The energy released when one nucleus of \({ }_{92} \mathrm{U}^{235}\) undergoes fission is \(188…
The energy released when one nucleus of \({ }_{92} \mathrm{U}^{235}\) undergoes fission is \(188 \mathrm{MeV}\). The energy released when \(100 \mathrm{~g}\) of \({ }_{92} \mathrm{U}^{235}\) undergoes fission is
\(3.55 \times 10^{12} \mathrm{~J}\)
\(7.71 \times 10^{12} \mathrm{~J}\)
\(3.55 \times 10^{13} \mathrm{~J}\)
\(7.71 \times 10^{13} \mathrm{~J}\)
Solution
Given, energy released, \(E_0=188 \mathrm{MeV}\) per nucleus fission and mass, \(m=100 \mathrm{gm}\)
As number of nucleus in \(100 \mathrm{gm}\) of \({ }_{92} \mathrm{U}^{235}\)
\(N=\frac{100}{M} \times N_A\)
where, \(N_A=\) Avogadro number and
\(M=\) molecular mass
\(\Rightarrow N=\frac{100}{235} \times 6.022 \times 10^{23}=2.56 \times 10^{23}\) nucleus
Energy released in fission of one nucleus
\(\begin{aligned}
E_0^{\prime} & =E_0=188 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J} \\
& =300.8 \times 10^{-13} \mathrm{~J}
\end{aligned}\)
So, Energy released in \(\mathrm{N}\)-nucleus fission,
\(\begin{aligned}
& E=E_0^{\prime} N=2.5610^{23} \times 300.8 \times 10^{-13} \mathrm{~J} \\
\Rightarrow \quad & E=7.71 \times 10^{12} \mathrm{~J}
\end{aligned}\)
Hence, the correct option is (b).