Question

(a) To release 100 MW of power, approximately how many uranium fissions must occur every second?

(b) How many kilograms of U-235 would have to fission in 1 year?

Short answer

(a)$N_Y=9.859\times {10}^{25}\text{decays/year}$

(b)$M=38.5kg$

Step-by-step solution

Given data

Given that the power released by the fission reaction is :

$P=100MW=100\times {10}^{6}W$

Then energy is:

$Q=200MeV=200\times 1.6\times {10}^{-13}J=320\times {10}^{-13}J$

Find the Number of uranium-235 fissions that will occur

(a)

The Number of fission required for the release of 100 MW of energy is:

$$\begin{gathered} N=\frac{P}{Q} \\ =\frac{{10}^{8}W}{320\times {10}^{-13}J} \\ =3.12\times {10}^{18}\text{decays/second} \end{gathered}$$

Let's calculate the Number of fission that would happen in a year,

$$\begin{gathered} N_y=N\times 3.16\times {10}^7 \\ =9.859\times {10}^{25}\text{decays/year} \end{gathered}$$

Therefore, the Number of fission will be ${\text{N}}_{\text{Y}}{\text{= 9.859×10}}^{\text{25}}\text{decays/year}$.

Finding the Mass of uranium-235.

Mass of uranium-235 that would be decayed to release the power of 100 MeV for a year is:

$\begin{array}{rcl}M& =& N_Y\times m_{U{r}_{235}}\\ & =& 9.89\times {10}^{25}\times 235u\times 1.66\times {10}^{-27}kg/u\\ & =& 38.5kg\end{array}$

Thus, the Mass of the uranium is$38.5kg$ .