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

(b) `Q = (sum B_r -sum B_p) c^2`
where `Sigma Br` = sum of the masses of reactions and `Sigma Bp` = sum of the masses of the products
`sum B_r = 2 xx 2.014741 am u`
`sum = B_p = (3.016977 am u)`
`sum B_p = (3.016977 + 1.008987)`amu
`=4.025964` amu
`sum B_r + sum B_p = (4.029482 - 4.025694)`amu
`=0.003218 `amu
Decrease in mass appears as equivalent energy
`:. Q = 0.003518 xx 931 MeV`
`=3.27 MeV`.