Q82PE.
Question
Unreasonable Results
A frazzled theoretical physicist reckons that all conservation laws are obeyed in the decay of a proton into a neutron, positron, and neutrino (as in \({\beta ^ + }\)decay of a nucleus) and sends a paper to a journal to announce the reaction as a possible end of the universe due to the spontaneous decay of protons. (a) What energy is released in this decay?
(b) What is unreasonable about this result?
(c) What assumption is responsible?
Step-by-Step Solution
Verifieda) The decay of proton into neutron, positron and neutrino is \( - 1.805{\rm{MeV}}\).
b) The result is irrational since the reaction is not spontaneous.
c) It is incorrect to assume that the reaction is spontaneous.
The proton decays into lighter subatomic particles such as a neutral pion and a positron in proton decay, which is a hypothetical type of particle decay.
a)
Let us solve the given problem.
The decay of proton into neutron, positron and neutrino is given by
\(P \to n + {B^ + } + {e_v}\)
So the energy released during decay is given by \(E = \left( {{m_p} - {m_n} - {m_e}} \right){c^2}\)
where \({m_p} = \)mass of proton, \({m_n} = \)mass of neutron and \({m_e} = \)mass of positron
Now plugin the value of \({m_e},{m_n},{m_p}\)in the above equation and solve for the value of
\(E\):
\(\begin{aligned}{\underline{\phantom{xx}}}E = (1.007276{\rm{u}} - 1.008665\,{\rm{u}} - 0.0005485\,{\rm{u}})\left( {931.5\,{\rm{MeV}}/{{\rm{c}}^2}} \right){c^2}\\ = - 1.805{\rm{MeV}}E\\ = - 1.805{\rm{MeV}}\end{aligned}\)
Therefore the calculation of \(E\) is \( - 1.805{\rm{MeV}}\).
The unreasonable about the result is the reaction is not spontaneous.
c)
The assumption that the reaction in spontaneous is wrong.