Q.Calculate the energy released in the following reactions, given the masses to be $_{88}^{223}$Ra: 223.0185 u, $_{82}^{209}$Pb: 208.9811 u, $_6^{14}$C: 14.00324 u, $_{92}^{236}$U: 236.0456 u, $_{56}^{140}$Ba: 139.9106 u, $_{36}^{94}$Kr: 93.9341 u, $_6^{11}$C: 11.01143 u, $_5^{11}$B: 11.0093 u. Ignore neutrino energy.

[!TLDR] (a) Ra-223 $\rightarrow$ Pb-209 + C-14: $Q\approx31.8$ MeV (book: 32.096 MeV). (b) U-236 $\rightarrow$ Ba-140 + Kr-94 + 2n: $Q\approx171.0$ MeV (book: 172.485 MeV). (c) C-11 $\rightarrow$ B-11 + $e^+$ + neutrino: $Q\approx1.47$ MeV (book: 1.485 MeV). [!ANSWER] (a) $\approx31.8-32.1$ MeV, ( …

(a) Radium-223 undergoing this decay emits a carbon-14 nucleus directly (an example of exotic 'cluster decay', rarer than ordinary alpha decay but a real process for a few heavy nuclei): $Q=[m_{Ra}-m_{Pb}-m_C]c^2=[223.0185-208.9811-14.00324]\times931.5$. The mass difference is $223.0185-208.9811-14.00324=0.03416$ u, giving $Q\approx31.8$ MeV, close to the book's more precisely rounded 32.096 MeV. (b) For the fission reaction $_{92}^{236}U\rightarrow\,_{56}^{140}Ba+\,_{36}^{94}Kr+2n$: $Q=[m_U-m_{Ba}-m_{Kr}-2m_n]c^2=[236.0456-139.9106-93.9341-2(1.00866)]\times931.5$. Working through: $236.0456-139.9106=96.1350$; $96.1350-93.9341=2.2009$; $2\times1.00866=2.01732$; $2.2009-2.01732=0.18358$ u; multiplying by 931.5 gives $Q\approx171.0$ MeV, close to the book's 172.485 MeV (the small residual difference is consistent with the limited number of significant figures given for the fragment masses). (c) For $_6^{11}C\rightarrow\,_5^{11}B+e^++\nu$: $Q=[m_C-m_B-m_e]c^2=[11.01143-11.0093-0.00055]\times931.5=[0.00158]\times931.5\approx1.47$ MeV, close to the book's 1.485 MeV. In all three parts, my own calculation usin …

Apply $Q=[\sum m_{reactants}-\sum m_{products}]c^2$ to each reaction using the given atomic masses (and the neutron/electron masses from the standard constants list), conve …

In part (b), forgetting the factor of 2 on the neutron mass (since the reaction releases 2 neutrons, not 1), which …