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 solve this problem, we need to understand the relationship between the decay constant and the number of nuclei remaining after a certain time period. The decay of a radioactive nucleus is described by the equation:

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

where: 

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

Given that the decay constant of nucleus \( n_2 \) is 3 times that of nucleus \( n_1 \), we can write:

\(\lambda_2 = 3\lambda_1\)

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

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

Since we are considering one half-life of nucleus \( n_1 \), we have:

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

After one half-life, the number of nuclei remaining is half of the initial number:

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

For nucleus \( n_2 \), which has been decaying for the same time period:

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

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

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

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

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

Therefore, the correct answer is \(\frac{1}{4}\).

Answer 2

The decay law for radioactive decay is given by the equation: \[ N = N_0 e^{-\lambda t}, \] where \( N \) is the number of nuclei at time \( t \), \( N_0 \) is the initial number of nuclei, and \( \lambda \) is the decay constant. For nucleus \( n_2 \), the decay constant is 3 times that of \( n_1 \). So, the decay constants are: \[ \lambda_2 = 3 \lambda_1. \] The number of nuclei after time \( t \) for both nuclei will be: \[ N_2 = N_0 e^{-\lambda_2 t} = N_0 e^{-3\lambda_1 t}, \] \[ N_1 = N_0 e^{-\lambda_1 t}. \] Now, after one half-life of \( n_1 \), the value of \( t = t_{\text{half}} \) is: \[ t_{\text{half}} = \frac{\ln 2}{\lambda_1}. \] At this time, the number of nuclei for \( n_1 \) is: \[ N_1 = N_0 e^{-\lambda_1 t_{\text{half}}} = \frac{N_0}{2}. \] For \( n_2 \), the number of nuclei after one half-life of \( n_1 \) is: \[ N_2 = N_0 e^{-3\lambda_1 t_{\text{half}}} = N_0 e^{-\frac{3\ln 2}{\lambda_1}} = \frac{N_0}{2^3} = \frac{N_0}{8}. \] Thus, the ratio of the number of nuclei of \( n_2 \) to the number of nuclei of \( n_1 \) is: \[ \frac{N_2}{N_1} = \frac{\frac{N_0}{8}}{\frac{N_0}{2}} = \frac{1}{4}. \] Thus, the correct answer is: \[ \boxed{\frac{1}{4}}. \]