Question

The isotope ${\mathbf{}}^{\mathbf{235}}\mathbf{U}$decays by alpha emission with a half-life of $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{y}$. It also decays (rarely) by spontaneous fission, and if the alpha decay did not occur, its half-life due to spontaneous fission alone would be $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{17}}\mathbf{y}$.

(a) At what rate do spontaneous fission decays occur in 1.0 g of${\mathbf{}}^{\mathbf{235}}\mathbf{U}$ ?

(b) How many ${\mathbf{}}^{\mathbf{235}}\mathbf{U}$ alpha-decay events are there for every spontaneous fission event?

Short answer

(a) The rate of spontaneous fission decay is 16 fission/day .

(b) The number of alpha decay for spontaneous fission is$4.3\times {10}^8$ .

Step-by-step solution

Identification of given data

• The half-life of alpha emission is $t_{1/2}=7\times {10}^{8}y$.
• The half-life for spontaneous fission is $T=3\times {10}^{17}y$.
• The mass of the uranium 235 is m = 1g

Radioactive decay is the process of decaying of protons from the nucleus and variation in the mass number of the atom. The number of protons increases by one unit due to beta decay and no change in the mass number of the atom, while the number of protons changes by two units and mass number by four units for alpha decay.

Determination of rate of spontaneous fission decay(a)

The rate of spontaneous fission decay is given as:

$R\mathit{=}\frac{\mathit{m}{\mathit{N}}_{\mathit{A}}\mathit{}\mathit{I}\mathit{n}\mathit{2}}{{\mathit{M}}_{\mathit{u}}\mathit{T}}$

Here $N_A$is the Avogadro constant and its value is $6.023\times {10}^{23}\mathrm{mol}/\mathrm{g}$.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{R}=\frac{\left(1\mathrm{g}\right)\left(6.023\times {10}^{23}\mathrm{mol}/\mathrm{g}\right)\mathrm{In}2}{\left(235\mathrm{mol}\right)\left(3\times {10}^{17}\mathrm{y}\right)\left({\displaystyle \frac{365\mathrm{day}}{1\mathrm{y}}}\right)} \\ \mathrm{R}=16.22\mathrm{fission}/\mathrm{day} \\ \mathrm{R}\approx 16\mathrm{fission}/\mathrm{day} \end{gathered}$$

Determination of the number of alpha decay for spontaneous fission(b)

The number of alpha decay for spontaneous fission is calculated as:

$n=\frac{T}{t}$

Substitute all the values in the above equation.

$$\begin{gathered} n=\frac{3\times {10}^{17}y}{7\times {10}^{8}y} \\ n=4.3\times {10}^8 \end{gathered}$$

Therefore, the number of alpha decay for spontaneous fission is $4.3\times {10}^8$.