Question

In the reaction ${}_7{N}^{14}+$ ${}_{2}H{e}^4{\to }_8{O}^{17}+$ ${}_1{H}^1$. The minimum energy of $\alpha$-particle is
$M_N=$ $14.00307$ amu $M_{He}=$ $4.00260$ amu
$M_o=$ $16.99914$ amu $M_H=$ $1.00783$ amu

• A. $1.21MeV$
• B. $1.62MeV$
• C. $1.89MeV$
• D. $1.96MeV$

Answer 1

The correct option is A $1.21MeV$

Mass on reactant side $=(14.00307+4.00260)amu$

Mass on product side $=(16.99914+1.00783)amu$

So, mass defect $=1.3\times {10}^{-3}amu.$

Energy of the $\alpha$ particle $=931.5\times 1.3\times {10}^{-3}MeV$

$=1.21MeV$