When subatomic particles undergo reactions, energy is conserved, but mass is not necessarily conserved. However, a particle’s mass “contributes” to its total energy, in accordance with Einstein’s famous equations, E=mc2
In this equation, E denotes the energy a particle carries because of its mass. 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 → proton + electron + ???
Table 1 summarizes some data from a single neutron decay. An MeV (mega electron volt) is a unit of energy.
Table 1
Data from a single neutron decay
Column-1 shows the rest mass of the particle times the speed of light squared and
Column-2 shows its kinetic energy.
Column-1Column-2particlemass×c2(MeV)Kinetic energy (MeV)Neutron940.970.00Proton939.670.01Electron0.510.39
Assuming table 1 contains no major errors, what can we conclude about the mass ×c2 of the undetected third particle?
When subatomic particles undergo reactions, energy is conserved, but mass is not necessarily conserved. However, a particle’s mass “contributes” to its total energy, in accordance with Einstein’s famous equations, E=mc2
In this equation, E denotes the energy a particle carries because of its mass. 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 → proton + electron + ???
Table 1 summarizes some data from a single neutron decay. An MeV (mega electron volt) is a unit of energy.
Table 1
Data from a single neutron decay
Column-1 shows the rest mass of the particle times the speed of light squared and
Column-2 shows its kinetic energy.
Column-1Column-2particlemass×c2(MeV)Kinetic energy (MeV)Neutron940.970.00Proton939.670.01Electron0.510.39
Assuming table 1 contains no major errors, what can we conclude about the mass ×c2 of the undetected third particle?
The correct option is D It is less than or equal to 0.39 MeV; but we cannot be more precise
According to the passage, subatomic reaction do not conserve mass. So, we cannot find the third particle’s mass by setting m_{neutron} equal to mproton+melectron+mthirdparticle.
By contrast, the total energy - in this case, the sum of “mass energy” and kinetic energy - is conserved. If E denotes total energy, then Eneutron=Eproton+Eelectron+Ethirdparticle
The neutron has energy 949.97 MeV. The proton has energy 939.67 MeV + 0.01 MeV = 939.68 MeV. The electron has energy 0.51 MeV + 0.39 MeV = 0.90 MeV. Therefore, the third particle has energy Ethirdparticle=Eneutron–Eproton–Electron=940.97MeV−939.68MeV−0.90MeV=0.39MeV
We just found the third particle’s total energy, the sum of its mass energy and kinetic energy. Without more information, we cannot figure out how much of that energy is mass energy.
According to the passage, subatomic reaction do not conserve mass. So, we cannot find the third particle’s mass by setting m_{neutron} equal to mproton+melectron+mthirdparticle.
By contrast, the total energy - in this case, the sum of “mass energy” and kinetic energy - is conserved. If E denotes total energy, then Eneutron=Eproton+Eelectron+Ethirdparticle
The neutron has energy 949.97 MeV. The proton has energy 939.67 MeV + 0.01 MeV = 939.68 MeV. The electron has energy 0.51 MeV + 0.39 MeV = 0.90 MeV. Therefore, the third particle has energy Ethirdparticle=Eneutron–Eproton–Electron=940.97MeV−939.68MeV−0.90MeV=0.39MeV
We just found the third particle’s total energy, the sum of its mass energy and kinetic energy. Without more information, we cannot figure out how much of that energy is mass energy.
