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

200 KeV

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2 MeV

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200 MeV

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2000 MeV

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Solution

The correct option is C 200 MeV
$Δ E = 8.5 \times 234 - 7.6 \times 236 = 195.4 M e V = 200 M e V$

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Similar questions

In a fission reaction
$_{92}^{238} U \to^{117} X +^{117} Y + n + n$
the binding energy per nucleon of X and Y is 8.5 MeV whereas that of $^{236} U$ is 7.6 MeV. The total energy liberated will be about

Q. In a fission reaction

_{92}^{236} U \to

^{117} X +

^{117} Y + n + n

the average binding energy per nucleon of

X

and

Y

8.5 M e V

whereas that of

^{236} U

7.6 M e V

.
The total energy liberated will be about :

Q. In a fission reaction

_{92}^{236} U \to^{117} X +^{117} Y + n + n

, the binding energy per nucleon of X and Y is 8.5 MeV whereas of 236 U is 7.6 MeV. The total energy liberated will be about

In a fission reaction $_{92} U^{236} \to X^{117} + Y^{117} + 2_{0} n^{1} + ε,$ The binding energy per nucleon of X and Y is 8.5 MeV where as of $_{92} U^{236}$ is 7.6 MeV. The total energy liberated will be about

Q. The binding energy per each nucleon in the neighborhood of medium nuclei is

8.5 M e V

and the binding energy per each nucleon is about

7.6 M e V

in the neighborhood of Uranium. The energy released in the fission of

U^{236}

is roughly

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

The correct option is C 200 MeVΔE=8.5×234−7.6×236=195.4 MeV=200 MeV

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

The correct option is C 200 MeV
$Δ E = 8.5 \times 234 - 7.6 \times 236 = 195.4 M e V = 200 M e V$