In a fission reaction $_{92}^{236} U \to^{117} X +^{117} Y + n + n$ . The binding energy per nucleon of $X$ and $Y$ are $8.5 M e V$ whereas of $^{236} U$ is $7.6 M e V$ . The total energy liberated will be about

400 M e V

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200 M e V

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300 M e V

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200 k e V

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Solution

The correct option is B $200 M e V$
$\begin{matrix} Total no. of nucleus in " X " and "Y" would be \begin{matrix} 117 + 117 = 234 Total B.E of X and Y = 8.5 \times 234 = 1989 MeV \end{matrix} \end{matrix}$

$\begin{matrix} ^{236} U h a v e B E = 7.6 \times 236 = 1793.6 M e V △ E & = 1989 - 1793.6 Δ E & = 195.4 M e V △ E & \approx 200 M e v \end{matrix}$
$Approximately$

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In a fission reaction 23692U→117X+117Y+n+n. The binding energy per nucleon of X and Y are 8.5MeV whereas of 236U is 7.6MeV. The total energy liberated will be about

The correct option is B 200MeV Total no. of nucleus in " X " and "Y" would be 117+117=234 Total B.E of X and Y=8.5×234=1989 MeV 236U have BE=7.6×236=1793.6MeV△E=1989−1793.6ΔE=195.4MeV△E≈200MevApproximately

In a fission reaction $_{92}^{236} U \to^{117} X +^{117} Y + n + n$ . The binding energy per nucleon of $X$ and $Y$ are $8.5 M e V$ whereas of $^{236} U$ is $7.6 M e V$ . The total energy liberated will be about