IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# Energy Calculation in Fission and Fusion Processes

## Problems from IIT JEE

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: $m(\text{H}^2)={2.014}\;\mathrm{u}$; $m(\text{n})={1.008}\;\mathrm{u}$; $m(\text{p})={1.007}\;\mathrm{u}$; $m(\text{He}^4)={4.001}\;\mathrm{u}$.)

1. $10^{6}$ s
2. $10^{8}$ s
3. $10^{12}$ s
4. $10^{16}$ 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}