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# Question and Answer

## QuestionPhysicsClass 12

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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