Question

The heavy nucleus 252 Cf has two decay modes: Its total half-life T12(252CD 2.64 y. decay (96.9%) and spontaneous fission (31%). 1. What is the ratio between the number of a particles and number of fission fragments created? 2. If252 Cf only decayed by a decay, what would the total half-life be? 3. If252Cfonly underwent spontaneous fission, what would the total half-life be?

Answer 1

Let the decay constant for alpha decay be $\lambda_1$ and for spontaneous fission be, $\lambda_2$ . Then we can write,

The total half life = 2.64 yr, Therefore,

$t_{1/2} = \frac{ln 2}{\lambda_{eq}} \Rightarrow \lambda_{eq} = \frac{0.693}{2.64} = 0.262 \, yr^{-1}$

We can also write,

$\frac{\mathrm{d}N }{\mathrm{d} t} = N_0 e^{-\lambda_{eq} t} = 0.969N_0e^{-\lambda_1t} + 0.031N_0e^{-\lambda_2t}$

$\Rightarrow e^{-\lambda_{eq} } = 0.969e^{-\lambda_1} + 0.031e^{-\lambda_2} \Rightarrow e^{-0.262 } = 0.969e^{-\lambda_1} + 0.031e^{-\lambda_2}$

Since it is equivalent decay constant for the simultaneous process, we can write,

$\Rightarrow \frac{1}{\lambda_{eq}} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}\Rightarrow \frac{1}{0.262} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$

Now, we have to solve these simultaneous equations, then we get,

$\Rightarrow \lambda_1 = 2.97 \, yr^{-1} \,\,\, ; \,\,\, \lambda_2 = 0.287 \, yr^{-1}$

1. Now the ratio of particles is,

$\frac{N_\alpha}{N_{f}} = \frac{0.969N_0e^{-\lambda_1 t}}{0.031N_0e^{-\lambda_2 t}} = \frac{0.969e^{-\lambda_1}}{0.031e^{-\lambda_2}} = 2.14$

2. Half life if only alpha decay is there,

$\Rightarrow t_{1/2} = \frac{0.693}{\lambda_1} = 0.23 \, yr$

3. Half life if it only underwent spontatneous fission, then we get half life as,

$\Rightarrow t_{1/2} = \frac{0.693}{\lambda_2} = 2.41 \, yr$