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In the reaction .1^2 H + .1^3 H rarr .2^...

In the reaction `._1^2 H + ._1^3 H rarr ._2^4 He + ._0^1 n`, if the binding energies of `._1^2 H, ._1^3 H` and `._2^4 He` are respectively `a,b` and `c` (in MeV), then the energy (in MeV) released in this reaction is.

A

a+b+c

B

c+a-b

C

c-a-b

D

a+b-c

Text Solution

Verified by Experts

The correct Answer is:
C
Energy released in given reaction =BE of products -
BE of reactants =C-(a+b)=C-a-b
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