In the reaction `""_(1)^(2)H + ""_(1)^(3)H to ""_(2)^(4)He + ""_(0)^(1)n`, if binding energies of `""_(1)^(2)H,""_(1)^(3)H, ""_(2)^(4)He` are respectively a, b and c (in MeV), then the energy released in this reaction is :

Options

A. a + b' + c
B. a + b - c
C. `c - (a + b)`
D. `c + a -b`

Answer

C

Explanation

Mass decay ` = ("mass of " ""_(2)^(4)He+ "mass of " ""_(0)^(1)n) - ("mass of" ""_(1)^(2)H + "mass of" ""_(1)^(3)H) (because Deltam = E xx u^(2))`
`= (B.E "of" ""_(2)^(4)He + 0-"B.E of" ""_(1)^(2)H - "B.E of " ""_(1)^(3)H)/(u^(2))`. may decay ` = (c-a-b)/(u^(2))`
Now E = mass decay ` xx u^(2) = (c-a-b)/(u^(2)) = c-a-b`