Question

A radioactive nucleus \( n_2 \) has 3 times the decay constant as compared to the decay constant of another radioactive nucleus \( n_1 \). If the initial number of both nuclei are the same, what is the ratio of the number of nuclei of \( n_2 \) to the number of nuclei of \( n_1 \), after one half-life of \( n_1 \)?

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{8} \)
• C. 4
• D. 8

Hint:

For radioactive decay, remember the equation \( N = N_0 e^{-\lambda t} \) and use half-life relations to solve for the number of nuclei after a given time.

Answer 1

The Correct Option is A

To address this issue, the relationship between the decay constant and the quantity of nuclei remaining over time must be understood. Radioactive nuclear decay follows the equation:

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

where:

• \(N(t)\) represents the number of nuclei at time \(t\).
• \(N_0\) is the initial quantity of nuclei.
• \(\lambda\) is the decay constant.
• \(t\) is the elapsed time.

Given that the decay constant for nucleus \( n_2 \) is thrice that of nucleus \( n_1 \), it follows that:

\(\lambda_2 = 3\lambda_1\)

The half-life \((T_{1/2})\) of a radioactive sample is linked to its decay constant by the formula:

\(T_{1/2} = \frac{\ln 2}{\lambda}\)

Considering one half-life of nucleus \( n_1 \):

\(T_{1/2,1} = \frac{\ln 2}{\lambda_1}\)

After one half-life, half of the initial nuclei remain:

\(N_1(T_{1/2,1}) = \frac{N_0}{2}\)

For nucleus \( n_2 \), after the same duration:

\(N_2(T_{1/2,1}) = N_0 e^{-\lambda_2 T_{1/2,1}}\)

Substituting \( \lambda_2 = 3\lambda_1 \) and \( T_{1/2,1} = \frac{\ln 2}{\lambda_1} \):

\(N_2(T_{1/2,1}) = N_0 e^{-3\lambda_1 \cdot \frac{\ln 2}{\lambda_1}} = N_0 e^{-3\ln 2} = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}\)

The ratio of the number of \( n_2 \) nuclei to \( n_1 \) nuclei after one half-life of \( n_1 \) is thus:

\(\frac{N_2(T_{1/2,1})}{N_1(T_{1/2,1})} = \frac{\frac{N_0}{8}}{\frac{N_0}{2}} = \frac{1}{4}\)

The correct answer is \(\frac{1}{4}\).