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Nuclear fission is a fundamental process where the nucleus of a heavy atom splits into two or more lighter nuclei, releasing a vast amount of energy. This phenomenon is crucial not only in understanding atomic physics but also in practical applications such as nuclear power generation. The core idea behind harnessing this energy efficiently lies in what is called a chain reaction, where neutrons released from one fission event induce further fission events, sustaining a self-amplifying cycle.
For students preparing for competitive exams, understanding nuclear fission and chain reactions is vital due to its frequent appearance in questions related to energy production, radioactivity, and nuclear reactors. This topic forms a bridge between theoretical nuclear physics and its real-world applications, including safety controls and economic aspects in India's growing nuclear energy sector.
Let's start by understanding the nuclear fission process. Certain heavy nuclei, such as Uranium-235 (\(^{235}\)U), can absorb a slow-moving neutron. This absorption makes the nucleus unstable, causing it to split into two smaller nuclei (called fission fragments) while releasing:
The emitted neutrons can then strike other \(^{235}\)U nuclei, causing more fission reactions. This ongoing series of reactions initiating more reactions is called a chain reaction.
This diagram illustrates one neutron triggering the fission of one \(^{235}\)U nucleus, which splits into two fragments, releases energy, and emits three neutrons that continue the process. Over time, this can yield a large amount of energy from a small amount of material.
A key parameter in understanding chain reactions is the neutron multiplication factor, denoted as \(k\). It is defined as the ratio of the number of neutrons in one generation to those in the previous generation:
Based on the value of \(k\), the chain reaction behaves differently:
| Value of \(k\) | Reaction Type | Outcome |
|---|---|---|
| \(k < 1\) | Subcritical | Neutron population decreases; chain reaction dies out. |
| \(k = 1\) | Critical | Neutron population remains steady; chain reaction is self-sustaining. |
| \(k > 1\) | Supercritical | Neutron population increases exponentially; chain reaction grows rapidly. |
In a controlled nuclear reactor, the aim is to maintain \(k = 1\), so the reaction produces steady energy output without exploding or dying out. In nuclear weapons, \(k > 1\) leads to a rapid, uncontrolled energy release.
Step 1: Convert mass defect from atomic mass units to kilograms.
\(\Delta m = 0.2 \, \text{u} = 0.2 \times 1.66 \times 10^{-27} \, \text{kg} = 3.32 \times 10^{-28} \, \text{kg}\)
Step 2: Use Einstein's mass-energy equivalence formula:
\[ E = \Delta m \times c^2 = 3.32 \times 10^{-28} \times (3 \times 10^{8})^2 \, \text{J} \]
Step 3: Calculate energy:
\[ E = 3.32 \times 10^{-28} \times 9 \times 10^{16} = 2.99 \times 10^{-11} \, \text{J} \]
Answer: One fission event of Uranium-235 releases approximately \(3.0 \times 10^{-11}\) Joules of energy.
Step 1: Use the relation \(A = \lambda N\) to find \(\lambda\):
\[ \lambda = \frac{A}{N} = \frac{1 \times 10^{6}}{1 \times 10^{20}} = 1 \times 10^{-14} \, s^{-1} \]
Step 2: Use \(\lambda = \frac{\ln 2}{T_{1/2}}\) to find half-life \(T_{1/2}\):
\[ T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1 \times 10^{-14}} = 6.93 \times 10^{13} \, s \]
Step 3: Convert seconds into years for clarity:
\[ 6.93 \times 10^{13} \, s \times \frac{1\, \text{year}}{3.15 \times 10^{7} \, s} \approx 2.2 \times 10^{6} \, \text{years} \]
Answer: The half-life of the radioactive sample is approximately 2.2 million years.
Step 1: Find the number of neutrons in the next generation:
Neutrons surviving: \(4000 - 1000 = 3000\)
Step 2: Calculate \(k\):
\[ k = \frac{\text{Neutrons in generation } (n+1)}{\text{Neutrons in generation } n} = \frac{3000}{4000} = 0.75 \]
Step 3: Interpret \(k\): \(k = 0.75 < 1\)
This means the chain reaction is subcritical and will gradually die out unless corrected.
Answer: The neutron multiplication factor \(k = 0.75\); the reaction is subcritical and not self-sustaining.
Step 1: Convert energy per fission from MeV to Joules:
\[ 200 \, \text{MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-11} \, \text{J} \]
Step 2: Calculate the number of fissions per second needed to produce 3000 MW:
\[ \text{Power} = 3000 \times 10^6 \, \text{W} = 3 \times 10^9 \, \text{J/s} \]
\[ \text{Number of fissions per second} = \frac{\text{Power}}{\text{Energy per fission}} = \frac{3 \times 10^9}{3.2 \times 10^{-11}} \approx 9.375 \times 10^{19} \]
Step 3: Calculate energy in kWh generated per day:
\[ \text{Energy per day} = 3000\, \text{MW} \times 24\, \text{hours} = 72000\, \text{MWh} = 7.2 \times 10^7\, \text{kWh} \]
Step 4: Calculate cost saving at Rs.5 per kWh:
\[ \text{Cost saving} = 7.2 \times 10^7 \times 5 = 3.6 \times 10^8 \, \text{Rs. (INR)} \]
Answer: Approximately \(9.4 \times 10^{19}\) atoms of Uranium-235 undergo fission per second. The daily cost saving by using nuclear power instead of coal is about Rs.360 million.
Step 1: Calculate decay constant, \(\lambda\), from half-life.
Convert half-life to seconds:
\[ T_{1/2} = 10 \times 3600 = 36000\, s \]
\[ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{36000} = 1.925 \times 10^{-5} \, s^{-1} \]
Step 2: Calculate activity \(A = \lambda N\):
\[ A = 1.925 \times 10^{-5} \times 5 \times 10^{18} = 9.625 \times 10^{13} \, \text{Bq} \]
Answer: The activity of the sample is approximately \(9.6 \times 10^{13}\) becquerels.
k = 1 means a stable chain reaction.Progress tracking is paywalled — subscribe to mark subtopics as understood and save your streak.
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