The binding energy/nucleon of deuteron `(""_(1)H^(2))` and the helium atom `(""_(2)He^(4))` are 1.1 MeV and 7MeV respectively. If the two deuteron atoms fuse to form a single helium atom, then the energy released is

Options

A. 26.9MeV
B. 25.8MeV
C. 23.6MeV
D. 12.9 MeV

Answer

C

Explanation

Binding energy of deuteron `(""_(1)H^(2))`
Binding energy of helium atom `(""_(2)H^(4))`
= `4 xx 7` MeV = 28 MeV
`""_(1)H^(2) + ""_(2)H^(4) rarr ""_(2)H^(4)` + Energy released
Energy released = B.E - `2 xx BE` of deuteron
= 28 MeV - `2 xx 2.2` MeV
= 28 MeV - 4.4 MeV = 23.6 MeV