First slide

Nuclear Binding energy

Question

If average binding energy per nucleon of X and Y is 8.5 MeV whereas that of $_{92}^{236} U$ is 7.6 MeV. Then the total energy liberated in the following reaction is
$_{92}^{236} U \to^{117} X +^{117} Y + n + n$

Moderate

A

200 keV
B

2 MeV
C

200 MeV
D

2000 MeV

Solution

Average binding energy per nucleon of X and Y = 8.5 MeV
Binding energy per nucleon of U = 7.6 MeV
Where binding energy $\to$ B.E.
So total energy liberated
$= (B . E \cdot of X + B . E . of Y - B . E . of U$
$= (117 \times 8.5 + 117 \times 8.5 - 236 \times 7.6)$
$= 195.4 \approx 200 MeV$

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