A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:

Question

The average energy released per fission for the nucleus of $^{235_{92}U$ is 190 MeV. When all the atoms of 47 g pure $^{235}_{92}U$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ___. (Avogadro Number $= 6 \times 10^{23}$ per mole)}

Answer 1

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:

T_{1/2,1} = 3.0 hours
T_{1/2,2} = 4.5 hours

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.

Answer 2