Question:

A nuclear reactor contains uranium-235 as fuel and operates at a power output of 1 GW. The primary nuclear reaction that occurs in the reactor is the fission of uranium-235 into barium-144 and krypton-89, with the release of 3 neutrons and a total energy of 200 MeV. Assume that each reactor is 35% efficient at converting the released energy into electrical energy.

a) Calculate the rate at which energy is released in the reactor.

b) Determine the rate at which the reactor converts the released energy into electrical energy.

c) Find the number of uranium-235 atoms undergoing fission per second in the reactor.

d) If the decay constant for a radioactive nucleus is $3.5 \times 10^3$ s$^{-1}$, calculate the average lifetime of the uranium-235 nucleus in the reactor.

e) The reactor needs a critical mass of uranium-235 to sustain the fission chain reaction. If each uranium-235 nucleus fissions into 2 nuclei on average, calculate the mass of uranium-235 needed in the reactor.

Answer:

a) The rate at which energy is released in the reactor can be calculated using the power output formula:

$$\text{{Power}} = \frac{{\text{{Energy released}}}}{{\text{{Time}}}}$$

Given that the power output is 1 GW and the total energy released is 200 MeV, we need to convert the energy and power units:

$$\text{{1 GW}} = 10^9 \text{{ W}}$$

$$\text{{200 MeV}} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{{ J}}$$

Now, we can solve for the time:

$$\text{{Time}} = \frac{{\text{{Energy released}}}}{{\text{{Power}}}} = \frac{{200 \times 10^6 \times 1.6 \times 10^{-19} \text{{ J}}}}{{10^9 \text{{ W}}}}$$

$$\text{{Rate of energy release}} = \frac{1}{{\text{{Time}}}}$$

Substituting the values, we get:

$$\text{{Rate of energy release}} = \frac{10^9}{200 \times 10^6 \times 1.6 \times 10^{-19}} = \frac{10^9}{3.2 \times 10^7} = 31.25 \times 10^1 \text{{ J/s}}$$

Therefore, the rate at which energy is released in the reactor is 31.25 × 10¹ J/s.

b) The rate at which the reactor converts the released energy into electrical energy can be found using the given efficiency of 35%. We can calculate the rate of energy conversion as:

$$\text{{Rate of energy conversion}} = \text{{Rate of energy release}} \times \text{{Efficiency}}$$

Substituting the values, we get:

$$\text{{Rate of energy conversion}} = 31.25 \times 10^1 \text{{ J/s}} \times 0.35 = 10.9375 \times 10^1 \text{{ J/s}}$$

Therefore, the rate at which the reactor converts the released energy into electrical energy is 10.9375 × 10¹ J/s.

c) To find the number of uranium-235 atoms undergoing fission per second in the reactor, we need to relate the rate of energy release to the average energy released per fission:

$$\text{{Number of fissions per second}} = \frac{{\text{{Rate of energy release}}}}{{\text{{Energy released per fission}}}}$$

Given that each fission releases a total energy of 200 MeV and an average of 3 neutrons, we can calculate the energy released per fission:

$$\text{{Energy released per fission}} = \frac{{200 \times 10^6 \times 1.6 \times 10^{-19}}}{{3}}$$

Converting the energy unit:

$$\text{{Energy released per fission}} = \frac{{200 \times 10^6 \times 1.6 \times 10^{-19}}}{{3}} = \frac{{320 \times 10^6 \times 10^{-19}}}{{3}} = \frac{{320 \times 10^{-13}}}{{3}}$$

Now, substituting the values, we get:

$$\text{{Number of fissions per second}} = \frac{{31.25 \times 10^1 \text{{ J/s}}}}{{\frac{{320 \times 10^{-13}}}{{3}}}}$$

Simplifying further, we have:

$$\text{{Number of fissions per second}} = \frac{{31.25 \times 10^1}}{{\frac{{320 \times 10^{-13}}}{{3}}}} = \frac{{31.25 \times 10^1 \times 3}}{{320 \times 10^{-13}}} \times \frac{{10^{13}}}{{10^{13}}} = 29.296875 \times 10^{13} \text{{ fissions/s}}$$

Therefore, the number of uranium-235 atoms undergoing fission per second in the reactor is approximately 29.296875 × 10¹³ fissions/s.

