Question

What do you mean by Nuclear fission ? In the fission of \(U^{235}\) nucleus, 200 MeV energy is produced. Power of 4 MW is obtained by a reactor. How many nuclei are fissioned per sec in the reactor ?

Hint:

In nuclear physics calculations, unit conversion is critical. Always convert energy from MeV to Joules and power from MW (or GW) to Watts before applying the formulas. Remember the relationship: Power = (Number of events per second) \(\times\) (Energy per event).

Answer 1

1. Nuclear Fission:
Nuclear fission is a nuclear reaction in which the nucleus of a heavy, unstable atom (such as Uranium-235 or Plutonium-239) absorbs a neutron and splits into two or more smaller, lighter nuclei (fission fragments). This process also releases a few neutrons and a very large amount of energy. The released neutrons can then go on to cause further fissions, leading to a self-sustaining chain reaction if controlled, as in a nuclear reactor.
2. Calculation of Fissions per Second:

Step 1: List the Given Data and Convert to SI Units:
\begin{itemize} \item Energy released per fission, \(E = 200\) MeV. \item Power of the reactor, \(P = 4\) MW. \end{itemize} We need to convert these to Joules and Watts (Joules/second) respectively. \begin{itemize} \item \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \implies 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}\) \[ E = 200 \times 10^6 \text{ eV} = 200 \times 1.6 \times 10^{-13} \text{ J} = 3.2 \times 10^{-11} \text{ J} \] \item \(1 \text{ MW} = 10^6 \text{ W}\) \[ P = 4 \times 10^6 \text{ W} = 4 \times 10^6 \text{ J/s} \] \end{itemize}

Step 2: Relate Power, Energy, and Number of Fissions:
Power is the rate at which energy is produced. If \(N\) is the number of fissions occurring per second, then the total energy produced per second (Power) is the product of \(N\) and the energy per fission (\(E\)). \[ P = N \times E \] We need to find \(N\), the number of fissions per second. \[ N = \frac{P}{E} \]

Step 3: Calculate the Value of N:
Substitute the values in SI units into the equation: \[ N = \frac{4 \times 10^6 \text{ J/s}}{3.2 \times 10^{-11} \text{ J/fission}} \] \[ N = \frac{4}{3.2} \times 10^{6 - (-11)} \text{ fissions/s} \] \[ N = \frac{40}{32} \times 10^{17} \text{ fissions/s} = \frac{5}{4} \times 10^{17} \text{ fissions/s} \] \[ N = 1.25 \times 10^{17} \text{ fissions/s} \]

Step 4: Final Answer:
The number of nuclei fissioned per second in the reactor is \(1.25 \times 10^{17}\).