Problem 62 - Master Resource Book in JEE Main Physics

Question

Energy released in the fission of a single nucleus is \(200 \mathrm{MeV}\). The fission rate of a \({ }_{92}^{235} \mathrm{U}\) filled reactor operating at a power level of \(5 \mathrm{~W}\) is (a) \(1.56 \times 10^{-10} \mathrm{~s}^{-1}\) (b) \(1.56 \times 10^{11} \mathrm{~s}^{-1}\) (c) \(1.56 \times 10^{-16} \mathrm{~s}^{-1}\) (d) \(1.56 \times 10^{-17} \mathrm{~s}^{-1}\)

Step-by-Step Solution

Answer

The fission rate is \(1.56 \times 10^{11} \text{ s}^{-1}\), option (b).

Step 1: Understanding power and energy concepts

Power is defined as the rate of doing work or transferring energy. It can be expressed using the formula: \[ P = \frac{E}{t} \] where \( P \) is the power in watts, \( E \) is the energy in joules, and \( t \) is the time in seconds. In this exercise, the reactor's power is given as \(5\) watts.

Step 2: Converting MeV to joules

The energy given in the problem is in mega-electron volts (MeV). We need to convert this into joules for further calculations. 1 MeV is equal to \(1.602 \times 10^{-13}\) joules. Thus, \(200\) MeV is \[200 \times 1.602 \times 10^{-13} = 3.204 \times 10^{-11} \text{ joules}\]

Step 3: Calculating the fission rate

The fission rate is the number of fissions occurring per second. We find it by using the relation of power and energy per fission. The reactor generates \(5\) watts of power, which is \(5\) joules per second. To find the rate of fission (\( R \)), we use \[ R = \frac{P}{E_f} \] where \( E_f \) is the energy released per fission. \[ R = \frac{5}{3.204 \times 10^{-11}} \approx 1.56 \times 10^{11} \text{ s}^{-1} \]

Step 4: Selecting the correct option

From the calculation, the fission rate of the reactor operating at \(5\) watts is approximately \(1.56 \times 10^{11} \text{ s}^{-1}\). This matches option (b).

Key Concepts

Energy Conversion

When dealing with nuclear fission, understanding energy conversion is key. In this process, a uranium nucleus splits into smaller nuclei, releasing energy as part of the reaction. This energy is initially measured in a unit called Mega-electron Volts (MeV). Mega-electron Volts are convenient for atomic-scale processes because they represent a very small amount of energy. Yet, to use this energy in real-world applications, like running a power reactor, it needs to be converted into a more widely understood unit, like joules.

The conversion here is essential. We know that 1 MeV equals approximately \(1.602 \times 10^{-13}\) joules. Thus, when 200 MeV is released from a single fission reaction, it converts to \(3.204 \times 10^{-11}\) joules. This conversion is the foundation for calculating how energy released at an atomic level translates into usable power.

Fission Rate Calculation

The fission rate in a nuclear reactor tells us how many fission reactions occur per second. For nuclear reactors, knowing this rate is crucial for understanding the power generation capability.

To calculate the fission rate, we use the relationship between power and energy released per fission. The power level of the reactor, expressed in watts, represents how much energy is produced per second. Mathematically, the fission rate \( R \) is given by the formula:

• \( R = \frac{P}{E_f} \)

where \( P \) is the reactor's power in joules/second and \( E_f \) is the energy released per fission in joules. For example, if a reactor operates at 5 watts of power with each fission releasing \(3.204 \times 10^{-11}\) joules, substituting these values gives \( R = \frac{5}{3.204 \times 10^{-11}} \approx 1.56 \times 10^{11} \, \text{s}^{-1} \). This means that approximately 156 billion fissions occur each second to sustain the reactor's power level.

Unit Conversion in Physics

Unit conversion is a fundamental skill in physics, ensuring that measurements are consistent across calculations. For physics calculations, especially in energy measurements, it is important to convert between different units accurately. In nuclear physics, energies are often given in MeV, which must be converted to joules when calculating macroscopic quantities like power or the fission rate.

Here, the conversion task involves using specific conversion factors. Remember that 1 MeV equals \(1.602 \times 10^{-13}\) joules. Performing such conversions ensure that all parts of an equation use compatible units, which is essential for deriving meaningful results.

• This principle extends beyond energy, touching any physical measurement involving different unit systems.
• Reliable conversions lay the groundwork for accurate and interpretable results.

Mastery of unit conversion allows students and scientists alike to bridge microscopic interactions (like those in nuclear fission) with macroscopic phenomena such as electrical power generation.