The following deutruim reactions and corresponding raction energies are found to occur. S

Options

A. `0.5eV`
B. `0.5MeV`
C. `0.05MeV`
D. `0.05 eV`

Answer

c

Explanation

The `Q` value of the first reaction implies that
`.^(14)N +d=^(15)N =P+8.53 MeV`
Where `.^(14)N` represents the mass of `.^(14)N` nucleus in energy units.
This can be rearranged to give
`.^(14)N-^(15)N =p +d+8.53 MeV`
The second reaction similarly implies as
`.^(15)N-^(13)C =alpha -d +7.58 MeV`
and the third reaction gives
`.^(13)C -^(11)B= alpha -d +5.16 MeV`
Adding these three equations, we have
`.^(14)N-^(11)B=p+2alpha -3d +21.27 MeV`
`.^(11)B(alpha,n).^(14)N=.^(11)B-.^(14)N+alpha-n`
`=3d -alpha-p-n-21.27 MeV`
Now,
`3d-alpha-p-n=(3 xx 2.014 -4.0020-1.0078 -1.0087) a.m.u.`
`:. Q =(0.0229 xx 931 -21.27)MeV=0.05MeV`.