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\vspace{3mm} \begin{abcliste} \abc Show that the nucleus \isotope[8][4]{Be} ($\SI{8.005305}{u}$) is unstable and will decay into two $\upalpha$ particles (\SI{4.002603}{u}). \abc Is \isotope[12][6]{C} (\SI{12.00}{u} by definition) stable against decay into three $\upalpha$ particles? Show why or why not. \end{abcliste}
(a) $\SI{0.09}{MeV}$, unstable \quad (b) $\SI{7.3}{MeV}$, stable
\begin{abcliste} \abc The mass difference between left and right is: \begin{align} \Delta m &= M(\isotope[8][]{Be})  2\cdot M(\isotope[4][]{He})\\ &= \SI{8.005305}{u}  2\cdot \SI{4.002603}{u}\\ &= \SI{9.9e5}{u} >0 \end{align} Because the massdifference is positive, i.e. the left side is heavier than the right side, it means mass got lost in the reaction, i.e. energy was produced, i.e. the reaction tend to happen spontaneous, i.e. Beryllium is unstable. \abc The binding enregy of this Carbon isotope considered consisting of Helium atoms is: \begin{align} \Delta m &= M(\isotope[12][]{C})  3\cdot M(\isotope[4][]{He})\\ &= \SI{12.000000}{u}  3\cdot \SI{4.002603}{u}\\ &= \SI{7.809e3}{u}\\ \end{align} Because the massdifference is negative, i.e. the left side is lighter than the right side, it means mass got produced in the reaction, i.e. energy is needed, i.e. the reaction can't happen spontaneous, i.e. Carbon is stable. \end{abcliste}
13:43, 4. June 2017  u  Urs Zellweger (urs)  Current Version 
13:43, 4. June 2017  si text  Urs Zellweger (urs)  Compare with Current 