Question

In the nuclear reaction ${}_1{H}^2{+}_1{H}^2{\to }_{2}H{e}^3{+}_0{n}^1$ if the mass of the deuterium atom $=2.014741amu$, mass of ${}_{2}H{e}^3$ atom $=3.016977amu$, and mass of neutron $=1.008987amu$, then the Q value of the reaction is nearly

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

Answer 1

The correct option is B 3.27 MeV

$Q=\Delta m{c}^2$
$\Delta m=2m_D-m_T-m_n$
$=2\left(2.014741\right)-3.016977-1.008987$
$=3.51\times {10}^{-3}u$
For $1amu,Q=931Mev$
Hence, for $\Delta m=3.51\times {10}^{-3}u$
$Q=931\times 3.51\times {10}^{-3}MeV=3.27MeV$