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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  \((Z=91).\) 
Show that  \({_{92}^{238}}{{}}{{}}{U}{}\)  can not spontaneously emit a proton.

Answer

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Solution

One atomic mass unit is  \(1u=1.6605\times 10^{-27}\mathit{kg}\) .
The mass,  \(m=1.6605\times 10^{-27}\mathit{kg}\) .
The speed of light,  \(c=3\times 10^8m/s\) 
Convert one atomic mass unit into energy E as follows:
 \(E=mc^2\) 
 \(E=(1.6605\times 10^{-27})\times \left(3\times 10^8\right)^2J\) 
 \(E=1.4924\times 10^{-10}J\times \frac{10^6\mathit{MeV}}{1.602\times 10^{-19}J}\) 
 \(E=931.5\mathit{MeV}\) 
 \(1u=931.5\mathit{MeV}/c^2\) 
 \(c^2=931.5\mathit{MeV}/u\) 
Consider  \({_{92}^{238}}{{}}{{}}{U}{}\) spontaneously emits a proton, the decay process will be
 \({_{92}^{238}}{{}}{{}}{U}{}\rightarrow {_{91}^{237}}{{}}{{}}{\mathit{Pa}}{}+{_1^1}{{}}{{}}{H}{}\) 
 Find the  \(Q\)  for this process to occur as follows:
 \(Q=\left[M_U-M_{\mathit{Pa}}-M_H\right]\times c^2\) 
 \(Q=\left[238.05079-237.05121-1.00783\right]\times c^2\) 
 \(Q=-0.00825u\times \left(c\right)^2\) 
 \(Q=-0.00825u\times \left(931.5\mathit{MeV}/u\right)^2\) 
 \(Q=-7.68\mathit{Mev}\) 
Thus, the  \(Q\)  of the process is negative and hence it cannot proceed spontaneously. We have to supply energy of  \(7.68\mathit{MeV}\)  to a  \({_{92}^{238}}{{}}{{}}{U}{}\)  nucleus to make it emit a proton.
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