Question

The thermal energy generated when radiation from radio nuclides is absorbed in matter can serve as the basis for a small power source for use in satellites, remote weather stations, and other isolated locations. Such radio nuclides are manufactured in abundance in nuclear reactors and may be separated chemically from the spent fuel. One suitable radionuclide is 238Pu $\left(T_{1/2}=87.7y\right)$, which is an alpha emitter with Q = 5.50 Me V. At what rate is thermal energy generated in 1.00 kg of this material?

Short answer

The rate of thermal energy generated is 557 w .

Step-by-step solution

Introduction

Thermal energy is the source of the temperature of a body. How hotter or cooler the body will be, depends on the thermal energy.

Find at what rate thermal energy is generated

The total mass of the material,M = 1.00 kg.

The half-life of the 238 pu nuclear can be calculated as:

$$\begin{gathered} {\mathrm{T}}_{\frac{1}{2}}=\left(87.7\mathrm{y}\right)\left(3.156\times {10}^7\mathrm{s}/\mathrm{k}\right) \\ =276.7812\times {10}^7\mathrm{s} \end{gathered}$$

The energy produced by each alpha decay is:

$$\begin{gathered} \mathrm{Q}=5.50\mathrm{Me}\mathrm{V} \\ =\left(5.50\mathrm{Me}\mathrm{V}\right)\left(\frac{{10}^6\mathrm{eV}}{1\mathrm{Me}\mathrm{V}}\right) \\ =5.50\times {10}^6\left(1.6\times {10}^{-19}\mathrm{J}\right) \\ =8.8\times {10}^{-13}\mathrm{J} \end{gathered}$$

The mass of the single 238 pu nucleus can be calculated as:

$$\begin{gathered} m=\left(238u\right)\left(1.661\times {10}^{-27}\mathrm{kg}/\mathrm{u}\right) \\ =393.318\times {10}^{-27}\mathrm{kg} \end{gathered}$$

The power output can be expressed as:

P = RQ ............ (1)

Here, the decay rate is R.

The relation between decay rate and the disintegration constant is:

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

Substitute this equation in, we get,

$p=\left(\frac{MQ}{M{T}_{{\displaystyle \frac{1}{2}}}}\right)\mathrm{In}\left(2\right)$ .............. (2)

Using the equation (2) , the rate of thermal energy generated is calculated as:

$$\begin{gathered} p=\frac{\left(1.00\mathrm{kg}\right)\left(8.8\times {10}^{-13}\mathrm{J}\right)}{\left(395.318\times {10}^{-27}\mathrm{kg}\right)\left(276.7812\times {10}^7\mathrm{s}\right)} \\ =557.4\mathrm{w} \\ \approx 557\mathrm{w} \end{gathered}$$

Therefore, the rate of thermal energy generated is 557 w.