has a decay constant \(\lambda\) and its radioactive daughter nucleus has a decay constant 10\(\lambda\). At time t = 0, N₀ is the number of parent nuclei and there are no daughter nuclei present. N₁(t) and N₂(t) are the number of parent and daughter nuclei present at time t, respectively The ratio \(\frac{N_2(t)}{N_1(t)}\) is

• A. \(\frac{1}{9}[1-e^{-9\lambda t}]\)
• B. \(\frac{1}{10}[1-e^{-10\lambda t}]\)
• C. \([1-e^{-10\lambda t}]\)
• D. \([1-e^{-9\lambda t}]\)

Answer 1

The Correct Option is A

To find the ratio \(\frac{N_2(t)}{N_1(t)}\), let's analyze the decay process of the parent and daughter nuclei. We are given that the decay constant of the parent nucleus is \(\lambda\) and that of the daughter nucleus is \(10\lambda\).

Initially, at \(t = 0\), the number of parent nuclei is \(N_0\), and there are no daughter nuclei, \(N_2(0) = 0\).

The number of parent nuclei at any time \(t\) is given by the exponential decay formula:

\(N_1(t) = N_0 e^{-\lambda t}\)

The daughter nuclei are produced as the parent nuclei decay. The rate of change of the parent and daughter nuclei can be described by these differential equations:

• \(\frac{dN_1}{dt} = -\lambda N_1\)
• \(\frac{dN_2}{dt} = \lambda N_1 - 10\lambda N_2\)

The solution to the first equation is the standard decay formula we have stated above. For the daughter nuclei, the solution involves solving the differential equation. The solution follows this formula:

\(N_2(t) = \frac{\lambda N_0}{10\lambda - \lambda} \left( e^{-\lambda t} - e^{-10\lambda t} \right)\)

Simplifying the expression, we have:

\(N_2(t) = \frac{N_0}{9} \left( e^{-\lambda t} - e^{-10\lambda t} \right)\)

The ratio of daughter to parent nuclei becomes:

\(\frac{N_2(t)}{N_1(t)} = \frac{\frac{N_0}{9} ( e^{-\lambda t} - e^{-10\lambda t})}{N_0 e^{-\lambda t}} = \frac{1}{9} [1 - e^{-9\lambda t}]\)

Thus, the correct answer is: \(\frac{1}{9}[1-e^{-9\lambda t}]\)