Question

Consider ${}_{13}^{25}\text{Al}$ $\to$ ${}_{12}^{25}\text{Mg}$ + ${}_{+1}^0\text{e}$+ $v$.
$m_{Al}$ = $24.990423\mu$; $m_{mg}$
= $24.485839\mu$.
Find the Q value of reaction

• A. $3.3$ $Mev$
• B. $1.3$ $Mev$
• C. $2.3$ $MeV$
• D. $5.3$ $MeV$

Answer 1

The correct option is A $3.3$ $Mev$

$Q$ $=[24.990432\mu -24.98583\mu -2m_e]c^2$
$=0.004593\left(931.5\right)-1.102$

$=3.3$ $MeV$