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

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: \[ \boxed{\frac{1}{4}}. \]