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We are given the following atomic masses:
\({_{92}^{238}}{{}}{{}}{U}{}=238.05079u\)
\({_{90}^{234}}{{}}{{}}{Th}{}=234.04363u\)
\({_{91}^{237}}{{}}{{}}{\mathit{Pa}}{}=237.05121u\)
\({_2^4}{{}}{{}}{\mathit{He}}{}=4.00260u\)
\({_1^1}{{}}{{}}{H}{}=1.00783u\)
Here the symbol \(\mathit{Pa}\)  is for the element protactinium \((Z=91).\)
Calculate the energy released during the alpha decay of \({_{92}^{238}}{{}}{{}}{U}{}\) .

Answer

One atomic mass unit is \(1u=1.6605\times 10^{-27}\mathit{kg}\) .
The mass, \(m=1.6605\times 10^{-27}\text{ kg }\) .
The speed of light, \(c=3\times 10^8\text{ m / s }\)
Convert one atomic mass unit into energy E as follows:
\(E=mc^2\)
\(E=(1.6605\times 10^{-27})\times \left(3\times 10^8\right)^2J\)
\(E=1.4924\times 10^{-10}J\times \frac{10^6\mathit{MeV}}{1.602\times 10^{-19}J}\)
\(E=931.5\mathit{MeV}\)
\(1u=931.5\mathit{MeV}/c^2\)
\(c^2=931.5\mathit{MeV}/u\)
The alpha decay of \({_{92}^{238}}{{}}{{}}{U}{}\)  is given by equation
\( \ ^{A}_{z} X \rightarrow \ _{z-2}^{A-4} Y+_{2}^{4} H e\)
The energy released in this process is given by
\(Q=\left[M_U-M_{Th}-M_{\mathit{He}}\right]\times c^2\)
\(Q=\left[238.05079-234.04363-4.00260\right]\times c^2\)
\(Q=0.00456u\times \left(c\right)^2\)
\(Q=0.00456u\times \left(931.5\mathit{MeV}/u\right)^2\)
\(Q=4.25\mathit{Mev}\)
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