In the nuclear reaction `._1H^2 +._1H^2 rarr ._2He^3 +._0n^1` if the mass of the deuterium atom `=2.014741 am u`, mass of `._2He^3` atom `=3.016977 am u`, and mass of neutron `=1.008987 am u`, then the `Q` value of the reaction is nearly .

Options

A. `0.00352 MeV`
B. `3.27 MeV`
C. `0.82 MeV`
D. `2.45 MeV`

Answer

B

Explanation

`Q =(Sigma B_(r) -Sigma B_(p))c^(2)`
where `Sigma B_(r) =` sum of the masses of reactants
and `Sigma B_(p) =` sum of the masses of the products
`Sigma B_(r) =2 xx 2.014741 a.m.u. =4.0294892 a.m.u.`
`Sigma B_(p) =(3.016977 + 1.008987)a.m.u.`
`=4.025964 a.m.u.`
`Sigma B_(r) -Sigma B_(p) =(4.029482 -4.025694) a.m.u.`
`=0.003518 a.m.u.`
Decreases in mass appears as equivalent energy.
`:. Q=0.003518 xx 931 MeV`
` = 3.27 MeV`.