(See Problem 21.) Among the many fission products that may be extracted chemically from the spent fuel of a nuclear reactor is ${}^{\mathbf{90}}\mathbf{Sr}\left(T_{1/2}=29y\right)\mathbf{}$. This isotope is produced in typical large reactors at the rate of about 18 kg/y. By its radioactivity, the isotope generates thermal energy at the rate of 0.93 W/g. (a) Calculate the effective disintegration energy ${\mathbf{Q}}_{\mathbf{eff}}$associated with the decay of a ${}^{\mathbf{90}}\mathbf{Sr}$nucleus. (This energy includes contributions from the decay of the ${}^{\mathbf{90}}\mathbf{Sr}$daughter products in its decay chain but not from neutrinos, which escape totally from the sample.) (b) It is desired to construct a power source generating 150 W (electric power) to use in operating electronic equipment in an underwater acoustic beacon. If the power source is based on the thermal energy generated by 90Sr and if the efficiency of the thermal–electric conversion process is 5.0%, how much${}^{\mathbf{90}}\mathbf{Sr}$is needed?

The half life of the ${}^{90}\mathrm{Sr}$nucleus is

$$\begin{gathered} T_{1/2}=\left(29y\right)\left(3.156\times {10}^7\mathrm{s}/\mathrm{y}\right) \\ =91.524\times {10}^7\mathrm{s} \end{gathered}$$

The mass of the angle ${}^{90}\mathrm{Sr}$nucleus is

$$\begin{gathered} m=\left(90\mathrm{u}\right)\left(1.661\times {10}^{-27}\mathrm{kg}/\mathrm{u}\right) \\ =149.49\times {10}^{-27}\mathrm{kg} \end{gathered}$$

The total mass of the material is

$$\begin{gathered} m=1.00\mathrm{g} \\ =\left(1.00\mathrm{g}\right)\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right) \\ =1.00\times {10}^{-3}\mathrm{kg} \end{gathered}$$

The power output p = 0.93 w

The power output can be expressed as

$$\begin{gathered} p=R{Q}_{eff} \\ {Q}_{eff}=\frac{P}{R}.....\left(1\right) \end{gathered}$$

Here, the delay rate is R

The relation between decay rate and the disintegration constant is

$$\begin{gathered} R=N\lambda \\ =\left(\frac{M}{m}\right)\frac{\mathrm{In}\left(2\right)}{{\mathrm{T}}_{1/2}} \end{gathered}$$

Substitute the given data in equation (2), we get

$Q=\frac{pm{T}_{1/2}}{M\mathrm{In}\left(2\right)}$

Substitute this equation in equation (1), we get

$$\begin{gathered} \mathrm{Q}=\frac{\left(0.93\mathrm{w}\right)\left(149.49\times {10}^{-27}\mathrm{kg}\right)\left(91.524\times {10}^7\mathrm{s}\right)}{\left(1.00\times {10}^{-3}\mathrm{kg}\right)\mathrm{In}\left(2\right)} \\ =1.84\times {10}^{-13}\mathrm{J} \\ =1.84\times {10}^{-13}\mathrm{J}\left(\frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right) \\ =1.15\times {10}^6\mathrm{eV} \\ =1.15\mathrm{MeV}\approx 1.2\mathrm{MeV} \end{gathered}$$

Hence, the effective disintegration is 1.0 MeV

The rate of thermal energy is 0.93 W/g

The electric power p = 150 W

The efficiency of the thermal-electric

Conversion process is

$$\begin{gathered} n=5.0\% \\ =\left(\frac{5.0}{100}\right) \\ =0.05 \end{gathered}$$

The amount of ${}^{90}\mathrm{Sr}$needed is

$$\begin{gathered} \mathrm{M}=\frac{\left(150\mathrm{W}\right)}{\left(0.05\right)\left(0.93\mathrm{W}/\mathrm{g}\right)} \\ =3225.80\mathrm{g} \\ =3.2\times {10}^3\mathrm{g} \\ =3.2\times {10}^3\mathrm{g}\left(\frac{1\mathrm{kg}}{{10}^3\mathrm{g}}\right) \\ =3.2\mathrm{kg} \end{gathered}$$

Hence, the amount of 3.2 kg is needed.