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

See analysis below
4.6     4.6     ## 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.          