
To determine the effective half-life of a radioactive nucleus decaying through two processes, the concept of parallel decay processes is used. When a nucleus decays via two simultaneous processes with half-lives T_{1/2,1} and T_{1/2,2}, the effective decay constant \lambda_{eff} is the sum of the individual decay constants:
\lambda_{eff} = \lambda_1 + \lambda_2
where \lambda_1 and \lambda_2 are the decay constants for each process, given by:
\lambda_1 = \frac{\ln(2)}{T_{1/2,1}} and \lambda_2 = \frac{\ln(2)}{T_{1/2,2}}
The effective half-life T_{1/2,eff} can then be calculated using:
T_{1/2,eff} = \frac{\ln(2)}{\lambda_{eff}}
Given:
First, find \lambda_1 and \lambda_2:
\lambda_1 = \frac{\ln(2)}{3.0} \approx 0.231 \text{hour}^{-1}
\lambda_2 = \frac{\ln(2)}{4.5} \approx 0.154 \text{hour}^{-1}
Add the decay constants to find \lambda_{eff}:
\lambda_{eff} = 0.231 + 0.154 = 0.385 \text{hour}^{-1}
Now, calculate the effective half-life T_{1/2,eff}:
T_{1/2,eff} = \frac{\ln(2)}{0.385} \approx 1.80 hours
Therefore, the effective half-life of the nucleus is 1.80 hours.
The electric potential at the surface of an atomic nucleus \( (z = 50) \) of radius \( 9 \times 10^{-13} \) cm is \(\_\_\_\_\_\_\_ \)\(\times 10^{6} V\).
In a nuclear fission reaction of an isotope of mass \( M \), three similar daughter nuclei of the same mass are formed. The speed of a daughter nuclei in terms of mass defect \( \Delta M \) will be: