The binding energies per nucleon for deuteron (`._1H^2`) and helium (`._2He^4)` are `1.1 MeV` and `7.0 MeV` respectively. The energy released when two deutrons fuse to form a helium nucleus (`._2He^4`) is

Options

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

Answer

C

Explanation

Total binding energy of helium atom `(._2He^4)` is
`4 xx 7=28 MeV`
Total binding energy of deutron `._1H^2(1p+1n)` is
`2 xx 1.1 =2.2 MeV`
Hence, binding energy of 2 deutrons is
`2 xx 2.2 =4.4MeV`
So, the energy released in forming helium nucleus from two deutons is
`28-4.4 MeV =23.6MeV`.