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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

`c+ a- b`

B

`c- a -b`

C

`a+b+c`

D

`a+b-c`

Text Solution

Verified by Experts

The correct Answer is:
B
Energy released `=c - (a+b) = c-a-b`
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