Question

Find the minimum kinetic energy (in $eV$ ) of an $\alpha$-particle to cause the reaction ${ }^{14} N (\alpha, p)^{17} O$. The masses of ${ }^{14} N ,{ }^{4} He ,{ }^{1} H$ and ${ }^{17} O$ are respectively $14.00307\, u$, $4.00260\, u , 1.007 .83\, u$ and $16.99913\, u$.

Answer 1

Since the masses are given in atomic mass units, it is easiest to proceed by finding the mass difference between reactants and products in the same units and then multiplying by $931.5\, MeV / u$. Thus, we have
$Q=(14.00307\, u +4.00260\, u$
$-1.00783\, u-16.99913\, u )\left(931.5 \frac{ MeV }{ u }\right)$
$=-1.20\, MeV$
$Q$ value is negative. It means reaction is endothermic.
So, the minimum kinetic energy of $\alpha$-particle to initiate this reaction would be,
$K_{\min }=|Q|\left(\frac{m_{\alpha}}{m_{N}}+1\right)$
$=(1.20)\left(\frac{4.00260}{14.00307}+1\right)$
$=1.54\, MeV$