Fission and Fusion Processes
By Nuclear fission and fusion involve the transformation of the atomic nucleus and the release of large amounts of energy.
Nuclear Fission
Nuclear fission is a process in which the nucleus of an atom is split into two or more smaller nuclei, along with the release of a large amount of energy. This process is typically initiated by bombarding the nucleus with a neutron, which causes it to become unstable and split into two smaller nuclei, along with the release of additional neutrons and energy.
Nuclear fission is used in nuclear power plants to generate electricity, where the energy released by the fission of uranium or plutonium nuclei is used to heat water and generate steam to power turbines.
Nuclear Fusion
Nuclear fusion is a process in which two lighter nuclei are combined to form a heavier nucleus, along with the release of a large amount of energy. This process requires extremely high temperatures and pressures, such as those found in the cores of stars, where hydrogen nuclei are fused to form helium nuclei. Nuclear fusion is the source of the sun's energy.
Problems from IIT JEE
Problem (IIT JEE 2015): A fission reaction is given by $\sideset{_{\ 92}^{236}}{}{U}\to \sideset{_{\ 54}^{140}}{}{Xe} + \sideset{_{38}^{94}}{}{Sr}+x+y$, where $x$ and $y$ are two particles. Considering $\sideset{_{\ 92}^{236}}{}{U}$ to be at rest, the kinetic energies of the products are denoted by $K_\text{Xe}$, $K_\text{Sr}$, $K_\text{x}({2}\;\mathrm{MeV})$ and $K_\text{y}({2}\;\mathrm{MeV})$, respectively. Let the binding energies per nucleon of $\sideset{_{\ 92}^{236}}{}{U}$, $\sideset{_{\ 54}^{140}}{}{Xe}$, and $\sideset{_{38}^{94}}{}{Sr}$ be ${7.5}\;\mathrm{MeV}$, ${8.5}\;\mathrm{MeV}$ and ${8.5}\;\mathrm{MeV}$, respectively. Considering different conservation laws, the correct option(s) is (are)
- $x=n$, $y=n$, $K_\text{Sr}={129}\;\mathrm{MeV}$, $K_\text{Xe}={86}\;\mathrm{MeV}$
- $x=p$, $y=e^\text{-}$, $K_\text{Sr}={129}\;\mathrm{MeV}$, $K_\text{Xe}={86}\;\mathrm{MeV}$
- $x=p$, $y=n$, $K_\text{Sr}={129}\;\mathrm{MeV}$, $K_\text{Xe}={86}\;\mathrm{MeV}$
- $x=n$, $y=n$, $K_\text{Sr}={86}\;\mathrm{MeV}$, $K_\text{Xe}={129}\;\mathrm{MeV}$
Solution: Given fission reaction is, \begin{align} \sideset{_{\ 92}^{236}}{}{U}\to \sideset{_{\ 54}^{140}}{}{Xe} + \sideset{_{38}^{94}}{}{Sr}+\sideset{_{Z_x}^{A_x}}{}{x}+\sideset{_{Z_y}^{A_y}}{}{y}. \end{align} Apply conservation of charge and mass to get, \begin{align} \label{fzb:eqn:2} &92=54+38+Z_x+Z_y \\ \label{fzb:eqn:3} &236=140+94+A_x+A_y. \end{align} These equations gives $Z_x+Z_y=0$ and $A_x+A_y=2$. From the given options, these conditions are satisfied if $\sideset{_{Z_x}^{A_x}}{}{x}=\sideset{_{0}^{1}}{}{n}$ and $\sideset{_{Z_y}^{A_y}}{}{y}=\sideset{_{0}^{1}}{}{n}$.
