Question

Two radioactive materials that alpha decay,${}^{\mathbf{238}}\mathit{U}$and ${}^{\mathbf{232}}\mathit{T}\mathit{h}$, and one that beta decays${}^{\mathbf{40}}\mathit{K}$, are sufficiently abundant in granite to contribute significantly to the heating of Earth through the decay energy produced. The alpha-decay isotopes give rise to decay chains that stop when stable lead isotopes are formed. The isotope${}^{\mathbf{40}}\mathit{K}$has single beta decay. (Assume this is the only possible decay of that isotope.) Here is the information:

In the table Qis the totalenergy released in the decay of one parent nucleus to the finalstable endpoint and fis the abundance of the isotope in kilograms per kilogram of granite;means parts per million. (a) Show that these materials produce energy as heat at the rate of$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathit{W}$for each kilogram of granite. (b) Assuming that there is$\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathit{k}\mathit{g}$of granite in a 20-km-thick spherical shell at the surface of Earth, estimate the power of this decay process over all of Earth. Compare this power with the total solar power intercepted by Earth,$\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{17}}\mathit{W}$1.

Short answer

a) The materials like uranium, thorium, and potassium produce energy as heat at the rate of $1.0\times {10}^{-9}W$for each kilogram of granite.

b) The power of the decay process over all of Earth is $2.7\times {10}^{13}W$which is very small compared to what is received from the Sun.

Step-by-step solution

The given data

Mass of granite,$m=2.7\times {10}^{22}kg$

The thickness of the spherical shell,$t=20km\left(\frac{{10}^{3}m}{1km}\right)=2.0\times {10}^{4}m$

Total solar power intercepted by Earth,$P=1.7\times {10}^{17}W$

The table of abundance and decay of all the nuclides present in granite is given.

Understanding the concept of energy transfer in decay

The total heat produced by three radioactive decays can be calculated by the summation value of the product of their rate of decay or disintegrations per second and the heat generated by the nuclear reaction of the individual process. Now, the power generated because of heating through radioactive decay should always be very small than the receiving limit by the Earth to maintain the existence of the planet.

Formulae:

The rate of decay,$R=\left(\frac{\ln 2}{T_{1/2}}\right)N$ (1)

where,$T_{1/2}$is the half-life of the isotope. N is the number of nuclei.

The number of nuclei in a given mass of an atom,

$N=\frac{m}{A}{N}_A,where{N}_A=6.022\times {10}^{23}nuclie/mol$ (2)

where A is the molar mass of the substance.

Calculation of the energy produced as heat

The total rate of energy produced due to the alpha and beta decays of the radioactive nuclides in a given m = 1.00 kg mass of granite can be given using equations (1) and (2) with the given tabular data as follows:

$$\begin{gathered} \frac{\mathrm{dE}}{\mathrm{dt}}=\sum _{\mathrm{i}=1}^3{\mathrm{R}}_{\mathrm{i}}{\mathrm{Q}}_{\mathrm{i}} \\ =\sum _{\mathrm{i}=1}^3\left(\frac{\ln 2}{{\mathrm{T}}_{1/2}}\right)\frac{{\mathrm{m}}_{\mathrm{i}}}{{\mathrm{A}}_{\mathrm{i}}}{\mathrm{N}}_{\mathrm{A}}{\mathrm{Q}}_{\mathrm{i}} \\ =\sum _{\mathrm{i}=1}^3\left(\frac{\ln 2}{{\mathrm{T}}_{1/2}}\right)\frac{{\mathrm{mf}}_{\mathrm{i}}}{{\mathrm{A}}_{\mathrm{i}}}{\mathrm{N}}_{\mathrm{A}}{\mathrm{Q}}_{\mathrm{i}} \end{gathered}$$

(mass of every isotope in granite (f is thier abundance ) ,$m_i=m{f}_i$

$$\begin{gathered} \frac{dE}{dt}=\frac{m\ln 2}{\left(1/N_A\right)}\sum _{i=1}^3\left(\frac{f_i{Q}_i}{A_i{T_{1/2}}_i}\right) \\ =\frac{\left(1.00kg\right)\left(\ln 2\right)\left(1.6\times {10}^{-13}J/MeV\right)}{\left(3.15\times {10}^{7}s/y\right)\left(1.661\times {10}^{-27}kg/u\right)} \\ \left[\frac{\left(4\times {10}^{-6}\right)\left(51.7MeV\right)}{\left(238u\right)\left(4.47\times {10}^{9}y\right)}+\frac{\left(13\times {10}^{-6}\right)\left(42.7MeV\right)}{\left(232u\right)\left(1.41\times {10}^{10}y\right)}+\frac{\left(4\times {10}^6\right)\left(1.31MeV\right)}{\left(40u\right)\left(1.28\times {10}^{9}y\right)}\right] \\ =1.0\times {10}^{-9}W \end{gathered}$$

Hence, the materials like uranium, thorium, and potassium produce energy as heat at the rate of $1.0\times {10}^{-9}W$for each kilogram of granite.

b) Calculation of the power of the decay process

The above part gives the energy produced by one kilogram of granite, thus the power produced by the total weight of granite$m=2.7\times {10}^{22}kg$can be given as follows:

$$\begin{gathered} P=\left(2.7\times {10}^{22}kg\right)\left(1.0\times {10}^{-9}W/kg\right) \\ =2.7\times {10}^{13}W \end{gathered}$$

Hence, the total power of the decay process is $2.7\times {10}^{13}W$which is very small compared to what is received from the Sun.