When subatomic particles undergo reaction energy is conserved, but mass is not necessarily conserved. However, a particle's mass 'contributes' to its total energy, in accordance with Einstein'f famouns equation, `E= mc^(2)`
In this question, `E` denotes the equivalent energy when a particle of mass `m` is converted into energy. The particle can also have additional energy due to its motion and its interactions with other particles.
Consider a neutron at rest, and well separated from other particles. It decays into a proton, an electron, and an undetected third particle: `"Neutron"rarr"proton"+"electron"+"third particle"`
The table below summarizes some data from a single nuetron decay. Column `2` shows the rest mass of the particle times the speed of light squared.
`{:("Particle",Mx C^(2),,"Kinetic Energy",),(,(MeV),,(MeV),),("Neutron",940.97,,0,),("Proton",939.66,,0.02,),("Electron",0.51,,0.42,):}`
Assuming the table contains no major errors, what can we conclude about the (mass `x c^(2)` of the undetcted third particle ?

Options

A. It is `0.90 MeV`
B. It is `0.36 MeV`
C. It is less than or equal to `0.90 MeV` , but we cannot be more percise
D. It is less than or equal to `0.36 MeV` , but we cannot be more percise

Answer

D

Explanation

From tha passage, sub-atomic reactions do not conserv mass. So, we cannot find the third particle's mass by setting `m_("neutron")` euqal to `m_("proton")+m_("electron")+_("third particle")`. By constrast, the total energy in this case, the sum of of 'mass energy' and kinetic energy, is conserved. If E denotes total energy, then
`E_("neutron")=E_("Proton")+E_("electron")++E_("third particle")`
The neutrons has energy `940.97 meV`. The proton has energy `939.66 MeV+0.02 MeV+0.42 MeV=0.93 MeV`. Therefore, the third particle has energy
`E_("third particle")=E_("neutron")-E_("proton")-E_("electron")`
We just found the third particle's total energy, the sum of mass energy and kinetic energy. Without more information, we cannot figure out how much of that energy is mass energy.