The Q value of the given reaction is, \begin{align} Q&=\mathrm{BE}_\text{products}-\mathrm{BE}_\text{reactants}\nonumber\\ &=(140\times 8.5+94\times8.5)-236\times7.5 \\ &={219}\;\mathrm{MeV}. \end{align} The energy released in the reaction (Q value) is equal to the kinetic energy of the products i.e., $K_\mathrm{Xe}+K_\mathrm{Sr}+K_\mathrm{x}+K_\mathrm{y}=Q$, which gives, \begin{align} K_\mathrm{Xe}+K_\mathrm{Sr} &=219-(2+2) \\ &={215}\;\mathrm{MeV}.\quad(\because K_\mathrm{x}\!=\!K_\mathrm{y}\!=\!{2}\;\mathrm{MeV}). \end{align} The linear momentum of a particle of mass $m$ and kinetic energy $K$ is given by $p=\sqrt{2mK}$. Since masses and kinetic energies of $x$ and $y$ are very small in comparison to that of $\sideset{_{\ 54}^{140}}{}{Xe}$ and $\sideset{_{38}^{94}}{}{Sr}$, we can neglect the linear momentum of these particles. Initially, the linear momentum of the $\sideset{_{\ 92}^{236}}{}{U}$ is zero (at rest). Finally, the products $\sideset{_{\ 54}^{140}}{}{Xe}$ and $\sideset{_{38}^{94}}{}{Sr}$ will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus, \begin{align} \sqrt{2M_\mathrm{X_e} K_\mathrm{Xe}} &=\sqrt{2M_\mathrm{Sr} K_\mathrm{Sr}},\quad\text{i.e.,}\\ 140 K_\mathrm{Xe}&=94 K_\mathrm{Sr}. \end{align} Solve above equations to get $K_\mathrm{Xe}={86}\;\mathrm{MeV}$ and $K_\mathrm{Sr}={129}\;\mathrm{MeV}$.
Problem (IIT JEE 1993): A star initially has $10^{40}$ deuterons. It produces energy via the processes ${}_1\!\text{H}^2 +{}_1\!\text{H}^2\rightarrow {}_1\!\text{H}^3 +p$ and ${}_1\!\text{H}^2 +{}_1\!\text{H}^3\rightarrow {}_2\!\text{He}^4 +n$. If the average power radiated by the star is ${10^{16}}\;\mathrm{W}$, the deuteron supply of the star is exhausted in a time of the order of, (The nuclei masses are: $m(\text{H}^2)={2.014}\;\mathrm{u}$; $m(\text{n})={1.008}\;\mathrm{u}$; $m(\text{p})={1.007}\;\mathrm{u}$; $m(\text{He}^4)={4.001}\;\mathrm{u}$.)
- $10^{6}$ s
- $10^{8}$ s
- $10^{12}$ s
- $10^{16}$ s
Solution: Given reactions can be written as $3\sideset{_1}{^2}{H}\to \sideset{_2}{^4}{He}+p+n$. The Q-value of this reaction is \begin{align} Q&=[3m(\sideset{_1}{^2}{H})-m(\sideset{_2}{^4}{He})-m(p)-m(n)]c^2\nonumber\\ &=[3(2.014)-4.001-1.008-1.007]c^2\nonumber\\ &=0.026 c^2={0.026\times 931}\;\mathrm{MeV} \\ &={3.87\times{10}^{-12}}\;\mathrm{J}. \end{align} Thus, energy released by consumption of three deuteron is ${3.87\times{10}^{-12}}\;\mathrm{J}$. Hence, total energy released by consumption of ${10}^{40}$ deuteron is, \begin{align} E&=\frac{3.87\times{10}^{-12}}{3}\times{10}^{40} \\ &={1.29\times{10}^{28}}\;\mathrm{J}. \end{align} At the rate of radiated power $P={10^{16}}\;\mathrm{W}$, the star will continue to give energy for time, \begin{align} t &=\frac{E}{P}=\frac{1.29\times{10}^{28}}{{10}^{16}} \\ &={1.29\times{10}^{12}}\;\mathrm{s}.\nonumber \end{align}
