The binding energy per nucleon of deuterium and helium atom is `1.1 MeV` and `7.0 MeV`. If two deuterium nuclei fuse to form helium atom, the energy released is.

Options

A. `19.2 MeV`
B. `23.6 MeV`
C. `26.9 MeV`
D. `13.9 MeV`

Answer

B

Explanation

(b) `._1 H^2 + ._1 H^2 rarr ._2 He^4 +` energy
Binding energy of a `(._1 H^2`) deuterium nuclei
=`2 xx 1.1 = 2.2 MeV`
Total binding energy of two deuterium nuclei
=`2.2 xx 2 = 4.4 MeV`
Binding energy of a `(._2 He^4`) nuclei
=`4 xx 7 = 28 MeV`
So, energy released in fusion `= 28 - 4.4 = 23.6 MeV`.