d) The average lifetime of the uranium-235 nucleus in the reactor can be calculated using the relation between decay constant and half-life:

$$\text{{Decay constant}} = \frac{{0.693}}{{\text{{Average lifetime}}}}$$

Rearranging the equation, we have:

$$\text{{Average lifetime}} = \frac{{0.693}}{{\text{{Decay constant}}}}$$

Given the decay constant of $3.5 \times 10^3$ s⁻¹, we substitute this into the equation:

$$\text{{Average lifetime}} = \frac{{0.693}}{{3.5 \times 10^3}}$$

$$\text{{Average lifetime}} = \frac{{0.693}}{{3.5 \times 10^3}} = \frac{{0.693 \times 10^{-1}}}{{3.5}} = \frac{{69.3}}{{3.5}}$$

Calculating the average lifetime, we get:

$$\text{{Average lifetime}} = \frac{{69.3}}{{3.5}} = 19.8 \text{{ s}}$$

Therefore, the average lifetime of the uranium-235 nucleus in the reactor is approximately 19.8 seconds.

e) In order to sustain a fission chain reaction, the reactor needs a critical mass of uranium-235. The mass of the uranium-235 needed can be calculated using the relation:

$$\text{{Critical mass}} = \frac{{\text{{Mass per fission}} \times \text{{Number of fissions per second}}}}{{\text{{Average lifetime}}}}$$

Given that each uranium-235 nucleus fissions into 2 nuclei on average, we can calculate the mass per fission:

$$\text{{Mass per fission}} = \frac{{\text{{mass of uranium-235}} \times \text{{percentage of uranium-235 in the reactor}}}}{{\text{{number of atoms per mole}}}}$$

Substituting the given values:

$$\text{{mass per fission}} = \frac{{\text{{mass of uranium-235}} \times 100\%}}{{6.022 \times 10^{23}}}$$

To find the mass of uranium-235 needed in the reactor, we assume a critical mass of the fuel:

$$\text{{mass of uranium-235 needed}} = \text{{critical mass}}$$

Substituting the values into the equation:

$$\text{{mass of uranium-235 needed}} = \frac{{\text{{mass of uranium-235}} \times 100\%}}{{6.022 \times 10^{23}}} \times \text{{Number of fissions per second}} \times \text{{Average lifetime}}$$

Simplifying further:

$$\text{{mass of uranium-235 needed}} = \frac{{\text{{mass of uranium-235}}}}{{6.022 \times 10^{23}}} \times 100\% \times \text{{Number of fissions per second}} \times \text{{Average lifetime}}$$

Substituting the given values:

$$\text{{mass of uranium-235 needed}} = \frac{{\text{{mass of uranium-235}}}}{{6.022 \times 10^{23}}} \times 100\% \times 29.296875 \times 10^{13} \times 19.8$$

Now, we can calculate the mass of uranium-235 needed:

$$\text{{mass of uranium-235 needed}} = \frac{{\text{{mass of uranium-235}} \times 100 \times 29.296875 \times 19.8}}{{6.022 \times 10^{23}}}$$

$$\text{{mass of uranium-235 needed}} = \frac{{297.85625 \times \text{{mass of uranium-235}}}}{{6.022 \times 10^{23}}}$$

Given that the reactor needs a critical mass, we can deduce that the mass of uranium-235 needed is equal to the critical mass:

$$\text{{Critical mass}} = \frac{{297.85625 \times \text{{mass of uranium-235}}}}{{6.022 \times 10^{23}}}$$

Now, we can solve for the mass of uranium-235 needed:

$$\text{{mass of uranium-235 needed}} = \frac{{\text{{Critical mass}} \times 6.022 \times 10^{23}}}{{297.85625}}$$

Substituting the given critical mass:

$$\text{{mass of uranium-235 needed}} = \frac{{\text{{Critical mass}} \times 6.022 \times 10^{23}}}{{297.85625}} = \frac{{\text{{critical mass}} \times 6.022 \times 10^{23}}}{{297.85625}}$$

Therefore, the mass of uranium-235 needed in the reactor is approximately $\text{{mass of uranium-235 needed}} = \frac{{\text{{critical mass}} \times 6.022 \times 10^{23}}}{{297.85625}}$.

Note: The numerical values used here are fictional for illustrative purposes. In practice, actual values would differ based on specific reactor designs and conditions.